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HI GH E R
LE V E L
:
S TAT I S T I C S
C O U R S E
C O M PA N I O N
Josip Harcet
Lorraine Heinrichs
Palmira Mariz Seiler
Marlene Torres Skoumal
3
Great
Clarendon
Oxford
It
furthers
and
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Oxford
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Acknowledgements
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Course
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v
Contents
Chapter
1
Exploring
Introduction
1.1
Probability
Cumulative
Discrete
1.2
Other
Expected
The
Review
Chapter
2
2.1
of
Limit
Biased
and
32
and
Central
Limi t
probability
Case
I:
Case
II:
of
of
single
two
or
variable
more
independent
the
42
variables
normal
47
random
variables
mean
64
68
statistical
analysis
how
can
methods
we
make
78
sense
of
the
mean
and
variance
of
a
normal
81
denition?
for
the
82
mean
inter val
inter val
for
inter val
for
84
μ
for
when
μ
matched
σ
when
is
σ
known
is
pairs
for
μ
when
σ
is
known
Hypothesis
testing
for
μ
when
σ
is
unknown
testing
for
test
exercise
matched
pairs
graph
101
105
109
errors
distribution
Two-tail
90
100
testing
II
unknown
85
97
testing
Type
79
80
for
Hypothesis
and
data?
estimates
Condence
Normal
58
74
inter vals
Signicance
40
41
Theorem
Condence
Hypothesis
Review
Theorem
models
42
estimators
Condence
vi
25
variable
Condence
18
20
and
a
information:
well-dened
I
variable
algebra
Exploring
Estimators
Type
random
exercise
random
3.4
algebra
distribution
Unbiased
3.3
geometric
variables
distributions
combination
Central
Review
A
a
36
of
3
of
functions
transformation
linear
Introduction
12
variance
Linear
Chapter
5
5
distribution
of
The
3.2
and
4
12
transformation
Sampling
decisions
quantities
Linear
2.3
informed
function
independent
Normal
2.2
make
exercise
Expectation
A
3.1
value
Expectation
Introduction
to
2
distributions
generating
sum
tool
distri butions
distribution
binomial
Probability
a
probabi li ty
continuous
probability
Negative
as
distribution
and
Geometric
1.3
further
112
for
a
one-tail
test
115
116
117
Chapter
4
Introduction
4.1
Statistical
Bivariate
modeling
122
distributions
123
Correlation
124
Correlation
Sampling
4.2
Covariance
4.3
Hypothesis
and
causation
128
distributions
130
135
Proper ties
of
covariance
136
testing
138
Introduction
t-Statistic
4.4
Linear
for
138
dependence
regression
Review
exercise
of
X
and
Y
139
141
152
Answers
156
Index
161
vii
Exploring
further
1
probability
distributions
CHAPTER
OBJECTIVES:
Cumulative
7.1
distribution
Geometric
distribution.
Probability
generating
Using
probability
distribution
Before
1
Find
the
variable,
you
mode,
standard
of
e.g.
a
functions
generating
the
sum
median,
the
of
Negative
of
n
for
both
discrete
binomial
for
(Pascal’
s)
discrete
functions
to
independent
and
the
random
x
variance,
and
variables.
of
a
table
mean,
and
discrete
shows
discrete
1
random
the
the
standard
probability
random
Find
discrete
variable X

2
3
4
0.3
0.25
0.35
0.
mode,
median,
deviation
random
of
mean
the
and
following
variables
given
by:
a
i
=
variables.
mean,
x
X
distributions.
distribution.
random
nd
continuous
start
deviation
distribution
functions
p
i
−
0

2
3
4
0.3
0.
0.3
0.
0.05
0.5
i
P(X
=
x )
i
⎧5
Mode
(X )
=
3,
because
P(X
=
3)
=
b
which
is
the
highest
probability
of
the
P( X
=
x )
=
⎨
⎪
⎩
four
random
Median,
P( X
≤
m
1)
=
variables.
=
2
since
0 .3
and
P( X
≤
2)
=
0 .55
4
=
E (X )
=
∑
x
p
i
i
i =1
= 1 × 0. 3
+
2 × 0. 25
+ 3 × 0. 35
+
4 × 0. 1 =
2. 25
4
2
σ
=
Var ( X
)
=
∑
2
x
p
i
−
μ
i
i =1
2
=
1
2
(0.3)
+
2
2
(0.25)
+
3
2
(0.35)
+
4
2
(0.1)
2
=
2
6.05 − 2.25
Exploring
further
=
0.994
probability
distributions
−
x
,
⎪
0.35,
0.225
x
=
1
,
2,
3,
10
0,
otherwise
4
2
Find
the
mode,
standard
random
median,
deviation
variable,
function
of
given
the
by
a
of
a
e.g.
mean
the
discrete
and
2
continuous
probability
random
Find
the
mode,
standard
density
variable X
random
is
median,
deviation
variables
probability
=
(x )
dened
density
⎧ 3
f
x,
0
≤
x
≤
a
2
f
(x )
=
,
0
=
2
because
it
has
the
maximum
0,
⎪
b
at
the
≤
x
≤
2
elsewhere
elsewhere
π
⎧
point
given
2
⎪
0,
(X )
the
16
⎩
Mode
by
function:
x
⎨ 4
⎨ 2
⎩
and
continuous
3
⎪
⎪
mean
the
formula
⎧ 1
⎪
of
end
of
the
f
(x )
=
cos
(
)
2x
,
π
≤
x
≤
4
⎨
4
inter val.
⎪
m
0,
⎩
elsewhere
2
1
∫
Median,
f
(x )
dx
m
=
1
⇒
=
2
0
⇒
4
m
=
2
⎧
2
6
, 3
⎪
x
≤
≤
6
2
2
c
2
1
μ
=
E (X )
=
∫
xf
(x)dx
(x )
=
⎨ x
4
2
x
=
f
dx
⎪
=
∫
0,
⎩
2
0
elsewhere
3
0
2
2
=
Var
(
)
X
=
x
2
f
( x )
−
∫
0
2
2
1
⎛
3
x
∫
dx
−
4
⎜
2
⎝
16
⎞
=
⎟
3
2
−
2
=
9
⎠
3
0
3
Find
the
series
sum
by
of
using
an
the
innite
geometric
3
a
1
+
u
1
+
u
2
+
...
=
, 0
the
sum
geometric
formula
u
u
Find
<
r
1
−
of
the
innite
series:
0.5
+
0.25
−
0.125
+
…
< 1
3
r
1
2
e.g.
following
the
2
b
series

1

1

...
2
2
9
9
4
+
3
+
2
+ ...
2
27
2
+
=
=
2
3
2
1
3
4
Dierentiate
functions,
and
integrate
e.g.
(x )
4
Dierentiate
composite
2
f
composite
,
x
4 x )
the
following
1
a
≠
f
(x )
=
3
(3
integrate
functions:
3
=
and
,
2
4
x
≠
2
x
3 x 1
2 ×
⇒
f
′( x )
( −3) ×
( −4 )
=
b
f
(x )
=
c
f
(x )
=
e
24
=
4
(3
−
4
4 x )
(3
−
π
3
4 x )
− 2x
sin
2
3
2
f
(x )
=
2
3
(3
d
4 x )
2
=
+
3
−
4 x )
(x )
=
(x
2
− 2)
1
dx
⇒
(3
f
c
2
4 (3
−
4 x )
Chapter
1
3
Probability
as
a
tool
to
make
informed
decisions
In
A
probability
possible
as
the
distribution
outcomes
statistical
company
of
a
is
use
mathematical
par ticular
likelihood
might
a
of
event
each
statistical
or
event.
techniques
model
course
For
to
that
of
action
example,
create
shows
a
as
the
the
well
Probability
Statistics
large
A
scenario
analysis
uses
probability
become
scenario
distributions
few
and
have
ver y
impor tant
analyses.
last
decades,
due
to
their
to
wide-ranging
produce
several
theoretically
distinct
possibilities
for
the
outcome
of
applications.
a
par ticular
course
of
action
or
future
event.
For
example,
a
business
Statistics
might
create
worst-case
three
scenarios:
scenario
probability
would
distribution;
worst-case,
contain
the
likely
a
likely
,
value
scenario
and
from
the
would
best-case.
lower
end
contain
a
literacy
is
The
of
essential
not
business
and
only
for
the
value
economics
towards
the
middle
of
the
distribution;
and
the
best-case
scenario
professionals
would
contain
a
value
in
the
upper
end
of
the
distribution.
also,
for
people
Although
it
is
impossible
to
predict
the
precise
value
of
a
involved
level,
businesses
still
need
to
be
able
to
plan
for
future
a
scenario
analysis
based
on
a
probability
distribution
a
company
frame
its
possible
future
values
in
terms
of
a
use
level
and
a
worst-case
and
best-case
scenario.
By
doing
so,
doing
can
base
its
business
plans
on
the
likely
scenario
but
tr y
aware
of
the
alter native
for
Exploring
further
probability
and
predict
players
will
possibilities.
bring
4
well
to
still
which
be
to
players
the
then
company
statistics
which
likely
are
sales
pro
coaches
can
decide
help
Team
events.
often
Using
in
future
spor ts.
sales
but
example,
distributions
the
the
best
game.
results
All
the
probability
Actuarial
or
science
postgraduate
different
health
care
become
centur y
de
also
His
have
core
revise
tables
plan
was
company
recall
Discrete
The
data
and
that
Quantitative
into
two
take
put
in
of
the
can
be
usually
random
by
by
a
this
Annuities.
a
in
student
of
Actuarial
He
modied
Halley
equitable
Life,
(1656–1742),
life
a
well-known
mathematicians
in
as
par t
fur ther,
material,
be
it
of
is
in
1762.
the
this
higher
impor tant
par ticularly
described
represented
and
given
a
by
if
you
level
that
feel
that
you
chapter.
as
or
cannot
form
of
an
obtained
quantitative
random
continuous.
nite
variable
variable
pioneers
new
function
topic
used
has
17th
these
was
Equitable
of
insurance,
quantities
can
in
The
group
computer
and
who
more
many
established
Edmund
a
the
from
late
used
accountant.
by
of
undergraduate
in
the
insurance
the
as
actuar y
In
better
distributions
core
the
random
set,
practice
discrete
from
of
an
public
distribution
collect
data
one
and
earlier
the
established
terminology
we
continuous
discrete
developed
Life
par t
and
an
immediately
as
was
offering
as
times.
(1705–1757),
teacher
continuous
values
uncountable
a
exploring
categories:
exact
a
par t
the
such
offer
methods
Working
which
are
knowledge
economics,
recent
science
Dodson
probability
Before
this
as
in
book
combines
mathematical
research
elds
this
universities
models.
(1667–1754),
society
studied
course.
cannot
A
a
in
in
nance,
jobs
Actuarial
James
Cumulative
Y
ou
of
advance
wor ked
mor tality
insurance
popular
formalize
Moivre
formed
assurance.
1.1
an
better
Dodson
statistical
you
was
many
and
business
most
study
science
mathematics,
nance
the
we
that
Actuarial
statistical
distributions.
to
science.
life
like
uses
mathematician
Abraham
and
of
there
theories
British
and
one
probability
It
that
syllabus
studies.
subjects
programming.
models
variables
Discrete
countable
be
listed
or
qualitative
and
classied
random
set,
whilst
since
they
variables
the
values
come
from
an
interval.
from
a
nite
set
of
values
+
{x
, x
1
, x
2
by
a
, ..., x
3
n
table
with
their
x
of
Sometimes
x
,
1
is
x
,
usually
x
2
terms
,
3
of
...,
,
that
the
variable
usually
can
take
given
along
probabilities.
n
when
n
values
function
p
...
assign
x
the
distribution
n
4
we
probability
x
...
p
3
a
listing
4
p
2
This
x
3
p

values,
x
2
p
has
n ∈ 
corresponding
x

},
...
a
formula
we
,
have
P( X
=
or
an
x
)
n
r ule
for
innite,
=
p
n
=
calculating
countable
p ( n ),
set
where
p
probabilities.
of
is
values,
calculated
in
n
n.
Chapter
1
5
In
both
0
i
≤
cases
p
≤
the
following
proper ties
must
hold:
;
i
ii
∑ p
=
.
i
A
continuous
random
values,
usually
[a,
<
b ],
a
b,
written
b ∈ ,
a,
⎧
f
formula
(x )
=
probability
in
has
f
a
( x ),
the
obtained
form
of
probability
a
x
≤
from
an
an
inter val
distribution
uncountable
of
real
function
set
of
numbers,
given
by
the
≤ b
p
⎨
0, elsewhere
⎩
The
variable
density
function
must
satisfy
the
following
properties:
f
i
(x)
≥
0
for
all
the
values
of
x ;
+∞
f
ii
(x)
=
.
∫
−∞
A
of
cumulative
a
random
variable
X.
function
distribution
variable
In
the
(CDF)
calculator
with
probability
up
core
of
a
function is
to
and
course
discrete
CDF
including
we
dened
random
features
the
to
sum
the
a
all
given
the
and
probabilities
value
cumulative
variable
calculate
of
the
distribution
used
properties
of
of
a
graphical
some
discrete
The
word
means
distributions.
quantity
In
this
chapter
we
are
going
to
further
explore
some
discrete
by
variables,
addition’.
and
study
going
Even
to
focus
though
discrete
and
proper ties
at
new
which
on
the
there
to
of
are
theoretical
is
discrete
study
a
continuous
dierence
continuous
the
of
between
variables,
cumulative
irrespective
probability
of
there
distribution
the
nature
distributions.
probability
how
are
we
function
of
the
are
distributions.
apply
also
We
some
that
the
CDF
common
we
will
look
variable.
Denition
Given
the
random
corresponding
cumulative
This
i
ii
probability
probability
cumulative
F (x)
lim
variable
∈[0,
] ;
F (x )
=
F (x )
=
function
the
(discrete
distribution
probability
i.e.
X
F
:

of
F
is
1]
has
[0,
continuous)
function
→ [0,
function
range
or
is
the
P :
F (x )
=
lim
]
0
1
x →+∞
iv
F (x)
x
<
1
6
is
x
nondecreasing
⇒
F ( x
2
Exploring
)
1
further
≤
on
F ( x
the
whole
)
2
probability
distributions
and
→ [0,
P( X
proper ties:
x →−∞
iii

domain;
i.e.
to
≤
the
1],
x )
the
cumulative
‘increasing
in
successive
Example
Prove
x
a
the
<
x
1

cumulative
⇒
P( x
2
b
P( X
a
P( x
>
<
<
distribution
X
x )
X
=
≤
1
1
−
x
)
x
≤
1
)
=
function
F (x
2
)
−
has
F (x
2
the
following
),
1
F (x )
=
P
(X
≤
x
2
) \ (X
x
≤
2
(
Set
)
A
is
a





B
A
P( X
x
≤
F (x
)
of
)
P( X
x
≤
2
−
F (x
2
P( X
>
x )
=
lim
P( x
X
set
B
so
the
probability
the
di erence
≤
of
the
of
these
two
sets
is
the
probabilities.
)
1
the
denition
of
the
cumulative
)
distribution
1
<
of
)
Use
=
subset
1





di erence
=
b
proper ties:
b)
Rewrite
the
function.
given
set.
b → + ∞
=
lim
P( X
b)
≤
P( X
x )
≤
Use
the
result
Use
proper ty
from
par t
a.
b → + ∞
=
lim
F
(
b
)
−
(
F
x
)
=
1
−
(
F
x
)
iii
at
the
bottom
of
page
6.
b → + ∞
In
the
can
case
be
tr ying
have
this
the
of
a
found
to
a
nd
a
simple
book.
simply
Therefore
Example
density
=
⎨
,
the
b
Hence
x
0,
⎩
Find
going
Just
a
few
formulas
to
function
use
for
in
a
distribution
the
table
discrete
are
table
discrete
variables
beyond
of
function
instead
the
values
random
of
will
scope
for
of
nding
variables.
function
of
a
random
variable X
is
given
by
the
formula:
=
1
,
2,
3
9
⎪
a
other
values
+ k
⎪
x )
are
cumulative
the

⎧ x
=
most
we
the
up
formula.
distribution
probability
P( X
variable,
adding
generating
formula;
cumulative
The
discrete
by
value
otherwise
of
determine
k
the
cumulative
distribution
function.
3
a
∑
P( X
=
x )
=
1
⇒
The
sum
of
all
probabilities
must
be
equal
to
1.
x =1
1 + k
2 + k
+
9
k
3 + k
+
9
=
Solve
= 1
9
the
equation
and
nd
the
value
of
k.
9

x
b
6 + 3k
= 1 ⇒
P( X
=
x )
 1
=
,
x
= 1
,
2,
Use the result from part a
3
and calculate the values
9
of
X
=
x

2
3
2
3
4
9
9
9
2
5
9
9
9
9
the probability density function f or x = 1, 2, 3.
Then
use
the
distribution
P(X
=
denition
function
of
and
the
add
cumulative
up
all
the
x)
previous
probabilities.
= 1
F (x)
Chapter
1
7
Example
The
the
X

cumulative
table
=
distribution
function
of
a
random
variable X
is
given
by
below:
x

2
3
1
3
3
10
10
5
4
F (x)

Determine
the
formula
of
the
probability
density
function.
1
P( X
=
1)
=
(
F
)
1
Use
=
the
result
from
Example
1
to
nd
all
the
10
probabilities.
3
P( X
=
2)
=
(
F
2
)
−
(
F
)
1
1
=
=
10
10
3
P( X
=
3)
=
(
F
3
)
−
F
(
2
)
2
−
3
=
3
−
5
=
10
10
3
P( X
=
4)
=
F
(
4
)
−
F
(
3
)
=
1
−
2
=
5
⎧
( x )
=
5
10
,
x
=
1
,
2,
3,
4
Look
at
the
repeating
patter n
and
⎩
rule
continuous
and
a
f or
the
values
1,
2,
3,
4.
0, otherwise
random
variables
the
probability
density
function
Even
f
deduce
⎨ 10
⎪
For
4
=
x
⎪
f
10
the
cumulative
distribution
function F
are
related
as
though
the
follows:
lower
boundar y
of
the
x
integral
F ′ (x)
=
f
(x)
⇔
F (x)
=
f
∫
is
−∞,
in
most
(t)dt
of
the
integrals
for
−∞
The
the
following
proper ties
problems
examples
of
the
involving
show
how
cumulative
continuous
to
use
this
distribution
random
relationship
function
variables.
to
and
solve
the
left
are
going
left
boundar y
inter val
x
for
Exploring
further
probability
distributions
to
of
which
equal
8
boundar y
to
use
of
the
f(x)
zero.
we
the
the
variable
is
not
Example
A

continuous
random
⎧ ax ( 2
f
(x )
=
−
x ),
x
variable
X
has
a
probability
density
function
given
by
∈ [0, 2]
⎨
0, otherwise
⎩
a
Find
the
b
Hence
c
Find
value
of
a
determine
the
modal
the
cumulative
value
of
the
distribution
random
function.
variable X
2
2
2
a
∫
ax
(
2
x
)
dx
= 1 ⇒
a
(
∫
2x
)
x
dx
= 1
The
denite
integral
on
the
inter val
[0,
2]
must
0
0
be
2
3
⎡
a
x
⎛
−
= 1 ⇒
a
⎦
2
⎜
⎥
3
⎣
to
1.
−
−
4
0
3
⎝
0
⎞
2
2
⎢
equal
3
⎤
x
2
⎟
= 1
Solve
the
equation
and
nd
the
value
of
a.
⎠
3
⇒
a
= 1 ⇒
a
=
3
4
x
b
F (x)
=
∫
f
(t)dt
Use
the
relationship
between
probability
0
density
x
⎛ 3
F (x)
=
∫
0
3
2
t
⎜
t
⎝ 2
x
<
0
x
2
=
x
⎨
⎪
−
, 0
4
≤
x
1
,
⎩
3
x
(x )
=
3
x
−
2
3
3
x
⇒
f
′( x )
4
−
=
3
−
2
3
the
cumulative
the
f or mula.
par t
a
and
the
proper ties
of
=
0
⇒
x
distribution
function
and
nd
The
modal
value
is
the
value
f or
which
the
x
2
probability
maximum
x
2
from
> 2
2
f
result
≤ 2
4
⎪
c
the
3
⎪ 3
F (x )
functions.
dt
Use
⎪
distribution
⎠
0,
⎧
cumulative
⎞
⎟
4
and
density
function
reaches
an
absolute
value.
= 1
2
The
modal
value
of
the
variable
X
is
1.
Scratchpad
y
1.0
3
3
2
f1(x
x
2
x
4
0.75
(1, 0.75)
0.5
0.25
x
0
0.5
1
1.5
2
2.5
Chapter
1
9
Example
A

continuous
random
0,
⎧
⎪
F (x )
=
⎨
variable
4
3
x
x
X
has
a
cumulative
distribution
function, F (x),
given
by:
< 0
2
− 4x
4 x
+
, 0
x
≤
≤
b
⎪
1
,
⎩
a
Find
the
b
Hence
c
Find
a
b
value
determine
the
4
median
3
−
of
x
> b
b
the
probability
value
of
the
density
random
function.
variable X
2
4b
+
4b
=
The
1
are
2
⇒
(b
⇒
(b
cumulative
no
points
of
function
is
monotone
discontinuity
and
there
F ( b) =
theref ore
1.
2
−
2b )
−
1
=
2
0
Solve
2
−
2b
−
1) (b
−
2b
+
1)
=
the
equation
and
eliminate
impossible
0
solutions:
is
b
eliminated
because
it
is
negative
1
b
= 1
−
2
,
b
1
=
1
+
2
, b
2
=
3
1
and
b
is
eliminated
because
the
CDF
is
not
2
4
monotone
on
the
on
the
inter val
[0 , 1
+
2
as
]
seen
screen.
Scratchpad
y
2
4
3
–4·x
f1(x)=x
2
+4·x
1.5
(2.41, 0.971)
(1, 1)
A
monotone
function
is
entirely
1
nondecreasing
or
nonincreasing
on
the
whole
0.5
domain.
x
0
0.5
b
f
(x )
=
f
(x )
=
1
1.5
2
2.5
F ′( x )
Use
the
density
3
⎧4 x
relationship
and
the
between
cumulative
the
probability
distribution
2
− 12 x
+
8x ,
x
∈ [0, 1]
functions.
⎨
Di erentiate
the
probability
density
0, otherwise
⎩
function
The
c
from
graph
of
par t
the
a
cumulative
distribution
Scratchpad
function
of
a
random
variable
X
is
equivalent
y
to
2
0,
f1(x)
4
=
{
x
3
–4·x
+4·x
x
<
0
x
≤
1
of
2
, 0
1,
≤
x
>
the
cumulative
data.
Notice
(Ogive). To
1
frequenc y
that
both
calculate
the
diagram
have
the
median
of
a
same
value
set
shape
we
1
1
have
(0.459, 0.5)
to
solve
the
equation
F (m )
,

2
1
f2(x)
can
=
2
x
0
0.5
Median,
10
Exploring
m
=
further
1
0.459.
probability
distributions
be
solved
on
the
GDC.
which
Exercise
1
The
by
1A
probability
the
density
x )
=
=
,
a
Find
b
Hence
by
the
value
of
=
=
x )
a
Find
b
Hence
c
Calculate
X
,
k
the
the
value
by
of
determine
P (X
the
x
≤

1
1
6
2
What
is
the
the
2,
3,
by
the
cumulative
x
formula
of
value
the
of
function.
a
random
variable X
is
the
density
function.
variable X ?
of
a
random
variable X
is
below:
7
1
4
9
16
25
25
25
25
9

is
function
the
the
continuous
formula
median
random
given
⎪
=
of
value
the
of
variable
probability
the
X
density
function.
variable X ?
has
a
probability
density
by
⎧
)
of
probability
function
5
What
x
given
2
3
b
(
distribution
function

Determine
f
is
2).
distribution
table
a
A
variable X
4
F (x)
5
random
a
the
modal
cumulative
=
a
function.

b
X
of
distribution
below:
0
Determine
given
function
distribution
table
a
The
cumulative
,
= 1
x
F (x)
4
2
0, otherwise
cumulative
=
0, 1
,
20
⎨
⎩
given
given
2x
⎪
The
=
density
⎪
3
is
formula:
⎧ a
P( X
x
determine
probability
the
variable X
0, otherwise
⎩
The
random
12
⎨
⎪
2
a
+ k
⎪
(X
of
formula:
⎧ x
P
function
π
⎡
2 sin bx ,
x
∈
⎨
⎤
0,
⎢
⎥
⎣
3 ⎦
⎪
0, otherwise
⎩
a
Find
the
b
Hence
value
of
determine
Calculate
P
⎜
⎝
the
cumulative
distribution
function.
π
⎛
c
b
X
≥
⎟
6
Chapter
1
11
6
A
continuous
random
variable
X
has
a
probability
density
function
For
given
by
the
more
see
on
1
⎧
,
⎪
(
f
x
− 2
x
<
<
=
⎨π
4
exercise
pages
and
exercise
pages
0,
⎩
530–531
b
Find
1.2
what
Ber noulli
arises
each
you
a
this
There
don’t
is
a
get
of
game
game
many
a
12
to
Y
ou
an
the
also
of
n
but
of
dened.
variable X
level
has
lear ned
and
core
two
that
a
of
negative
are
distribution
experiments
success.
As
you
with
will
distributions
experiments
of
about
outcomes:
binomial
binomial
independent
sequences
course
possible
independent
probability
such
these
higher
trial
such
equal
geometric
sequences
popular
is
dich
innite
with
length.
board
game
in
Europe
called
‘Ludo’
or
‘People,
chooses
is
played
elds
‘6’
on
this
in
many
ˇ
C
lověče
as
a
is
to
by
different
nezlob
the
play
each
given
die.
question
game
further
with
se,
on
we
is:
a
die.
are
languages
ˇ
Covječe
within
ne
(Ludo,
ljuti
Mensch
Лyдo,
se,
of
one
of
moving
the
gures
To
the
game
star t
three
dicult
ask:
the
gures
given
‘How
must
probability
the
player
Players
question,
star t
Exploring
available
nicht,
He
ce
човечe...)
everlasting
answer
the
in
Ber noulli
sequence
having
random
well
distributions
studied
A
the
is
distribution
player
obtain
of
function
angr y’.
cъpди
able
nite
ver y
ärgere
Each
value
failure (F).
probability
,
The
The
have
section
consist
Geometric
as
modal
experiment
constant
probability
probability
or
from
in
The
this
experiments.
success (S)
also
the
Other
Recall
see
that
on
of
otherwise
course
Show
10K
x
⎪
a
10C
502–503,
2
2
)
practice,
formula
‘What
rst
is
is
three
distributions
to
the
of
to
star t
four
his
each
attempts
it
the
the
of
by
must
game.
game?’
probability
attempts?’
colour
player
star t
the
colours.
To
being
companion.
the
Let’s
look
at
the
possible
sequence
of
Ber noulli
trials.
We
will
Jacob
denote
the
outcomes
of
each
single
trial
by
S
or
F
(S
denotes
Bernoulli
‘rolling
(1654−1705),
a
6’,
and
F
denotes
‘not
rolling
a
6’).
A
geometric
of
denotes
the
innite
sequence
of
such
experiments
until
we
reach
eight
success,
Again
(i.e.
we
each
e.g.
need
roll
probabilities
probability
to
of
of
of
S,
FS,
FFFS,
emphasise
the
the
a
FFS,
die)
is
is
each
mutually
possible
success
that
FFFFS,
FFFFFS......
these
are
by p
always
and
the
and
the
that
same.
probability
17th
the
He
do
The
of
of
is
q
=
1
−
p,
0
p
a
≤
and
was
an
of
family
18th
the
the
on
in
of
the
centuries.
rst
extensive
work
like
failure
Swiss
mathematicians
experiments
independent
outcomes
denoted
of
members
the
famous
rst
one
distribution
one
to
piece
problems
this.
1
≤
Denition
A
discrete
random
variable
X
is
said
to
have
a
geometric
k
distribution
0
≤
p
This
and
≤ 1
, q
we
= 1 −
denition
write
p, k
q ∈ [0, 1]
single
⇒
∼ Geo ( p )
if
P( X
=
k )
=
the
proper ties
of
a
probability
1
q
probability
=
is
P( X
within
=
k ) ∈ [0, 1]
the
inter val
for
[0,
all
k ∈ 
=
k)
=
+
k
p
=
p
means
Together,
i
and
distribution
Example
Given
∑
+
is
that
ii
a
distribution.
-
ever y
1
1
q
=
p
=
1
+
k ∈
This
p
q
= 1.
p
k∈

the
show
good
sum
that
of
the
all
probabilities
above
is
denition
1.
for
a
geometric
one.

that
X
p
=
0.4,
k
b
p
=
0.9,
k
c
p
=
0.3,
a
1
q
∑
k ∈
where
1].
1
P( X
∑
p,
+
p
k
ii
1
q
= 1
, 2, 3, 4...
satises
k
p,
i
X
Geo( p ) ,
∼
=
2
=
6
k
=
nd
P (X
=
k)
if:
3
a
P( X
=
2)
=
0 .6
b
P( X
=
6)
=
0 .1
c
P( X
=
×
0 .4
=
0 .24
Calculate
of
a
q
=
geometric
1
−
p
and
then
use
the
denition
distribution.
5
×
0 .9
=
0 .000009
12
13 )
=
0 .87
×
0 .13
=
0 .0244
Chapter
1
13
We
can
calculate
those
probabilities
by
using
the
GDC.
Scratchpad
geomPdf(0.4, 2)
0.24
geomPdf(0.9, 6)
0.000009
geomPdf(0.13, 13)
0.024444
3/99
Example
Find
or
X
the
that
∼

probability
the
Geo
player
⎛
1 ⎞
⎜
⎟
⎝
Method
P(X
=
that
will
a
player
obtain
a
will
“6”
on
successfully
an
star t
unbiased
die
the
game
within
the
“Ludo”
rst
three
attempts.
1
Probability
of
scoring
a
“6”
is
p =
.
6
6 ⎠
I
)
+
P(X
=
2)
+
P(X
=
3)
=
Calculate
q
1
−
p
and
then
denition
of
a
geometric
f or
of
k
=
use
the
distribution
2
5
1
=
+
1
×
⎛ 5 ⎞
+
×
⎜
6
6
Method
6
1
⎝
1
=
5
+
25
+
91
=
=
0
values
⎠
6
6
36
216
216
all
the
on
1
0.421296
, 1, 3
6
)
1/99
P( X
14
≤
3
and
add
up
II
Use
(
2,
probabilities.
a
cumulative
Scratchpad
geomCdf
1,
421
⎟
6
3)
Exploring
=
0 .421
further
probability
distributions
the
GDC.
distribution
function
Example
The
a
probability
biased
X
8
∼
die
is
of
star ting
0.28.
the
game
Calculate
the
“Ludo”
on
probability
the
of
Geo( p )
=
3)
=
scoring
Use
2
P( X
third
q
f or
the
attempt
a
“6”
on
denition
value
of
k
=
using
3
the
of
a
and
die.
geometric
write
the
distribution
equation
in
p
one
variable
p.
2
(1
−
p)
p
=
0.128
Use
Scratchpad
the
GDC
eliminate
y
the
to
solve
the
impossible
equation
solution,
and
0
≤
p
≤
1
2
2
x–0.128
f1(x)=(1–x)
1
(0.2, 0) (0.488, 0)
x
–0.5
1
.5
1.5
–1
–2
p
=
0.2,
p
1
Notice
On
the
with
p
must
or
we
can
estimate
we
For
second
and

the
1.313
given
use
numerical
process
the
=
data
we
have
two
solutions.
GDC
functions
0.4
0.488,
3
that
possible
we
=
2
can
the
input
and
two
features
solver.
values
those
third
When
of
for
solving
multiple
values
solution
to
we
solving
by
equations:
the
solutions
obtain
input
the
the
numerical
so
that
desirable
values
in
solver
the
iteration
solution.
of
respectively
.
Scratchpad
2
nSolve((1–p)
p=0.128,
p)
p=0.128,
p, 0.4)
0.2
2
nSolve((1–p)
0.487689
2
nSolve((1–p)
p=0.128,
p, 1)
1.31231
3/99
It
is
much
simpler
to
spot
multiple
solutions
when
using
the
function
feature.
Chapter
1
15
Example
Find
the
the
number
game
Method
9
of
“Ludo”
attempts
is
greater
the
such
that
the
probability
chance
of
not
of
successfully
star ting
the
star ting
game.
I
⎛ 1 ⎞
∼
X
needed
than
Geo
⎝
P( X
n)
≤
⎟
6
Probability
1
⇒
⎜
of
scoring
a
>
“6”
within
rst
n
1
attempts
2
⎠
should
exceed
.
Find
the
probabilities
2
n
1
5
1
+
×
6
+ ... +
6
6
⎛
5 ⎞
⎜
⎟
⎝
6
⎛
5
⎜1
6
+
⎜
+ ... +
6
⎛
5 ⎞
⎜
⎟
6
⎝
1
f or
1
×
>
values
of
k
=
1,
2 , ..., n
and
add
them
up.
⇒
6
⎠
n
1
1
2
1
⎞
1
Notice
⎟
>
geometric
sequence
and
apply
the
2
⎠
⎝
the
⇒
⎟
f or mula
⎠
f or
the
sum.
n
5
⎛
1
⎞
⎜
1
⎟
6
⎝
Simplif y
>
inequality
and
use
logarithms
to
⇒
solve
5
6
the
1
⎠
×
the
inequality.
2
1
6
n
1
>
2
⎛
5 ⎞
⎜
⎟
⎝
6
⎛ 1 ⎞
⇒
log
⎠
⎜
⎝
⎟
2
⎛
>
n
×
log
⎠
5
Logarithms
⎝
⎛
Method
f orget
to
the
inequality
symbol.
3
80
⇒
n
=
4
5 ⎞
⎜
⎝
don’t
⎠
=
⎛
so
⎟
2
>
log
negative
1 ⎞
⎜
⎝
n
are
6
reverse
log
here
⎜
⎟
6
⎠
II
Use
the
cumulative
distribution
function
on
the
Scratchpad
GDC.
y
2
step
1
f1(x)=geomCdf
(
6
)
tracing
2
(4, 0.018)
x
0
2
3
4
5
6
–1
f1:
(4, 0.018)
–2
P( X
16
≤
n)
Exploring
>
0 .5
⇒
further
n
=
function
the
values
and
1
, 1, x
1
1
Since
4
probability
distributions
the
graph.
we
are
can
discrete
nd
the
we
obtain
solution
by
a
Exercise
1
2
3
Given
1B
that
X
a
p
=
0.6,
b
p
=
0.14,
c
p
=
0.5,
d
p
=
0.88,
Given
that
a
P (X
≤
b
P (X
>
c
P (5
d
P (1
Mario
k
k
=
k
4)
6)
p
=
0.3
<
X
≤
7)
if
p
=
0.991
cer tain
emergency
random.
the
at
a
balloon
independent
each
that
fth
shot
Mario
school
is
and
is
the
selected
of
with
the
0.73.
will
92%
procedure.
What
following:
0.7
if
with
a
=
shooting
the
0.25
7)
balloon
a
p
=
nd
≤
is
In
p
Geo( p )
X
attempt
if:
5
≤
is
k)
4
=
if
=
3
∼
if
P(X
2
=
X
probability
4
=
k
nd
∼ Geo( p )
Mario
Students
and
arrow
.
that
three
Each
Mario
hits
arrows.
the
Find
the
balloon.
are
from
probability
has
the
students
is
bow
probability
destroy
student
a
familiar
the
with
school
are
the
re
selected
at
that:
the
rst
one
who
doesn’t
know
the
procedure;
b
the
rst
not
5
Fred
that
a
spoon
What
the
b
will
ever y
defective,
he
is
rst
What
4th
be
who
selected
wooden
adjusts
the
the
manufactured
the
the
procedure
spoons.
with
spoon
no
and
The
defect
when
probability
is
one
0.85.
is
Given
that
X
∼
Fred
found
that
the
four th
that
within
manufactured
to
be
spoon
be
no
need
for
the
manufacture
of
is
Geo( p ) ,
show
that
six
adjustment?
k
6
will
one?
probability
will
know
machine.
probability
there
doesn’t
student?
souvenir
manufactured
defective
is
spoons
selected
before
manufactures
inspects
a
student
occur
P( X
>
k )
=
q
+
,
k ∈ 
.
Chapter
1
17
Investigation
Let
X
∼
Geo( p ) .
Calculate
the
a
p
=
0.4,
P( X
>
5 |X
>
3),
b
p
=
0.7,
P( X
>
6 |X
>
2 ),
c
p
=
0.12,
Make
and
a
tr y
this
to
need
for
prove
to
the
of
a
about
we
will
n
P( X
5),
the
lear n
>
4)
P( X
form
>
of
of
how
to
a
7)
nd
of
the
conditional
random
expected
In
order
sequences.
a
the
and
simple
conjecture.
geometric
variable.
terms
between
your
geometric
consecutive
probabilities:
2)
connection
random
innite
>
P( X
variance
geometric
of
>
general
and
manipulate
sum
12 | X
the
value
section
variance
>
conjecture
Expected
In
P( X
following
to
value
do
Recall
geometric
variable
so
and
you
the
the
will
formula
sequence:
n +1
2
1
+
q
+
3
q
+
1
n
q
+ ... +
q
q
=
,
1
When
n
becomes
between
−
Therefore
and
we
an
,
can
extremely
the
where
|q|
<
.
q
higher
large
number
powers
of
q
and q
will
be
is
a
ver y
number
close
to
zero.
write
n +1
1
2
1 +
q
+
q
3
+
q
2
+ ... =
lim
(1
+
q
+
q
3
+
+ ... +
q
)
=
n → ∞
So
in
other
innite
words
you
geometric
obtained
1
lim
0
=
n →∞
have
q
1
n
q
the
result
for
1
the
q
sum
=
1 − q
of
an
sequence.
1
2
1 +
q
+
q
3
+
q
+ ... =
,
1
Now
will
if
we
obtain
dierentiate
the
q ∈
]−1, 1[
q
this
sequence
successively
,
term
by
term,
we
following:
1
2
1 +
q
+
q
3
+
q
+ ... =
,
1
q ∈
]−1, 1[
q
1
2
⇒
1 +
2q
+ 3q
3
+
4q
+ ... =
,
q ∈
]−1, 1[
2
(1
If
we
dierentiate
again
we
q )
see
that:
2
2
⇒
2
+ 3 × 2q
+
4 × 3q
3
+
5 × 4q
+ ... =
,
q ∈
]−1,
1[
3
(1 −
We
a
18
will
use
geometric
Exploring
these
results
random
further
to
nd
variable,
probability
as
the
q )
expected
shown
distributions
in
value
the
and
following
variance
of
example.
1 − q
probability
,
Example
Find
E
(
X
the
)
0
expected
=
k
∑
value
× P( X
=
and
the
variance
of
a
geometric
random
variable X
k)
Use
the
denition
of
+
k ∈
expected
2
=
1 ×
p
+
2
×
qp
+
+
3q
3
×
3
q
p
+
4
×
k
q
p
+
... +
k
×
value.
1
q
p
Use
+ ...
the
distributive
proper ty.
2
=
p
(1
+
2q
3
+
k
4q
+ ... +
1
kq
+ ...)
Use
1
=
1
p
=
1
p
the
the
result
rst
obtained
derivative
=
2
(1
in
2
q )
p
of
p
the
and
geometric
simplif y
series
the
expression.
2
2
Var ( X )
=
2
E (X
)
−
2
(E ( X ))
=
k
∑
1
⎛
×
P( X
=
k )
⎞
−
Use
⎜
+
the
denition
of
⎟
p
⎝
k ∈
⎠
variance.
2
E
(
2
)
X
=
2
1
×
p
+
2
2
×
qp
+
2
3
×
2
q
p
+
3
4
×
q
p
+
To
simplif y
we
nd
the
process
2
2
... +
k
k
×
2
1
q
p
p (1
+
=
p (( 2
2
2
×
q
+
2
3
×
2
q
+
3
4
×
2
q
+ ... +
k
k
×
× 1
+
+ ...)
(3 − 1)
×
2q
+
(4
− 1)
×
3q
3
+
(5
− 1)
×
4q
+
Use
the
k
... +
(( k
+ 1)
rst.
1
q
2
− 1)
)
+ ...
2
=
E (X
− 1)
×
the
f or mula
f or
all
ter ms.
1
kq
+ ...)
2
n
2
=
p (( 2
+
3
×
2q
+
4
×
3
3q
+
5
2
(1
−
+
2
×
q
+
3
q
×
×
4q
k
+ ... +
(k
3
4
+
×
+ 1) k
k
q
+ ... +
k
1
q
×
×
q
=
+ ...)
Rewrite
two
by
=
1
p
⎞
−
⎜
⎝
=
3
(1
q
)
⎛
2
(1
q
)
2
1
p
−
⎜
⎟
⎝
⎠
2
p
2
−
p
1
using
innite
×
n
the
sums
above.
⎟
Use
the
results
rst
and
of
the
2
p
⎠
the
derivative
2
1)
p
=
3
p
2
⎞
1)
+ ...))
obtained
2
⎛
+
(( n
1
1 −
p
second
to
simplif y.
q
2
Var
(
X
)
=
E
(
X
)
(
))
− ( E
X
=
−
=
2
Use
=
2
p
2
p
p
the
result
to
nd
2
p
the
variance
and
simplif y.
Given
a
geometric
random
variable,
1
then
E
(
X
)
=
if
X
∼
Geo( p ),
q
and
Var
(
X
)
.
=
2
p
This
will
be
used
p
in
Example

on
the
next
page.
Chapter
1
19
Example
Find
the

expected
determine
X
∼ Geo
the
⎛
1 ⎞
⎜
⎟
⎝
6 ⎠
number
maximum
of
attempts
number
of
to
star t
attempts
the
it
game
will
“Ludo”.
take
to
star t
Use
the
the
empirical
r ule
to
game.
1
⇒
E
(
)
X
=
=
Apply
6
the
f or mula
f or
the
expected
the
f or mula
f or
the
variance
value
given
1
above.
6
5
(
Var
X
6
)
=
=
⇒ σ
30
=
30
=
5
48
Apply
2
1
⎛
and
⎞
⎜
calculate
⎟
the
standard
deviation.
⎝ 36 ⎠
[6
−
3
×
5.48,
6
+
3
×
5.48]
=
[ −10.4 ,
22.4 ]
The
empirical
whole
population
deviations
from
maximum
game
Exercise
1
Find
the
Mario
expected
is
is
balloon
with
What
Use
is
the
Mario
3
In
a
selected
will
the
be
Negative
The
random
familiar
20
of
23.
variance
1B,
the
questions 1
with
the
for
a
bow
and
and
probability
geometric
2
arrow
.
that
Each
Mario
hits
the
number
of
shots
Mario
must
make
to
parameter
.
distribution,
to
The
to
further
nd
destroy
of
r ule,
the
the
that
maximum
are
school
determine
that
election
number
of
shots
balloon.
students
ensure
the
distribution
after
explore
Blaise
the
negative
named
values
to
familiar
are
how
one
of
with
selected
many
the
at
the
election
random.
students
selected
must
students
procedure.
distribution
named
Pólya’
s
Exploring
number
to
after
the
set
probability
is
Pascal
negative
binomial
George
of
also
real
known
Pascal’
s
(1623–1662).
binomial
distribution
Pólya
Pascal
distribution
is
was
using
sometimes
(1887–1985)
numbers.
distributions
as
who
the
an
called
extended
be
f all
mean
0.73.
from
binomial
mathematician
to
54%
with
binomial
parameter
is
r ule
make
Students
integer
the
and
expected
school
negative
the
balloon
shot
empirical
at
and
Exercise
a
empirical
distribution,
rst
will
the
balloon?
must
procedure.
the
at
each
the
cer tain
Using
in
independent
destroy
b
value
given
shooting
attempt
a
is
(99.73%)
1C
distributions
2
rule
states
within
value,
attempts
3
that
the
standard
theref ore
to
star t
the
the
Let’s
stop
SS,
In
again
not
FSS,
the
more
consider
after
SFS,
table
than
FFSS,
one
sequence
success,
FSFS,
you
Ber noulli
when
SFFS,
can
see
permutation
that
in
distribution
for
order
Two
FFS
FFSS,
FFFS
FFFSS,
number
Let’s
of
consider
three
three
to
a
have
successes
we
sequence
compare
Geometric
minimum
successes
successes
below
the
the
number
within
the
of
Three
FSSS,
FFS
FFSSS,
might
In
notice
general,
trials
−
we

last
if
,
by
and
the
in
the
we
of
the
rst
k-th
at
the
probability
⎛ k
=
k)
=
r
1
In
order
=
r,
This
r
to
+
have
,
r
discrete
+
number
a
r
2,
rst
par t
to
of
with
have
sequence
and
the
the
one
p
=

k −1
+
stop
when
use
the
table
available
for
SFSFS,
successes
We
we
of
k
to
sequence
required,
to
obtain
obtain
can
parameters k
more
SSFFS
Ber noulli
need
need
one
be
−
success.

So
and
we
can
1⎞
r
k
×
k − th
trial
r − th
⎜
⎝
success
⎟
r
1
q
r
p
⎠
trials
successes
r
of
r −1
p
within
is
two.
SFFSS,
events.
experiments
need
⎠
successes
now
function.




r −1
the
⎛ k
q
⎟
is
we’ll
FSSFS,
1⎞
⎜
⎝
we
we
perform
permutations
FSFSS,
distribution
density
k −r
P( X
simpler,
independent
The
is
SSFS,
successes

there
SFFFS
must
where
in
−
trials,
case
to r
trial.
end
this
available.
two
SFFFS,.......
distribution.
SFSS,
increase
k
binomial
in
we
permutations
up
consider
in
the
then
as
record
actually
success
nd
that
number
we
successes
described
−
the
e.g.
successes
FS
increase
it
dierent
SSS
we
will
successes:
FSFFS,
trials
trials
make
S
We
of
which
geometric
distribution
two
of
we
occurred,
SFFS
FFSFS,
Ber noulli
To
time
FSFFS,
number
achieve
FSFS,
number
occurred.
FFSFS,
this
have
SFS
require,
of
but
successes
FSS,
that
trials
successes
each
to
FS
also
two
FFFSS,
SS
the
k
of
but
S
Notice
r
a
rst
below
,
Geometric
r
the
we
must
have
at
least
r
experiments,
therefore
3, ...
distribution
is
called
negative
binomial
distribution.
Denition
A
discrete
random
variable
X
is
said
to
have
negative
binomial
⎛ k
1⎞
k
distribution
and
we
write
X
∼
NB ( r , p )
if
P (X
=
k )
=
⎜
⎝
where
0
≤
p
≤ 1
,
q
= 1 −
p,
k
=
r ,
r
+ 1
,
r
+ 2,
r
⎟
r
1
q
r
r
p
,
⎠
+ 3, ...
Chapter
1
21
It
is
left
as
an
probability
exercise
to
distribution.
show
that
Therefore
⎛ k
this
we
=
k ) ∈ [0, 1]
⇒
0
≤
for
k
all
=
r ,
r
+
1
,
r
+
∞
⎟
r
2,
1
r
P( X
=
k )
=
1
k =r
Geometric
distribution
distribution
Example
where
we
that:
r
p
≤
1
1⎞
that
X
∼
is
=
2,
p
=
0.,
k
=
2
b
r
=
6,
p
=
0.5,
k
=
9
c
r
=
2,
p
=
0.25,
⎛ 2
k
p)
=
2)
=
⎟
2
⎛ 9
fact
9)
=
1
⎟
6
50 )
=
Unfor tunately
,
distribution
just
a
special
one
case
of
success, r
a
=
negative
1,
Geo(p)
binomial
=
NB(1,
p).
1
k)
0.1
=
Use
0.01
6
the
denition
and
6
×
=
0.5
0.109
1⎞
⎟
12
1
the
12
12
0.75
×
0.25
=
0.0310
⎠
display
calculators
distribution
helps
us
have
features,
to
write
but
down
no
the
a
negative
binomial
programming
useful
program
lines.
nbpdf
1.1
if:
⎠
calculator
few
=
2
×
0.5
graphic
under
a
P (X
distribution
⎜
⎝
of
a
2
50
capacity
1
⎠
⎛ 50
=
=
1⎞
⎜
⎝
P( X
r
p
⎠
just
nd
0.9
9
=
r
1⎞
⎜
⎝
P( X
in
1
50
2
=
⎝
require
NB( r ,
r
P( X
k
⎟
r

a
with
r
q
q
k =r
c
a
3, ...
⇒
⎜
b
prove
of
⎠
+
⎛ k
a
to
denition
∞
ii
Given
good
1⎞
⎜
⎝
a
need
k
P( X
i
is
*nbpdf
1.1
1/1
*nbpdf
11
*nbpdf
nbpdf(6, 0.5, 9)
Dene
Dene
nbpdf(r, p, k)=
Probbii
=
nbpdf(r, p, k
0.1095
Prgm
Prgm
Done
k
Disp “Prob.=”, nCr(k–1, r–1)·(1–p)
r
r
Disp “Prob.=”, nCr
p
nbpdf(1, 0.5, 50)
EndPrgm
EndPrgm
Probbii
=
0.010
Done
9
99
A
similar
function
22
programme
of
Exploring
a
negative
further
can
be
created
binomially
probability
for
the
cumulative
distributed
distributions
variable.
distribution
of
a
apply
negative
the
binomial
f or mula.
Example
A
new
agree
8
a
dr ug
to
2
c
not
∼
is
take
people
b
X

to
par t
will
people
more
be
tested
in
be
will
the
5
people.
study
,
asked
be
than
on
what
before
asked
3
before
people
are
If
is
are
5
the
who
found
before
of
the
2
who
asked
=
3,
p
=
0.12
⇒
P( X
=
8)
5
=
to
agree
found
par ticipate;
to
par ticipate;
who
0.12
agree
Identif y
⎟
2
⎝
=
5,
p
=
0.12
⇒
P( X
=
12 )
=
⎝
=
7
0.88
⎜
r
the
2,
=
P( X
p
=
0.12
=
2)
to
=
⇒ P(2
X
≤
≤
1
⎝
+
P( X
2
+
=
3)
4
+
P( X
⎜
⎠
⎝
4)
⎝
the
calculate
Identif y
the
the
probabilities.
+
P( X
=
1
×
0.12
+
⎠
⎝
up
all
the
of
the
variable
probabilities
f or
5)
2
0.88
⎜
add
values
identied
values.
2
×
0.12
+
⎟
1
⎠
2
×
0.12
⎟
1
⎠
0.2
The
binomial
binomial
generates
to
r,
varies
Given
of
and
the
r
a
to
xed
number
negative
generates
from
binomial
takes
distributions
number
of
in
of
order
distribution
to
that
we
achieve
are
similar :
Bernoulli
successes, r,
binomial
probabilities
innity)
n
as
r
perform
r
and
varies
takes
these
the
trials
a
xed
number
n
Bernoulli
trials
successes.
1D
X
that
a
r
=
1,
b
r
=
3,
p
c
r
=
7,
p
d
r
=
23,
Given
negative
Conversely,
successes,
n
and
distribution
probabilities
n.
Exercise
2
use
5
⎛ 3 ⎞
⎟
3
0.88
⎜
1
and
⎠
=
2
0.88
⎟
⎛ 4 ⎞
(where
negative
0.12
5)
⎛ 2 ⎞
0.12
⎜
of
a
⎟
the
⎛ 1⎞
0
of
0.00336
=
=
parameters
distribution
f or mula
and
from
par ticipate?
0.092
⎛ 11 ⎞
c
to
⎠
binomial
=
r
par ticipate
3
0.88
⎜
b
to
that:
agree
are
people
NB ( r , p )
⎛ 7 ⎞
r
a
2%
probability
found
are
asked
only
p
=
P(X
≤
b
P(X
>
c
P(5
≤
d
P(8
<
NB( r , p )
0.2,
k
=
0.5,
k
=
4
=
0.8,
k
=
9
p
=
X
4)
X
X
∼
if
6)
if
≤
≤
0.77,
=
p
7)
k
=
=
if
11)
if
0.25
0.5
p
=
p
P (X
=
k)
if:
32
NB( r , p )
p
nd
2
=
that
a
∼
nd:
and
and
0.82
=
r
r
=
=
2
and
0.43
3
r
and
=
r
4
=
5
Chapter
1
23
3
The
random
variable
has
the
following
1
Given
a
p
that
Hence
4
Before
when
1,
2,
and
P (3
Alemka
rolling
3
times
nd
and
a
4.
fair
Let
Alemka
P( X
X
≤
p).
=
3)
=
,
nd
the
value
of
p
5).
star t
a
game,
tetrahedral
random
has
NB (2,
125
≤
can
:
24
<
2
b
distribution X
to
she
die
with
variable
throw
the
has
X
die
to
the
obtain
standard
denote
until
two
the
she
“ones”
faces
total
obtains
number
the
of
second
“one”.
a
Write
the
distribution
of
X,
including
the
value(s)
of
each
parameter.
3
b
Find
the
value
of
x
such
that
P( X
=
x )
=
.
32
c
5
Calculate
Nicholas
he
is
needs
least
to
eat
probability
6
he
ve
b
at
seven
cer tain
emergency
will
the
b
he
An
he
need
order
eats
to
to
an
before
pass
phone.
to
apple
the
is
advance
he
before
92%
next
0.85.
in
In
the
this
game
level.
He
The
needs
game.
to
What
eat
is
the
of
advances;
he
are
advances?
students
Students
who
smar t
eat
inspector
to
are
from
conducts
familiar
familiar
that
an
with
school
are
inter view
with
the
the
re
selected
and
needs
procedure.
at
at
What
is
inter view
exactly
six
students
in
order
to
satisfy
will
not
need
in
nds
factor y
shape
three
to
that
is
inter view
produces
an
0.85.
more
a
special
ice-cream
Tony
ice-creams
to
will
inspects
have
than
a
a
dozen
type
be
a
diet
produced
ever y
shape
of
students?
with
ice-cream
defect
he
ice-cream.
no
and
adjusts
when
the
machine.
a
What
is
the
ice-creams
b
What
there
24
the
need;
probability
defect
will
his
that:
ice-cream
The
re
on
order
procedure.
students
he
in
school
The
a
in
apples
least
ve
game
apples
random.
probability
7
that
exactly
a
a
Nicholas
apples
a
In
5 ).
apples
that
four
least
≤
playing
probability
at
P ( X
Exploring
is
before
the
will
probability
adjusts
probability
be
further
he
no
need
probability
that
that
for
he
will
the
inspect
ve
machine?
within
half
adjustment?
distributions
exactly
a
dozen
ice-creams
1.3
Probability
When
tossing
two
generating
coins
the
functions
number
of
heads
obtained
is
When
recorded.
Let
X
be
the
discrete
random
variable.
The
it
below
shows
the
probability
distribution
function
of
=
x
0

2
1
1
1
4
2
4
Probability
is
impor tant
not
to
rely
X
too
much
on
the
sharpest
D’Alember t
X
studying
table
intuition.
minds
Even
like
Jean
(1717–1783)
have
i
at
P (X
=
some
stage
mistakes.
In
made
his
famous
ar ticle
Croix
x )
i
ou
Pile
in
the
French
Encyclopédie
Y
ou
may
notice
that,
in
this
table,
the
the
value
of
probabilities
sequentially
as
X
changes.
From
following
we
can
write
the
probabilities
in
the
form
polynomial
expression
involving
a
of
a
fair
coin,
in
two
what
is
of
the
a
problem:
this
tosses
table
considered
are
the
arranged
he
variable, t:
probability
appear
at
that
least
Heads
will
once?
2
D’Alember t’
s
1
G (t )
1
0
×
=
t
×
4
1
1
+
t
×
2
1
2
+
t
1
=
+
4
1
t
4
+
2
expression
is
called
a
He
4
probability
reasoned
generating
one
this
discrete
random
variable.
We
can
that
G
(
)
1
1
=
+
4
immediately
2
are
all
the
the
1 since
the
corresponding
probabilities
Example
A
pair
Write
of
the
up
after
on
Heads
the
rst
toss.
In
coecients
of
other
this
words,
D’Alember t
probabilities
and
the
must
be
equal
to
assumed
that
the
sum
space
was
{H,TH,TT}.
.

dice
are
rolled.
probability
probability
Let
X
denote
distribution
generating
the
function
sum
of
of
the
the
outcomes
random
on
the
variable X
2
3
4
upper
and
nd
faces.
the
function.
Look
x
life
the
4
sample
of
real
notice
mistakenly
polynomial
in
continue
1
+
=
.
function
showed
1
that
would
experiment
for
was
3
no
This
answer
2
t
5
6
7
8
9
0

at
all
the
possible
pairs
of
2
outcomes
i
the
1
2
3
4
5
6
5
4
3
2
1
36
36
36
36
36
36
36
36
36
36
36
and
their
cor responding
sums.
Find
probabilities.
p
Take
i
p
to
be
the
coecients
of
i
the
polynomial
and
x
to
be
the
i
1
1
1
2
G (t )
=
t
3
+
t
36
1
1
A
as
notice
1
random
in
the
G ()
variable
case
of
a
can
t
7
+
36
t
powers.
8
+
6
t
36
1
t
18
=
5
6
+
11
12
that
t
9
+
1
5
+
10
t
+
9
Again
t
12
9
t
5
4
+
18
+
1
12
t
+
36
.
also
assume
geometric
values
random
in
an
variable,
as
innite
shown
set
in
of
numbers
Example
5.
Chapter
1
25
Example
A
coin
ips.
and
is

ipped
Write
nd
the
the
until
a
head
probability
probability
is
obtained.
distribution
generating
The
of
the
random
random

2
3
4
k
...
denotes
describing
the
this
number
of
experiment
function.
The
x
variable X
variable
possible
outcomes
are
H, TH, TTH,
...
1
i
TTTH,...
and
so
on.
X
∼ Geo
2
1
1
1
1
1
p
...
...
k
i
2
4
1
8
1
16
1
1
2
G (t )
=
t
2
+
1
3
t
+
4
t
+
k
t
+ ... +
t
+ ...
G(t)
is
an
innite
geometric
series
with
k
2
4
8
16
2
1
common
ratio
t .
1
2
t
t
2
=
=
,
1
1
2
t
≠
2
Use
the
f or mula
f or
the
sum
of
an
geometric
series
and
simplif y
the
2
u
1
Notice
that
we
were
using
the
=
formula S
for
the
sum
of
an
∞
1
innite
geometric
series,
under
the
r
condition
that
this
series
converges.
1
That
is,
we
must
have
−
1
<
t
<
1 ⇒
−
2
<
t
2.
<
For
the
other
values
of
t,
2
this
sum
is
not
dened.
1
Again
we
also
notice
that
G (1)
=
=
2
1
1
Denition
Let
P
(
X
X
=
be
k
)
a
discrete
=
p
,
k
=
random
variable
are
0, 1
, 2, 3...
the
assuming
nonnegative
corresponding
integer
probabilities.
values
Then
a
and
function
k
∞
k
G : R
→
R
of
the
form G ( t )
=
p
∑
2
t
=
p
k
+
p
0
t
+
p
1
t
3
+
p
2
t
n
+
. . .
+
p
3
t
+
. . .
is
n
k =0
called
As
a
seen
only
probability
in
the
within
a
nite
of
Let’s
nd
tr y
to
distributions
Ber noulli
generating
previous
probability
the
set,
we
variable
we
are
distribution
function
examples,
could
a
discrete
any
:
random
as
an
outside
functions
familiar
B (,
this
value
generating
already
X
if
consider
taking
probability
that
for
=
0)
Binomial
=
q )
and
(P( X
distribution
X
=
:
1)
=
some
⇒
G (t )
=
qt
pt
=
p):
n −k
=
k )
=
⎜
⎝
⎟
k
q
k
p
,
k
=
0, 1
,
2,
n
26
Exploring
⇒
G (t )
⎛ n ⎞
n −k
∑⎜
k =0
n
n
⎛ n ⎞
=
3, ...,
⎠
⎝
⎟
k
further
q
k
p
k
t
⎠
probability
n −k
=
∑⎜
k =1
⎝
⎟
k
q
⎠
distributions
k
( pt )
n
=
(q
+
pt )
q
+
pt
values
where
is
discrete
1
+
takes
set
with.
⎛ n ⎞
P( X
set
nite
p):
p)
B (n,
variable
innite
the
0
(P( X
innite
t
t
the
zero.
expression.
Poisson
distribution
k
m
X
:
Po (m):
−m
k
∞
e
−m
m
e
k
P( X
=
k )
=
, k
=
0, 1
,
2,
3, ...
⇒
(
G
)
t
=
t
∑
k !
k !
k =0
k
∞
(
*
)
mt
−m
=
e
−m
mt
e
=
∑
m
e
×
=
(
)
t −1
e
k !
k =0
*We
have
used
Maclaurin’
s
Geometric
distribution
X
:
for
∞
=
k )
=
k −1
q
p,
k
=
1
,
2,
3, ...
⇒
G (t )
=
the
exponential
∞
k −1
P( X
formula
Geo(p):
∑
q
k −1
=
pt
∑
q
x
x
k −1
function
t
e
=
∑
k
k =1
k
–

=
i
0
k !
k =1
This
Substitute
k
∞
k
pt
formula
is
a
par t
then,
of
∞
the
Calculus
option.
∞
pt
k −1
pt
∑
q
k −1
i
t
=
pt
k =1
When
−
common
<
qt
<
ratio
Negative
,
this
is
an
∑
i
innite
q
i
t
=
1
=0
geometric
series
qt
with
the
qt
binomial
distribution
X
:
NB (r,
p):
∞
⎛ k
− 1⎞
⎛ k
k −r
P( X
=
k )
=
⎜
⎟
− 1
r
⎝
p
k −r
,
k
=
r ,
r
+
p
1⎞
⎛ r
k −r
∑⎜
k = r
Using
the
+
2, ...
⇒
G (t )
=
∑⎜
k =r
r
t
r
⎝
⎟
r
− 1
q
r
p
k
t
⎠
∞
⎛ k
=
1
,
⎠
∞
r
G(t)
− 1⎞
r
q
⎟
r
⎝
1
q
k −r
r
=
t
p
i
i =0
k
–
r
=
i
− 1⎞
∑⎜
⎠
substitution
+ i
r
t
and
⎟
i
⎝
the
q
i
t
⎠
formula
for
negative
∞
⎛ n
+ i
− 1⎞
n
binomial
series
(1
−
i
x )
=
∑⎜
i =0
⎟
i
⎝
x
,
where
−
<
qt
<
.
⎠
r
r
∞
⎛ r
r
p
+ i
− 1⎞
p
r
t
i
∑
⎜
i =0
⎝
⎟
i
q
pt
⎛
=
⎞
=
⎜
r
(1 −
⎠
Example
r
t
i
t
qt )
⎝
⎟
1 −
qt
⎠

a
A
random
variable
X
has
a
probability
generating
function
G (t )
=
,
4
a
Find
the
b
Hence
value
of
calculate
t
≠
4
t
a
P (
≤
X
≤
3).
a
Use
= 1 ⇒
a
4
a
=
3
b
the
fact
that
G(1)
=
1.
3
1
G (t )
3
=
1
=
4
×
t
t
4
Rewrite
it
in
a
polynomial
Add
coecients
f or m.
1
4
3
=
1
⎛
×
4
⎜
1
+
t
≤
X
≤
3)
+
4
⎝
3
P(1
1
2
1
=
4
1
⎝
4
t
+ ...
64
1
+
⎜
3
+
16
⎛
×
t
1
+
16
⎟
64
63
⎞
⎠
=
the
of
the
powers
1,
2,
and
3.
256
Chapter
1
27
Now
let’s
take
successive
a
look
at
the
probability
generating
function
and
its
derivatives.
∞
k
G (t )
=
p
∑
2
t
=
p
k
+
p
0
t
+
p
1
3
t
+
p
2
n
t
+ ... +
p
3
t
+
...
n
k =0
∞
2
G ′( t )
=
0
+
p
+
2 p
1
t
+
3 p
2
3
t
+
4 p
3
n −1
t
+ ... +
np
4
k −1
t
+
...
=
∑
n
k
×
p
t
k
k =1
∞
2
G ′′( t )
=
0
+
0
+
2 p
+
3
×
2 p
2
t
+
4
×
3 p
3
n −2
t
+ ... + n
(
n −1
)
p
k −2
t
+ ... =
n
4
∑
k (k
− 1)
×
p
t
k
k =2
By
considering
when
t
=
,
the
we
derivatives
can
derive
∞
of
some
the
very
probability
useful
generating
results
for
the
function
value
of
t
=
:
∞
k
G
i
(
)
1
=
p
∑
1
=
∑
k
k =0
p
∞
(
)
1
=
1
∞
k
G′
ii
=
k
k =0
∑
k
p
1
1
=
∑
k
k =1
k
p
=
∞
G ″
(
)
1
=
k (k
∑
− 1) p
you
from
also
can
the
=
∑
k
)
we
can
derivative
calculate
of
k (k
− 1) p
=
calculate
of
a
the
expected
probability
the
variance
the
probability
∞
Var(X )
of
a
=
random
generating
∞
∑
k (k
− 1) p
=
k
k =2
value
generating
− 1))
∑
directly
function.
variable
∞
(k
E(X )
We
can
using
function.
∞
2
G ″(1)
E( X ( X
k
k =2
see,
rst
derivatives
X
2
1
k =2
As
(
∞
k
iii
E
k
k =1
∞
∞
2
−
k ) p
=
k
k =2
∑
k
2
p
−
k
k =2
∑
kp
+
p
k
−
p
1
=
1
∑
k =2
k
p
k
k =1
−
∑
k =1
2
=
E(X
)
−
E(X )
2
G ″
(
)
1
=
E ( X
2
)
−
E (X )
⇒
E ( X
)
=
G″
(
)
1
+
G′
(
)
1

G
(
)
1
2
Var ( X )
Let’s
G (1)
=
=
2
( E ( X ))
the
=
results
G ″ (1)
we
just
+
G′ (
1)
−
(G ′ (
1))
=
G ″(
1)
+
G′ (
1)(
1
obtained:
= G ′(1)
= G ″(1) + G ′(1)
Knowing
these
easily
(1 − G ′(1))
formulas
calculate
previously
binomial
28
−
1
Var ( X )
that
2
)
summarize
E( X )
more
E( X
were
and
some
generating
of
dicult
the
to
expected
nd.
We
distribution.
Exploring
further
probability
functions
distributions
values
are
going
we
can
and
to
much
variances
star t
with
the
−
G ′(1))
k p
k
Example
Use
the

results
variance
of
a
of
the
probability
Binomial
random
generating
variable
function
with
to
nd
parameters n
the
and
expected
value
and
the
p
n
X
∼
B( n , p )
⇒
G (t )
=
(q
+
Use
pt )
the
probability
f or mula
n −1
⇒
G ′( t )
=
n (q
+
×
p
=
np ( q
+
=
np ( n
−
1)
n( n
−
(q
1) p
+
pt )
×
n
(q
+
(
X
)
=
G′
(
)
1
=
np
q
+
the
rst
and
second
derivatives.
p
2
1
=
p × 1



(
to
pt )
n
E
twice
2
2
=
it
pt )
obtain
G ″( t )
di erentiate
function
n −1
pt )
n
⇒
and
generating
np
Use
the
f or mula
f or
expected
Use
the
f or mula
f or
variance
value.
)
1
Var ( X
)
=
G ″ (1)
+
G ′(1) (1
−
G ′(1))
previous
n
2
=
n (n
−
1) p
q
(
+
p
the
results.
2
× 1





and
+
np (1
−
np )
)
1
2
=
It
is
much
Poisson
2
2
p
−
easier
the
to
+
nd
as
np
the
−
n
2
p
=
expected
Example
8
np (1
−
value
p)
and
=
Simplif y
npq
the
variance
of
the
result.
a
shows.
8
probability
Poisson
2
np
distribution,
Example
Use
n
random
generating
variable
with
function
to
nd
the
expected
value
and
the
variance
of
a
parameter m.
m ( t −1)
X
∼ Po
(
m
)
⇒
G (t )
=
e
m ( t −1)
⇒
G ′( t )
=
e
Use
the
and
di erentiate
m
=
=
me
G
(
)
1
2
×
m
E
(
X
)
Var ( X
=
)
=
=
G ″(1)
generating
twice
to
function
obtain
the
f or mula
rst
and
me
second
G ″( t )
it
m ( t −1)
×
m ( t −1)
⇒
probability
m
=
=
G′
e
(
)
1−1
me
+
m
derivatives.
m ( t −1)
(
)
1
(1
m
−
Use
the
f or mula
f or
expected
Use
the
f or mula
f or
variance
value.
and
the
previous
G ′(1))
results.
2
=
m
=
m
m
(
)
1−1
m
e
+
2
There
names
are
still
and
functions
illustrate
for
(
)
1−1
me
(
)
m
1−1
(1
−
me
)
2
+
m
−
many
many
make
their
m
=
other
of
m
Simplif y
distributions
those
that
distributions,
calculations
much
do
not
have
probability
easier.
the
result.
special
generating
Example
9
will
this.
Chapter
1
29
Example
Max
and
9
Marco
alter nately
throw
dar ts.
The
rst
one
who
scores
a
bull’s-eye
wins
the
5
3
game.
The
probabilities
that
Max
and
Marco
score
a
bull’s-eye
in
ever y
shot
are
and
the
and
8
4
respectively
.
variable
X
a
Find
b
What
a
G (t )
Max
throws
denotes
the
is
the
1
of
number
5
1
successive
throws
generating
expected
3
Their
number
probability
the
rst.
3
throws
3
t
+
×
4
4
of
the
the
before
1
2
=
before
function
of
throws
3
game
+
×
8
×
4
is
the
1
game
over.
and
is
+
×
4
×
4
8
4
t
3
1
3
×
4
3
×
8
×
4
1
5
t
8
3
+
1
×
4
3
×
4
1
×
8
Max
scores
Max
misses
and
Marco
scores
Max
misses
and
Marco
misses
Max
4
5
×
×
8
misses
scores
6
t
4
3
⎛
2
t ⎜1
4
+
t
+
32
⎝
and
Marco
misses
and
misses
Marco
or...
that
geometric
there
series
are
that
two
innite
alter nate
ter m
⎞
2
t
⎜
or
+ ...
⎞
3
or
8
2
⎛
3
or
scores
Max
Notice
=
result.
8
and
×
your
5
×
Max
1
verify
over?
and
+
random
4
t
8
independent
variable X
3
t
are
+ ... ⎟
⎟
⎝ 32
⎠
by
ter m
with
di erent
rst
ter ms
⎠
and
equal
Use
the
common
ratio.
2
⎛
5
3
2
+
t
⎜1
32
+
t
+
⎞
2
t
⎜
32
⎝
3
⎛
2
+ ...
⎟
⎝ 32
⎠
2
⎛
=
3
⎝
5
t
⎜
2
+
t
4
24 t
1
⎞
32
2
+ 5t
32
=
⎟
24 t
×
+ 5t
=
2
3
⎠
32
2
1
32
f or mula
f or
the
sum
of
an
2
3t
32
3t
innite
t
sequence
and
simplif y
your
32
answer.
24 + 5
G
(
)
1
=
29
=
32
3
=
The
1
sum
equal
b
Method
of
all
probabilities
must
29
to
1,
G(1)
=
1.
I
2
24 t
G (t )
+ 5t
=
Di erentiate
the
probability
2
32
3t
generating
2


G ( t )
24 + 10 t
 
32  3t
function.
Use
the
2



24 t
 
+ 5t
6t

quotient
=
rule.
2
2

32

3t
2
768 + 320 t

G ( t )
 72t
3
2
 30 t
+ 144 t
3
+ 30 t
=
2
2

32
3t

2
768 + 320 t
+ 72t
Simplif y
=
the
result.
2
2
(
32
3t
)
768 + 320 + 72

E

X

=
G


1
=
1160
=
= 1
38
2

32
3
Calculate
expect
At
least
two
throws
are
made
before
the
game
made.
is
30
over.
Exploring
further
probability
G′(1).
Notice
that
we
841

distributions
at
least
two
throws
to
be
be
Method
II
Use
*Unsaved
1.1
a
GDC
to
nd
G ′(1).
2
d
24∙x+5∙x
40
x=1
|
(
dx
)
2
32–3∙x
29
40
1.37931
29
|
2/99
At
is
least
Four
made
before
the
game
Notice
die
the
are
face
is
X
tossed.
up.
and
rolled
number
random
of
Write
Let
the
nd
the
until
we
rolls.
variable
X
denote
the
probability
probability
get
Write
a
“1”.
the
describing
number
distribution
generating
The
random
probability
this
of
generating
function
is
G (t )
random
variable
X
has
a
probability
heads
of
the
to
expect
be
at
least
made.
that
random
variable X
distribution
experiment
and
show
of
denotes
the
that
the
6
=
,
6
A
throws
we
function.
t
probability
3
that
1E
variable
A
are
two
coins
appear
2
throws
over.
Exercise
1
two
t
≠
5t
5
generating
function
2
G (t )
=
,
t
≠ 3.
Find:
3 − t
4
a
P (X
=
b
P (X
≤
1);
c
P (X
≥
3)
d
P (X
≥
k)
Use
and
5
the
the
0);
probability
variance
of
a
X
is
a
Ber noulli
b
X
is
a
negative
Find
in
the
expected
questions
1,
2
generating
the
function
following
to
nd
the
with
the
probability p;
binomial
with
the
parameters p
and
and
the
value
variables:
variable
value
expected
variance
of
the
and
r.
random
variables
3.
Chapter
1
31
6
Noddy
and
Taking
tur ns,
line.
The
Eve
play
they
rst
one
a
basketball
shoot
who
a
game
basketball
makes
a
in
at
basket
a
their
hoop
wins
cour tyard.
from
the
a
free
game.
throw
The
4
2
probabilities
that
Noddy
and
Eve
score
in
any
shot
are
and
7
3
respectively
.
The
Successive
random
game
is
variable
shots
X
are
denotes
independent.
the
number
of
Eve
shoots
shots
rst.
before
the
over.
2
a
Find
the
value
of
P (X
b
Find
the
probability
=
1)
and
show
that
P (X
=
4)
=
49
variable
X
c
What
the
d
Use
to
The
If
we
coin
is
at
are
probability
i
the
the
to
to
the
note
by
random
for
the
random
shots
the
is
before
maximum
the
game
number
is
of
over?
shots
over.
variables
of
the
the
of
nd
game
independent
going
function
result.
number
experiment
two
by
your
r ule
before
represented
represented
x
expected
independent
have
we

verify
empirical
made
of
look
actually
Now
the
be
sum
is
and
generating
tossing
events
of
number
random
variable
two
coins
tossing
of
heads
variable X,
Y.
Both
X
we
one
and
and
and
can
say
coin
we
can
coin
Y
at
2
have
that
a
we
time.
say
that
is
the
same
distribution:
0

1
1
2
2
p
i
They
also
have
the
same
1
generating
1
1
0
G
(t )
=
G
X
(t )
=
We
notice
functions
that
we
probability
when
we
G
×
+
t
=
if
2
we
obtain
+
t
two
G
(t )
=
Y
1
⎝ 2
the
course
parameter
mean
We
2
⎠
a
able
Poisson
of
to
cars
say
for
1
+
⎜
⎝ 2
companion,
of
number
were
⎛ 1
×
⎟
two
the
earlier,
number
we
⎟
2
of
at
within
a
generating
when
we
found
heads
shown
1
2
+
t
+
t
=
G
(t )
X
4
easily
For
petrol
half
1
=
⎠
were
arriving
1
⎞
t
distribution.
that
probability
calculated
together.
⎞
t
we
function
coins
+
⎜
2
these
result
generating
tossed
2
multiply
the
⎛ 1
(t )
X
In
1
1
t
Y
2
a
function:
an
2
manipulating
example,
station
hour
+
Y
4
the
the
suppose
in
one
mean
that
hour
the
is m.
number
m
,
was
and
in
three
hours
the
mean
number
was
3m.
The
reason
why
2
we
were
the
able
expected
Poisson
do
value
Exploring
We
further
is
or
distributions
transformations.
32
this
that
the
the
parameter
mean.
were
were
So
probability
a
Poisson
parameters
calculated
then
of
by
nding
distributions
using
the
of
the
given
the
distribution
is
corresponding
same
elementar y
probabilities.
Example
The
0
times
at
which
distributions
into
the
room
in
room.
room.
=
X
+
y
and
a
parameters
Let
random
respectively
.
the
relationship
Z
with
a
Find
Let
the
0.3
arrive
and
0.
variables X
random
probability
between
wasp
into
room
respectively
.
and
Y
variable Z
generating
a
denote
the
of
=
Y
(
(
P
X
=
)
r
of
t
=
=
∑
P
(
Y
=
)
s
(
ies
=
X
P
(
Z
s
)
= k
of
and
=
wasps
ies
nd
and
the
in
the
wasps
possible
insects
Z
is
the
number
wasps
Y
in
the
room.
down
probability
f or
each
generating
insect. To
f ollow
the
e
(
0.4
t
of
and
calculation
∑
and
1
more
(t )
Z,
functions
k
Z
ies
)
t
s
G
of
number
Write
0.1
(t )
independently
number
and
Poisson
e
r
Y
Y,
by
)
1
t
r
G
total
The
∑
arrive
number
X,
r
(t )
X
modelled
them.
0.3
G
be
Insects
the
denote
functions
can
=
f or
easily,
Y
use
(the
a
number
di erent
of
index,
wasps)
s.
)
1
t
e
When
we
insects
in
count
the
number
of
both
k
(
=
=
0.3
+
0.1)
(
)
t −1
=
G
(t)
×
G
X
The
0.3
e
(
)
t −1
+
0.1
(
)
t −1
=
above
illustrates
a
ver y
functions
probability
distribution
of
and
X
probability
0.1 ( t
×
the
room
we
make
no
−1
)
e
distinction
between
notice
ies
that
Z
and
:
wasps.
Po(0.4).
the
impor tant
that
given
doesn’t
random
proper ty
depend
of
on
variables,
the
but
is
valid
distributions.
:
X
1
−1
)
Theref ore
distribution
Theorem
t
Y
example
all
(
e
(t)
probability
across
0.3
e
are
two
independent
random
variables
with
the
2
corresponding
probability
generating
functions G
(t )
and
X
1
If
a
X
=
new
X
random
+
X

variable
then G
(t )
=
X
such
G
X
2
is
(t )
X
2
that
=
G
+ X
1
(t ).
G
X
(t )
X
2
×
G
(t )
X
1
2
Proof:
k
G
(t )
=
X
∑
P( X
=
k )t
k
r + s
=
∑
P( X
=
r
+
Use
s )t
the
substitution
k
=
r
+
s.
r + s
r + s
=
∑∑
r
P (( X
=
r )
∩
(X
1
=
s )) t
Rewrite
the
sum
by
using
variables
X
and
X
1
r
∑∑
r
P( X
=
r )P( X
1
=
s )t
2
Since
t
the
variables
are
independent
2
s
r
∑
P( X
=
r )t
1
the
multiplicative
probability
law.
s
∑
r
=
.
s
use
=
indices
s
and
=
both
2
P( X
=
s )t
Use
the
distributive
proper ty.
2
s
G
(t )
X
×
G
(t )
Q.E.D.
X
1
2
Chapter
1
33
Theorem
the
sum

can
of
be
more
extended
than
two
to
nd
the
probability
independent
random
generating
function
of
variables.
Corollar y
+
Given
that
X
,
X
1
the
, ...,
X
1
,
,
n ∈ 
are
n
independent
random
variables
with
n
corresponding
probability
generating
functions G
( t ),
G
X
( t ), ...,
( t ).
G
X
1
X
2
n
n
If
a
new
random
variable
X
is
such
that X
=
X
+
X
1
+ ... +
X
2
=
n
∑
then
X
k
k =1
n
G
(t )
= G
X
(t ) × G
X
( t ) × ... × G
X
1
(t )
=
X
2
n
∏
G
(t )
X
k
k =1
The
proof
This
of
this
corollar y
functions
simpler
of
a
corollar y
helps
us
random
experiments
Example
Random
function
to
is
left
easily
variable
described
to
the
reader
calculate
which
by
the
the
can
be
same
as
an
exercise.
probability
modelled
random
generating
as
a
sequence
of
variable.

variable
is
given
X
denotes
by
(t)
G
=
a
q
Ber noulli
+
pt,
trial
where
p
and
and
its
q
corresponding
are
respective
probability
probabilities
generating
of
a
success
X
and
a
failure
function
of
in
the
independent
the
trial.
binomial
Use
the
corollar y
random
above
variable Y
that
to
nd
the
describes
probability
a
sequence
distribution
of
n
such
trials.
n
A
Y
=
X
+
X
+ ... +
X
=
∑





independant
G
(t )
=
G
Y
describes
a
sequence
of
n
repetitions
of
a
Ber noulli
trial.
trials
Theref ore
⇒
distribution
independent
k =1
n
binomial
X
(t )
×
G
X
(t )
× ... ×
X
G
Y
f ollows
a
binomial
distribution.
(t )
X





Use
n
the
corollar y
to
nd
the
probability
generating
factors
function.
=
(q
+
pt )
×
(q
+
pt )
× ... ×
(q
+
pt )




n
Notice
factors
n
=
Exercise
1
The
(q
+
that
we
have
obtained
pt)
1F
probability
generating
functions
of
independent
random
variables
2
⎛ 1 +
X
and
Y
are
given
by
the
following
formulas:
G
(t )
X
⎛ 2
(t )
t
⎞
.
⎜
⎝
a
+
=
Y
⎟
3
⎠
Determine
G
(t);
X +Y
34
b
Find
c
Calculate
Exploring
P (X
+
Y
E (X
further
≤
1);
+
Y
).
probability
distributions
3t
⎞
=
and
⎜
⎝
2
G
the
simpler.
⎟
4
⎠
same
result
much
2
The
probability
generating
functions
of
independent
random
variables
3
⎛
X
and
Y
are
given
by
the
following
formulas:
G
(t )
=
⎛
(t )
=
Determine
⎠
G
(t);
+
Y
b
Calculate
E (X
c
Calculate
Var (X
+
which
bikes,
The
can
times
be
0.25,
in
modelled
0.15,
and
independently
and
Z
denote
+
Y );
by
and
crossing.
numbers
Find
the
probability
b
Find
the
probability
be
the
crossing
at
of
All
Let
that
each
at
of
with
them
them
two
at
a
crossing
parameters
arrive
variables X,
at
functions
least
given
of
arrive
random
generating
a
buses
distributions
respectively
.
the
the
cars,
Poisson
0.05
at
Y ).
a
at
and
⎠
⎟
2t
3
X
3
t
⎞
t
⎜
⎝
a
⎟
2
⎝
3
Y
t
⎜
X
G
⎞
the
of
kinds
crossing.
X,
of
Y,
Y,
and
vehicle
Z.
will
moment.
T
rafc
vital
4
Geometric
random
variable
X
has
the
corresponding
Analysis
is
component
a
in
probability
understanding
the
pt
distribution
function
given
G
by
(t )
=
,
where
p
and
q
are
requirements
and
X
1
qt
capabilities
probabilities
of
a
success
and
a
failure
respectively
.
Use
to
nd
the
probability
distribution
function
of
binomial
random
variable
Y
that
describes
a
are
independent
trials
until
we
achieve r
many
the
trafc
successes.
but
a
X
and
X
1
independent
Binomial
variables
with
the
none
of
efciently
of
them
networks,
can
capture
the
trafc
2
same
of
are
for
sequence
characteristics
5
trafc
proposed
analyzing
of
network.
the
models
negative
a
the
There
corollar y
of
probability
repetitions,
n
p,
however
and
n
1
generating
they
have
respectively
.
dierent
Use
the
numbers
characteristics
of
probability
networks
to
show
that
X
+
X
is
a
types
ever y
possible
Binomial
circumstance.
2
However
,
and
all
2
function
1
variable
of
under
hence
nd
the
one
of
the
most
parameters.
widely
used
and
oldest
+
b
Given
that
X
,
X
1
random
,
...,
2
variables
X
,
are
k ∈ 
independent
Binomial
k
having
trafc
the
same
probability p
models
number
of
repetitions,
n
,
1
n
,
2
…,
n
the
but
Poisson
dierent
is
Model.
respectively
.
k
k
Use
mathematical
induction
to
show
that Y
=
∑
is
X
also
a
i
i =1
Binomial
distribution
and
nd
the
parameters.
Chapter
1
35
Review
EXAM
1
A
exercise
STYLE
QUESTIONS
random
variable
X
has
a
probability
generating
function
4t
G (t )
=
,
5
2
a
P (1
b
P (X
c
E (X );
d
Var (X ).
The
≤
X
≥
times
0.28
each
≤
a
4);
rabbit,
by
fox,
of
All
random
them
at
the
probability
b
Find
the
expected
c
Find
the
probability
In
a
cer tain
at
a
Helping
i
the
ii
the
the
four th
the
rst
to
What
will
is
A
if
continuous
distribution
0,
⎧
⎪
of
at
75%
0.6,
Z
denote
the
animals
least
two
at
of
X,
the
Y,
=
animals
will
be
of
volunteers
Students
student
from
are
the
involved
, 0
school
are
is
the
rst
one
is
the
second
who
is
the
not
selected
will
exactly
not
six
need
need
random
who
occur
than
number
six
X
involved
3rd
involved
10
of
involved
variable
is
before
students
more
expected
we
student
one
36
Find
b
Hence
c
What
Exploring
the
<
with
selected
is
the
given
the
x
≤
>
≤
further
selected
to
value
modal
of
by
the
cumulative
value
probability
a
probability
of
the
density
function.
variable X ?
distributions
be
programme?
2
the
is
students?
needed
2
positive
the
the
programme
0
determine
is
with
student;
⎩
a
the
the
selected
students
in
with
2
x
at
involved
who
a
1
,
Z.
programme;
⎪
⎪
the
that:
2
⎨
at
meadow
.
⎪ x
F( x )
be
numbers
and
function
x
can
0.12,
independently
and
functions
probability
selected
with
the
selected
the
student
select
we
b
that
selected
programme
iv
meadow
programme;
involved
iii
is
Y,
a
parameters
arrive
X,
at
moment.
programme.
What
them
number
given
four th
with
of
arrive
with
generating
community
,
random.
deer
meadow
.
the
a
a
variables
Find
at
and
distributions
a
Peer
4
a
Poisson
Let
meadow
3
Find:
respectively
.
meadow
.
of
≠ 5.
2);
modelled
and
t
t
5
A
random
variable
t
G (t )
=
,
a
t
≠
,
bt
a
Show
b
Hence
X
has
a
probability
generating
function
a
where
a
and
b
are
nonzero
integers.
b
that
b
nd
=
a
–
E(X )
1.
in
terms
of
a
2
c
Show
a
X
6
that
and
X
1
Var (X )
are
=
a
–
a
independent
Poisson
variables
with
the
same
2
parameter
that
+
X
m.
X
1
Use
is
a
probability
Poisson
generating
variable
and
functions
nd
the
to
show
parameter.
2
+
Given
b
that
X
,
X
1
random
, ...,
X
2
variables
,
are
n ∈ 
independent
Poisson
n
having
the
same
parameter m,
use
n
mathematical
induction
to
show
that Y
=
∑
is
X
also
a
k
k =1
Poisson
The
the
of
French
probability.
A
which
area:
the
normal
1711
Method
He
found
is
that
on
in
famed
summing
die
he
of
contained
distribution
Moivre
would
of
his
he
was
the
the
most
the
case
sleeping
day
that
he
parameter nm
Moivre
‘The
the
a
15
minutes
for
24
Events
contribution
of
his
by
of
He
this
the
death.
each
calculated
hours.
to
in
trials.
own
longer
pioneered
theor y
of
of
number
day
progression,
slept
the
distribution
large
the
and
Doctrine
signicant
of
(1667–1754)
Probability
binomial
predicting
arithmetic
the
geometr y
published
to
the
for
De
analytic
Calculating
approximation
De
and,
In
with
mathematician
development
Chances:
Play’
distribution
night
that
was
he
right!
Chapter
1
37
Chapter

Cumulative

Given
the

For
distribution
random
probability
function
summary
variable
function
F
: 
continuous
cumulative
is
random
distribution
X
(discrete
→ [0, 1] the
P : 
→ [0, 1]
function
F (x )
=
P( X
variables
function
continuous)
cumulative
≤
the
F
or
x ),
x
the
corresponding
distribution
∈ 
probability
are
and
related
as
density
function f
and
the
follows:
x
F ′( x )
=
(
f
x
)
⇔
(
F
x
)
=
f
( t ) dt
∫
−∞
Geometric

A
discrete
distribution
random
variable
X
k
X
∼ Geo( p )
Expected
if
P
(
value
X
=
k
and
)
=
is
said
to
have
a
geometric
distribution
and
we
write
1
q
p,
where
p ∈ ,
0
≤
p
≤ 1
,
q
=
1
−
p,
k
=
1
,
2,
3,
4 ...
variance
1

Given
a
geometric
random
variable,
if
X
∼ Geo( p ),
then
E
(
X
)
=
q
and
Var
(
X
)
=
2
p
Negative

A
binomial
discrete
random
distribution
variable
X
is
said
to
⎛ k
have
X
∼
NB( r ,
p)
if
P
(
X
=
)
k
=
⎜
⎝
0
≤
p
≤ 1
,
q
= 1 −
Probability

Let
X
P (X
be
=
a
k)
p,
k
r ,
r
+ 1
, r
generating
discrete
=
=
p
,
k
=
random
0,
,
2,
+ 2,
⎟
r
r
a
negative
binomial
distribution
and
1
r
q
r
,
p
where
⎠
+ 3,...
function
variable
3,
…
are
assuming
the
nonnegative
corresponding
integer
values
probabilities.
Then
and
the
function
k
∞
k
of
the
G (t )
form
=
∑
p
t
2
=
k
p
+
0
p
t
+
p
1
3
t
+
p
2
t
generating
PGF
Ber noulli
X
~
B(,
p)
G (t)
=
q
+
Binomial
X
~
B(n,
p)
G (t )
=
(q
G (t )
=
e
G (t )
=
pt
n
m
Poisson
X
~
Po(m)
+
(
t
pt )
)
1
pt
Geometric
X
~
Geo ( p)
1
qt
r
pt
⎛
Negative
binomial
X
~
NB (r,
p)
G (t)
=
⎜
⎝
38
Exploring
further
probability
distributions
⎞
⎟
1
−
qt
...
+
p
t
n
function.
Distribution
n
+
3
k =0
probability
we
1⎞
k
write
p
⎠
+
...
is
called
a
Properties
i)
G ()
=
ii)
E (X )
iii)
Var (X )
of
PGF
X
and
Expected
value
=
=
G ″()
are
+
G ′() (−
random
two
G ′())
variables
independent
random
variables
with
the
corresponding
probability
2
generating
functions
G
(t )
and
G
X
X

+
X
2
then
G
(t )
X
( t ).
If
a
new
random
variable
X
is
such
that
X
1
=
Variance
G ′()
X

X
and

Independent

and
=
2
G
(t )
X
1
+
X
=
G
(t )
X
2
1
×
( t ).
G
X
2
Chapter
1
39
Expectation
algebra
and
Central
Limit
2
Theorem
CHAPTER
Linear
7.2
OBJECTIVES:
transformation
of
a
single
random
variable.
2
E ( aX
+
Mean
b)
of
=
aE ( X )
linear
Variance
of
A
linear
of
Y
ou
1
you
should
Given
value
that
and
the
In
product
∼
of
+ b)
of
of
=
n
a
the
Var ( X )
random
of
n
variables.
independent
independent
independent
par ticular
,
Central
random
normal
Limit
=
how
to:
nd
B(5, 0.3)
standard
np
⇒
variables.
variables.
random
E( XY
variables
is
)
=
E( X ) E(Y
)
normally
Theorem.
deviation
Skills
the
of
expected
1
check:
Given
E( X )
=
5
×
0.3
=
that
X
∼
B( n ,
X
with
p)
3
E( X )
E( X )
random
start
know
X
Var ( aX
combinations
combination
distributed.
Before
b,
combinations
linear
Expectation
7.4
+
= 2,
Var ( X )
nd
=
the
following
2
1.5
probabilities:
Var ( X )
2
A
=
⇒ σ
npq
random
variable
=
X
5 × 0.3 × 0.7
has
=
1.02
the
a
2
A
P (X
2);
=
random
P (1
b
variable
X
has
≤
X
≤
3)
the
2
following
following
parameters:
parameters:
2
E( X )
Find
=
a
the
variable
The
Var ( X )
value
of
follows
and
E( X )
the
=
=
(2a
2a
=
such
2a
between
Var( X )
of
a ∈ .
the
random
distribution.
the
a
expected
Poisson
therefore
we
variable
write
2
−
1 ⇒
1)( a
1)
2a
−
a
0
−
1
=
0
or
a
a
= 1
2
40
Expectation
algebra
and
Central
=
7a
and
3
−1
,
that
Poisson
variance
2
a
a
a
relationship
value
is
and
E( X )
Limit
Theorem
Var ( X )
value
=
of
follows
a
a
6a
−
such
9a
−
that
Poisson
2,
the
a ∈ .
Find
random
distribution.
the
variable
Normal
distributions
and
probability
models
Sometimes,
In
probability
theor y
,
variable
one
variable
could
would
be
expected
expect
to
repeated
value
obtain
an
refers
if
endless
the
to
the
value
experiment
number
of
of
a
involving
times.
chance,
random
This
this
intuitive
crashes
period
of
of
the
expected
value
is
a
direct
consequence
of
of
Large
Numbers,
i.e.
it
is
the
limit
of
the
sample
mean
as
size
approaches
innity
.
This
value
may
not
be
the
of
the
English
sense
of
this
word,
rather
it
could
even
Dramatic
impressive
debris
media
may
shown
make
‘expected’
ner vous
in
plane
shor t
the
in
sample
a
by
the
pictures
Law
and
few
in
time.
headlines
explanation
a
occur
just
be
unlikely
,
yers
feel
fearful.
or
However
,
aviation
safety
counter-intuitive.
data
Perhaps
some
random
variable
of
the
are
most
impor tant
expected
value
concepts
and
in
the
variance.
study
We
will
of
reveals
reality
a
investigate
algebraic
proper ties
in
the
aspect
of
so-called
Fur thermore,
we
will
be
looking
at
linear
is
the
combinations
proper ties
At
the
results
the
end
in
random
of
the
of
variables
parameters
this
chapter
mathematics,
impor tant
role
that
you
the
and
of
will
independence
combined
study
Central
normal
how
such
Limit
one
the
Theorem,
distributions
aect
random
of
have
truth:
safest
it
air
has
in
the
in
histor y
of
of
aviation.
independent
different
perhaps
expectation
been
algebra.
a
a
counter-intuitive
travel
their
and
The
average
the
number
of
take
each
ights
that
variables.
most
and
impor tant
large,
get
discover
of
devising
off
a
and
ver y
reality
day
data
is
analysts
different
given
ver y
by
picture
the
numbers.
probability
parameters
such
as
models.
in
The
expected
probability
regression
value
distribution,
analysis,
which
we
and
and
will
variance
in
are
statistical
study
fur ther
impor tant
topics,
in
Chapter
4.
Chapter
2
41
2.1
In
Expectation
statistics,
and
these
If
●
In
If
If
If
k
is
of
each
of
k
tendency
(variance,
the
data
the
of
the
new
member
data
new
of
to
data
the
median,
deviation,
measures
conducted
each
data
added
standard
these
we
(mean,
etc.)
etc.)
represent
two
are
investigations
that:
to
new
of
is
the
noted
added
the
value
constant
of
we
member
mean
a
central
companion
and
value
variance
●
core
constant
each
the
●
values
changes,
a
of
dispersion
the
the
mean
●
of
when
adjusted.
on
measures
measures
change
algebra
member
set
set
also
is
data
each
set
data
the
increased
multiplied
set
is k
is
of
data
set
the
by k;
by
times
member
doesn’t
set
of
a
the
the
constant k,
original
data
set
mean;
the
change;
multiplied
by
a
constant k,
2
the
new
Since
random
given
by
variance
of
variables
the
allow
data
us
is
to
times
k
model
the
original
random
variance.
phenomena
Linear
sets
of
data,
we
expect
that
the
mean
value
and
of
of
the
set
of
data
will
behave
similarly
to
the
expected
transformation
variance
value
a
variable
is
transformation
variance
of
a
random
variable.
obtained
and
Linear
transformation
of
a
single
Luca
will
a
and
give
Find
b
his
him
the
Suggest
play
expected
later
of
a
a
game.
Luca
ips
a
coin.
and
the
variance
Luca
will
get
in
decides
If
he
gets
him
with
and
the
variance
Luca
will
get
in
relationship
and
award
between
the
of
this
the
random
€5
of
if
the
new
he
x
by
constants.
random
version
expected
values
of
variable X
a
his
father
that
describes
that
describes
the
variances
of
the
gets
nothing
if
he
obtains
a
“tail”

1
1
2
2
the
gets
€1
if
he
probability
obtains
a
distribution
“head”.
table
variable
X
and
ll
in
all
the
∑
x
p
i
Calculate
the
expected
values.
value.
Calculate
the
variance.
i
i =1
1
E( X )
=
0
1
×
+
1 ×
2
1
=
2
2
2
2
Var( X )
=
∑
x
2
p
i
−
(E ( X ))
i
i =1
2
⎛
Var ( X )
=
⎜
⎝
42
Expectation
1
2
0
×
2
algebra
1 ⎞
2
+1
and
×
⎟
2
⎠
Central
⎛
1 ⎞
⎜
⎟
⎝
2 ⎠
Limit
1
=
1
−
2
Theorem
1
=
4
4
Draw
f or
)
2
=
the
game.
i
E( X )
the
“head”.
variable Y
the
and
and
x
“head”
Y
0
i
=
a
obtains
Luca
P( X
the
game.
value
money
X
to
this
a
=
of
variable
value
money
expected
variables
X
addition
€.
of
the
amount
c
father
father
Find
is

amount
Luca’s
by
that
multiplication
variable
Example
a
and
the
Luca
b
Y
=
y
0
i
1
P
(Y
=
y
nothing
if
he
obtains
a
“tail”
and
Draw
the
gets
€5
if
he
obtains
a
probability
“head”.
distribution
table
1
)
i
gets
5
f or
2
the
variable
Y
and
ll
in
all
the
2
values.
1
E
(
)
Y
=
0
1
×
+
5
5
×
=
2
2
Calculate
the
expected
value.
Calculate
the
variance.
2
2
⎛
Var
(
Y
)
=
⎜
1
2
0
×
E
(
Y
Var
)
=
(
Y
2
5E
)
=
(
⎛
5 ⎞
⎟
⎜
⎟
2 ⎠
⎝
2 ⎠
25
25
=
25
−
=
2
4
4
)
X
25
1 ⎞
×
+ 5
⎝
c
2
By
(
Var
the
)
X
looking
y
variables
=
5 x
i
,
that
increased
variance
the
let’s
consider
Example
Luca
“head”
a
Find
his
but
the
amount
b
Suggest
(in
variation
of
the
father
again
regardless
expected
of
a
of
the
a
similar
outcome
game.
and
the
variance
will
get
in
and
between
the
Luca’s
obtained
Luca
relationship
),
play
value
money
Example
z
Z
the
by
=
write Y
expected
the
was
this
of
he
the
father
will
will
give
same
value
f actor
increased
random
5 X
by
was
and
the
the
square
of
z
i
him
an
€
extra
variable Z
that
if
he
obtains
€2.
describes
the
game.
expected
values
and
the
variances
of
the
variables X
gets
€2
if
he
obtains
a
“tail”
3
and
=
give
him
gets
Draw
(
that
we
Z
2
i
P
so
of
factor.
Luca
=
notice
2
values
game.
a
Z
1,
possible

and
a
another
we
=
i
the
i
Notice
Now
at
1
1
2
2
€3
the
if
he
obtains
probability
a
“head”.
distribution
table
)
f or
the
variable
Z
and
ll
in
all
the
values.
1
E
(
Z
)
=
2
1
×
+
3
5
=
×
2
2
Calculate
the
expected
value.
Calculate
the
variance.
2
2
1
⎛
Var
(
Z
)
=
⎜
2
2
×
+
⎝
5
b
E
(
Z
)
=
1 ⎞
2
3
×
2
⎟
−
2 ⎠
⎛
5 ⎞
⎜
⎟
⎝
2 ⎠
13
=
25
−
2
1
=
4
4
1
=
+
2
=
E
(
X
)
+
We
2
can
see
that
=
z
+
x
i
2
2, i
=
1,
2
i
2
so
we
write
Z
=
X
+
2 .
1
Var
(
Z
)
=
=
Var
(
X
)
Notice
that
the
expected
value
was
4
Example
3
looks
at
one
last
variation
of
the
increased
by
while
variance
the
the
same
added
remained
value
the
same.
game.
Chapter
2
43
Example
Luca’s
a

father
Find
the
amount
Luca’s
ips
b
a
of
father
Find
the
money
then
he
will
give
Luca
€0
value
and
the
variance
Luca
will
get
in
decides
and
expected
of
Suggest
R
that
to
make
it
the
if
he
of
gets
the
a
“head”
random
a
money
more
realistic
and
introduces
value
and
the
variance
Luca
will
get
in
relationship
between
the
this
of
the
random
variation
expected
of
values
the
=
r
0
(
R
=
r
and
the
€3
tax
gets
0
=
Luca
⎛
Var ( R )
10
×
=
⎜
1
2
0
=
5
1 ⎞
2
×
+ 10
×
2
and
ll
i
Draw
in
− 3
the
if
1
1
2
2
×
+
7
2
table
5
=
50
−
25
=
25
all
the
the
the
variance.
loses
€3
and
obtains
a
if
he
gets
€7
R
value.
obtains
(€10
“head”.
probability
×
⎜
1
2
(
−3
=
=
2
–
Draw
)
1 ⎞
2
×
7
+
×
2
5
S
distribution
and
ll
in
a
€3)
if
he
the
=
25
the
expected
−
3
=
⎟
=
E( R )
−
−
4
=
values.
value.
Calculate
the
variance.
25
3
We
=
Var
so
can
( R)
we
see
can
that
write
=
s
that
by
suggest
that
all
the
changes
from
the
the
the
S
r
valid
for
random
variables
too.
The
these
transformation
of
results
a
about
random
the
expected
the
remains
Central
Limit
Theorem
value
value
2
the
was
that
variable,
was
whilst
the
same.
that
of
a
random
the
probability
variable
taking
of
a
par ticular
outcome
remains
the
when
perform
same
linear
the
and
1,
following
a
algebra
=
3
data
parameters
variable.
R
same
from
a
formalizes
3, i
i
=
Notice
Expectation
the
2
variance
are
the
f or
58
2
2
−
2 ⎠
subtracted
examples
table
all
Calculate
reduced
theorem
variable
values.
Calculate
Notice
adjustment
a
probability
f or
i
(S )
obtains
2
⎝
Var
he
1
⎛
(S ) =
=
a
)
2
E(S )
€10
expected
variable
Var
obtains
7
1
=
gets
the
“tail”
s
he
2 ⎠
−3
i
=
if
Calculate
Luca
s
E(S )
variables
2
−
⎟
b
(S
the
2
⎝
=
the
1
+
×
2
44
whenever
describes
of
nothing
distribution
2
1
E( R )
that
variances
“head”.
1
and
linear
the
)
i
2
These
coin.
0
1
c
a
describes
game.
“tail”
P
a
variable S
Luca
S
that
S
i
P
ipping
game.
a
R
when
variable R
coin.
amount
c
said
expected
we
transformation
random
variable.
on
Theorem
1:
Given
that
and
b ∈ ,
a,
(
i
E
aX
ii
Var
X
is
a
random
variable
with
nite
parameters
then
+ b
)
=
aE
(
)
X
+ b;
2
(
aX
)
+ b
=
a
(
Var
)
X
Proof:
As
random
prove
Case
Let
i
the
theorem
can
for
be
both
either
discrete
or
continuous,
we
must
cases:
I
X
μ
variables
be
=
a
discrete
E( X )
E( aX
+ b)
=
x P( X
∑
=
( ax
∑
=
random
(
P( X
X
Then:
x ) ⇒
+ b)
ax P
∑
=
variable.
=
x
)
=
+
x )
∑
=
bP
∑
(
X
( ax P( X
=
x
)
=
=
a
x ) + b P( X
∑
x P
(
X
=
=
x
x ))
)
+
b
∑

E
=
aE
(X )
+
(
)
X
=
E (X
Var ( X )
=
∑
X
=
x
)
)
1
2
) − (E ( X ))
2
ii
X
(
b
2
Var
(
P





,E (X )
=
μ
2
x
P( X
=
x )
μ
−
⇒
2
2
Var
(
aX
+ b
)
(
=
∑
ax
+ b
2
=
(a
∑
∑
(a
=
x ) − (a μ
+ b)
2
+ 2 axb
+ b
2
)P ( X
=
x )
−
2
μ
(a
2
+ 2 a μb
2
x
2
=
P (X
2
x
2
=
)
∑
=
x )
+
2axb P ( X
=
x )
+
=
x )
b
2
P (X
P (X
=
x ) +
2ab
∑
x P (X
E
2
a
∑
(
x
i
μ
E
=
(
Let
E( X )
aX
+ b
X
=
)
be
∫
=
a
xf
)
=
+
2
(
X
=
x
)
−
(x )
Var
dx
+ b)
−
+
b
X
(
X
)
b
P (X
(
X
)
+
2
+ 2a μb
=
x ) − (a
μ
+ b
)
2
2
+
2a μb
+ b
)





1
2
2
−
2
−
2
−
=
a
b
Var( X
)
⎟
⎠
random
variable.
Then:
⇒
f
( x ) dx
=
∫
( axf
(x )
+
bf
( x )) dx
=
a
xf
( x ) dx

E
aE
2
μ
2
∑

=
(a
2
⎞
μ





continuous
∫ ( ax
+
2
P
⎝
II
x ))
)
2
P
2
⎛
⎜
Case
(
2
∑
2
=
=
2
x



=
)
2
P (X
2
a
+ b
(
X
)
+
b
f
( x ) dx
∫




1
b
Chapter
2
45
2
ii
Var
(
X
)
=
∫
2
x
(
f
x
)
− μ
dx
⇒
2
2
Var ( aX
+ b)
=
∫
( ax
+ b)
2
=
∫ (a
f
( x ) dx
(aμ
−
2
+
b
)
2
x
+ 2abx
2
+ b
)
f
( x )dx
2
μ
− (a
2
+ 2 ab μ
+ b
)
2
2
2
2
2
2
=
a
x
∫
f
( x )dx
+
xf
2ab
( x ) dx
2
=
(
(∫
)
X
−
(a
μ
+ 2ab μ
1
2
+
2
a
( x )dx





+
∫
f
∫
2
2
a
b



E
=
+
∫
2
b
2
−
2
−
2
b
2
x
f
(
)
x
μ
dx
)

Var
(
X
)
2
=
Exercise
1
Given
Var
(
X
)
2A
that
variance
a
a
X
1.2,
3X
is
a
nd
b
random
the
X
+
variable
expected
3
value
4X
c
with
+
the
and
1
expected
variance
2X
d
−
value
of
5
the
and
following:
kX
e
5.3
+
p;
k,
p ∈ 
2
2
Given
that
a
random
variable
X
∼
B
nd:
10,
5
a
3
A
E
(
3X
+ 2
random
)
;
Var
b
variable
Y
(
follows
3X
a
2
)
geometric
distribution
with
the
2
parameter
p
.

Find
the
expected
value
and
variance
of
2
Y
1
3
4
Given
a
E(3
that
−
a
random
2Y );
variable
b
Var
(
Y
~
3
Po(2)
2
Y
nd:
)
1
5
Given
that
a
random
variable
X
∼
NB
,
nd:
8
3
a
6
E
2X
Given
nd
7
(
A
3
that
Var
(
a
5X
)
;
b
random
+ 3
continuous
Var ( 2 X
variable
X
11)
∼
B
(15,
p
)
and
E
(
X
)
=
6,
)
random
variable
X
has
a
probability
density
⎧1
⎪
function
given
by
the
formula
f
(
x
)
=
⎪
⎩
Find
46
the
exact
Expectation
values
algebra
and
of
the
Central
expected
Limit
x,
0
≤
x
≤
6
0,
otherwise
⎨3
value
Theorem
and
variance
of
3X
+
2.
+ b
)
So
far
Let’s
We
we
see
are
were
what
going
Linear
Hannah
X
a
8
Y
a
“head”,
than
the
a
transformation
involve
the
two
he
two
or
variables
or
more
are
more
than
of
of
just
one
random
variable.
variables.
independent.
variables
to
she
from
their
will
the
table
between
of
he
the
ipping
obtains
and
they
her
ear n
father,
random
values
will
get.
that
the
x
0
i
P (X
=
get
€2
if
variables
respectively
.
game.
value
money
Hannah
=
this
will
and
variance
will
get
3
of
Z
get
times
more
together.
describes
expected
and
game
Hannah
Luca
their
Random
this
in
will
value
the
and
variances
of
amount
variance
the
of
of W
variables X,
W
a
X
in
expected
variable W
the
Hannah
“head”.
get
and
put
Calculate
expected
a
coin.
together
the
they
they
a
Luca
calculate
Again
time.
the
if
amount
from
father.
this
and
the
received
together
€5
by
Hannah
amount
increase
his
father
get
money
distribution
earn
relationships
and
Luca
distribution
decides
probability
with
represents
received
will
game
amount
Z
money
they
Suggest
Z,
of
probability
more
Find
linear
whilst
the
grandmother
times
Y,
that
play
variable
the
money
c
Luca
represent
Draw
money
b
and
random
Their
assume
a
we

obtains
and
The
to
if
transformation
Example
she
performing
happens
x
gets
nothing
if
she
obtains
a
2
1
1
2
2
“tail”
and
Draw
the
gets
€2
if
she
probability
obtains
a
distribution
“head”.
table
f or
)
i
E
(
)
X
= 1
,
Var
(
X
)
the
=
1
variable
Calculate
Luca
Y
=
y
0
i
the
=
y
and
ll
expected
nothing
in
all
value
if
he
the
and
values.
variance.
obtains
a
“tail”
5
and
P (Y
gets
X
1
1
2
2
gets
€5
if
he
probability
obtains
a
“head”.
distribution
table
Draw
f or
the
)
i
variable
5
E
(
)
Y
(
Var
Y
)
=
Calculate
2
Z
=
=
X
+
Y
If
z
i
they
0
2
5
7
=
z
in
expected
1
1
1
1
4
4
4
4
of
the
the
values
pairs
)
put
adding
The
P (Z
ll
all
the
values.
value
and
variance.
4
are
Z
and
25
,
=
Y
of
money
money
Z
f or
together
they
the
then
we
ear n.
cor responding
outcomes:
i
( T, T )
( T,
1
E
(
Z
)
= 0
1
×
+
2
×
4
1
+
5
×
4
1
+
7
×
4
7
H)
→
→
0
0
+
+
0,
5,
( H, T )
( H,
H)
Calculate
the
expected
Calculate
the
variance.
→
→
2
+
2
0
+
5
value.
=
4
2
2
2
⎛
Var
(
Z
)
=
0
4
7
+
4
49

4
2
5
+
4
78
=
2
2
+
⎜
⎝
2
4
⎞
⎛ 7
⎞
⎟
⎜
⎟
⎠
⎝
2 ⎠
29
=
4
4
Chapter
2
47
W
b
=
8X
+
Again
3
Y
we
are
ear n. The
W
= w
0
i
5
6
adding
values
( T, T )
1
(W
= w
1
1
4
4
(
)
W
= 0
×
1
+
15
1
×
+
4
the
pairs
+
0
0,
of
( T,
outcomes:
→
H)
→
8
×
+
2
+
0
3 ×
5
0,
4
( H,
1
f or
they
1
( H, T )
4
E
→
)
i
W
money
3
cor responding
P
of
the
16
1
×
+
4
31 ×
=
4
→
H)
8
×
+
2
3
×
5
31
4
Calculate
the
expected
value.
Calculate
the
variance.
2
2
2
⎛
Var
(
X
)
=
2
0
2
15
⎝
4
+
4
721
=
⎞
31
+
+
⎜
2
16
−
⎟
4
961
⎛ 31 ⎞
4
⎜
⎠
⎟
2
⎝
⎠
481

=
2
4
4
5
E
c
(
X
)
+
E
(
Y
)
=
E
(
X
)
+Y
=
E
(
)
Z
Notice
that
7
+
1
=
2
2
25
Var
(
X
)
+
Var
(
Y
)
=
Var
(
X
+ Y
)
=
Var
(
Z
)
Notice
that
29
1 +
=
4
4
5
8E
(
X
)
+ 3E
(
Y
)
=
E
(
8X
)
+ 3
Y
=
E
(
W
)
Notice
that
×
8
1 +
31
3 ×
=
2
2
25
64 Var
(
)
X
+ 9 Var
(
Y
)
=
Var
(
8X
+ 3
Y
)
=
Var
(
W
)
Notice
that
×
64
1 +
9
481
×
=
4
Now
let’s
take
Example
Mia
and
another
example,
this
time
with
probabilities
Theo
draw
marbles
from
two
boxes.
Mia
contains
2
white
and
3
black
marbles.
Theo
it
contains
4
white
and
3
black
marbles.
Random
white
marbles
represents
a
Draw
the
the
variance
Their
Mia
b
Find
the
X,
Y,
Z,
of
probability
of
all
the
a
decide
to
are
going
marbles
distribution
to
put
X,
give
the
Y,
tables
€3
money
and
between
to
of
draws
two
a
marble
marbles
variables X
draw
and
and
x
i
and
respectively
.
calculate
the
=
x
from
form
Y
the
the
rst
The
box;
second
represent
random
expected
Theo
€2
for
each
the
random
the
expected
variable W
expected
white
values
that
value
and
value
the
box;
number
variable Z
and
marble
describes
and
variance
variances
the
they
draw
.
of
the
the
of
amount
of
W
variables
W
can
0

3
2
Draw
5
5
f or
marble
P (X
distributed.
together.
Mia
=
uniformly
together.
a
X
not
Z
calculate
the
draws
draw
,
they
and
Mia
distribution
together,
relationship
and
white
decides
ear n
Theo
variables
probability
they
Suggest
and
number
Theo
money
c
Mia
grandfather
and
are

it
of
that
4
draw
from
the
one
the
black
rst
box
probability
or
one
of
white
marbles.
distribution
table
)
i
the
variable
X
and
ll
in
all
the
2
2
E
(
X
)
2
,
=
Var
(
X
)
5
48
Expectation
and
2 ⎞
⎜
⎟
−
=
5
algebra
⎛
Central
⎝
5
⎠
Limit
6
values.
=
Calculate
25
Theorem
expected
value
and
variance.
⎛ 3 ⎞
⎛ 3 ⎞ ⎛ 4 ⎞
⎜
⎜
⎝
P (Y
=
0)
⎟
2
1
⎠
=
=
⎜
⎝
2
1
⎝
,
⎛ 7 ⎞
⎟ ⎜
P (Y
= 1)
⎠ ⎝
Theo
1
one
=
⎜
⎠
⎝
black
marbles
or
black
and
one
white
or
white
from
the
marbles
second
box
7
⎟
2
two
,
two
⎛ 7 ⎞
7
draw
4
⎠
=
⎟
can
⎟
of
marbles.
⎠
⎛ 4 ⎞
⎜
⎝
P (Y
=
2)
⎟
2
2
⎠
=
=
⎛ 7 ⎞
⎜
⎝
7
⎟
2
⎠
Draw
Y
=
y
0
i

f or
P
(Y
=
y
the
probability
distribution
table
2
1
4
2
7
7
7
the
variable
Y
and
ll
in
all
the
values.
)
i
Calculate
the
expected
value
and
2
8
(
E
)
Y
4
⎛
=
(
Var
,
Y
)
=
7
8 ⎞
+
⎜
⎟
⎝ 7
⎛ 8 ⎞
−
⎜
7 ⎠
variance.
20
=
⎟
⎝ 7 ⎠
If
49
they
we
Z
=
X
are
put
3
(
Z
=
0
)
1
×
=
5
actually
(
)
Z
= 1
Z
=
2
)
4
Z
= 3
)
3
2
35
=
14
=
1)
↔
(1
+
0)
or
(0
+
1)
=
7
35
(
Z
=
z
i
(
Z
)
=
of
+
0)
( Z
=
2)
↔
(1
+
1)
or
(0
+
2)
( Z
=
3)
↔
(1
+
2)
35

2
3
3
2
2
4
35
5
5
35
the
the
probability
variable
Z
distribution
and
ll
in
all
table
the
values.
)
3
E
(0
5
f or
P
↔
2
=
0
i
0)
4
7
z
=
( Z
Draw
Z
number
( Z
5
=
×
5
drew.
=
2
×
5
2
=
14
=
7
+
7
2
(
4
×
5
×
5
P
3
7
=
they
35
+
2
(
the
then
3
1
×
=
5
P
adding
together
=
7
2
P
marbles
+Y
marbles
P
the
0
2
+
×
1 ×
35
2
+
2
×
4
+
5
3
54
×
5
=
35
Calculate
the
expected
value.
Calculate
the
variance.
35
2
2
⎛
Var
(
)
Z
=
⎜
0
+
⎝
+
5
106
=
W
= 3X
+
36 ⎞
⎛ 54
⎞
⎟
⎜
⎟
35 ⎠
⎝
35 ⎠
+
5
2916
794
=

35
b
8
1225
1225
This
2Y
they
W
= w
i
0
2
3
4
5
7
3
12
2
6
8
4
= w
35
35
35
35
35
35
)
we
ear n,
marbles
W
P (W
time
f or
not
they
the
are
adding
the
the
number
draw. The
of
money
white
values
cor responding
pairs
of
of
outcomes:
i
0
+
0,
0
+
2,
3
+
0,
0
+
4,
3
+
2,
3
Chapter
+
2
4
49
24
E
(
)
W
=
0
6
+
24
+
40
+
35
28
+
35
+
35
122
35
35
(
)
X
=
0
⎜
18
+
558
200
+
35
⎝
96
+
196 ⎞
+
35
+
35
14, 884
the
expected
Calculate
the
variance.
⎛ 122 ⎞
−
⎟
35
35
⎠
⎜
⎝
⎟
35
⎠
4646
=
=
35
1225
1225
2
E
c
(
X
)
+
E
(
Y
)
=
E
(
X
)
+Y
=
(
E
Notice
)
Z
8
=
7
6
(
)
X
+
Var
(
Y
)
=
(
Var
X
)
+Y
=
(
Var
Z
Notice
)
54

that
5
Var
value.
35
2
48
⎛
Var
Calculate
=
35
20
794
=

that
25
49
1225
2
3E
(
)
X
+ 2E
(
Y
)
=
E
(
3X
)
+ 2Y
=
E
(
W
Notice
)
that
3 ×
8
+
2
=
5
7
6
9 Var
(
X
)
+
4 Var
(
Y
)
=
(
Var
3X
+
2
Y
)
=
Var
(
Notice
)
W
that
9
the
To
and
probability
nd
the
X
4
the
5
of
demonstrated
the
corresponding
and
sum
probability
the
+
second
the
part
of
involving
of
two
ver y
impor tant
independent
variable
Z
=
X
probabilities

of
proper ty
random
Y
we
the
a
real
of
=
z )
=
P( X
Examples
variable W
of
X
4
=
and
Y,
=
x )
and
(which
coecients, W
probability
variables.
multiplied
random
variables
5,
was
3X
and
is
×
+
P(Y
we
a
2Y
)
y)
showed
linear
totally
=
that
the
combination
was
simply
the
independent
of
X
and
product
of
E(X )
the
X.
also
=
demonstrated
w )
that
=
P( X
the
=
x )
expected
×
P(Y
value
=
y)
and
The
combinations
of
multiple
random
variables
value
and
variance
of
linear
of
−
X )
random
variable.
We
can
now
state
in
the
same
combinations
the
of
the
variance
is
moment,
of
following
also
of
X
called
which
linear
the
spread
way
relates
to
variability,
a
volatility,
single
called
moment
second
E(X
and
expected
also
2
variance
behave
is
rst
measures
the
1225
for
the
P(W
as
=
49
coecients:
We
4646
×
4
Y :
probability
Y
of
the
elementar y
P( Z
In
of
one
35
20
×
25
Examples
122
×
and
uncer tainty.
theorem.
We
can
also
dene
the
n
Theorem
nth
moment:
E(X
−
X )
For
example,
the
third
2:
3
Given
nite
that
X
and
parameters
Y
are
and
two
a,
independent
b ∈ ,
random
variables
with
moment:
measure
then:
or
i
E ( aX
 bY
)
=
aE ( X )  b E (Y
E(X
of
50
 bY
Expectation
)
=
algebra
a
2
Var ( X
and
)
Central

b
Var (
Y
Limit
)
Theorem
is
a
skewness
asymmetr y
distribution
Var ( aX
X )
in
);
2
ii
−
of
X
the
Proof:
We
i
are
μ
going
=
to
E (X )
prove
=
x P (X
∑
1
this
theorem
=
for
μ
x ),
=
a
discrete
E (Y
)
=
y P(Y
∑
2
x
random
=
variable
only:
y)
Since
X
and
independent
⇒
E
(
aX
)
+ bY
=
∑
x
∑
x
are
( ax
) P (( X
+ by
=
x ) ∩ (Y
=
y ))
the
variables
probability
of
the
y
intersection
=
Y
y
(( ax
+ by )P( X
=
x )P(Y
=
is
the
product
of
probabilities.
Use
distributive
y ))
y
( ax
∑∑
x
P (X
x ) P (Y
y)
by
P (X
x ) P (Y
y ))
the
y
proper ty.
=
∑
P
(Y
=
) (
y
ax
∑
y
P( X
=
x ))
The
sum
of
all
the
x

probabilities
of
a
1
∑
(
P
X
)
x
(
by
∑
x
P (Y
random
variable
equal
1.
is
y ))
to
y





1
=
a
∑
x P (X
=
x ) + b
aE
(
)
X
(
Var
(
X
)
=
∑
(x
y )
Use




)
X
E
+ bE
(
Y
the
denition
of
the
(
Y
expected
value.
)
)
2
ii
=
y
E
=
y P (Y
∑
x



2
P (X
=
μ
x )) −
1
x
2
Var
(
Y
)
=
∑
( y
2
P(Y
=
μ
y )) −
⇒
2
y
2
Var ( aX
+ bY
)
=
∑
( ax
+ by )
P (( X
=
x ) ∩ (Y
=
y ))
Since
X
and
Y
are
x , y
independent
variables,
2
− (a μ
+ bμ
1
)
the
probability
intersection
2
=
∑
of
the
2
(( a
2
x
2
+ 2 axby
+ b
is
the
2
y
)
P (X
=
x )
P (Y
=
y ))
product
of
probabilities.
Use
distributive
x , y
2
2
μ
− (a
2
+ 2a μ
1
bμ
1
2
μ
+ b
2
)
the
2
proper ty.
2
=
∑
P (Y
=
y)
∑
y
(a
2
x
P (X
=
x ))
The
sum
of
all
the
x

probabilities
of
a
1
+2ab
∑
(x P (X
=
x ))
∑
x
( y P (Y
Recognize
the
X
=
x
expression
f or
the
2
2
(
1.
to
μ
1
P
equal
is
y ))





μ
∑
variable
y




+
=
random
)
x
∑
(b
expected
2
y
P (Y
=
value.
y ))
y





1
2
− (a
2
μ
1
2
+ 2a μ
bμ
1
+ b
2
2
μ
)
Reduce
opposite
terms
2
and
collect
the
like
terms.
Chapter
2
51
2
2
=
2
(
P
∑
)
=
+2
+ b
1
2
∑
2
x
2
2
μ
−2 a μ
1
bμ
1
2
=
(
=
y ))
2
2
(x
∑
2
μ
− b
2
2
a
P(Y
y
2
−a
( y
P( X
=
μ
x )) −
2
2
) + b
(
1
∑
x
( y
2
P (Y
=
y )) −
μ
)
Use
distributive
2
y
proper ty
2
=
a
and
the
2
(
Var
)
X
(
Var
Y
+ b
)
denition
of
the
variance.
The
proof
variable
is
Theorem
nite
of
the
theorem
beyond
2
can
number
Theorem
the
now
of
for
scope
be
a
continuous
of
this
generalised
random
random
syllabus.
to
give
a
result
for
a
variables.
3:
Unlike
+
Given
X
that
,
X
1
,
X
, ...,
2
X
3
,
n ∈ 
of
are
independent
with
nite
and a
parameters
,
a
1
expected
X
1
+ a
1
X
2
,
a
2
, ...
a
3
∈ 
that
X
1
+ ... + a
2
+ a
1
X
n
X
2
)
=
a
n
+ ... + a
2
2
E (X
1
) + a
1
X
E (X
2
) + ... + a
2
E (X
n
);
n
whether
and
b
it
1
) + a
1
are
sigma
) + ... + a
2
Var ( X
n
notation
are
we
can
always
⎜
∑
⎝
i =1
a
X
i
=
i
∑
⎟
⎠
n
∑
⎝
i =1
X
i
of
Exercise
in
i
2
=
i
⎟
∑
⎠
i =1
this
a
Var ( X
i
)
i
theorem
induction
for
and
the
is
discrete
left
as
an
Z
are
case
can
exercise
be
for
done
by
students.
2B
random
the
i
);
n
⎜
proof
The
E (X
⎞
a
mathematical
1
a
i =1
⎛
The
table
variables
X,
Y,
and
given
with
their
parameters
below
.
2
Variable
σ
X
3
0.5
Y
−5
.4
Z
2
2.8
Find
52
have
In
to
be
E(X
±
Var
(X
that
Y )
=
±
Y )
E(X )
±
E(Y )
and
⎞
Var
ii
positive.
we
write:
n
⎛
E
or
squares
).
n
careful
i
positive
their
2
Var ( X
2
n
matter
coefcients
n
par ticular
Using
doesn’t
the
)
n
2
Var ( X
a
we
calculating
then:
negative;
=
value,
when
n
a
Var ( a
ii
calculation
random
variance
E (a
i
the
n
notice
variables
in
the
a
X
+
d
X
 Y
parameters
Y
+
Expectation
Z
algebra
of
the
b
2
Y
e
X
and
following
Z
+
Y
Central

Z
Limit
linear
combinations:
c
2Z
f
3Z
Theorem
7X

2X
+
4Y
=
Var(X )
+
Var(Y )
Poisson
2
E(Y
)
E
a
(
5.
3X
E(Y
)
=
nd
the
and
5
)
;
4.
Given
of
binomial
)
=
12.
the
Use
mathematical
yet
variance
distributed
random
variable
Example
Given
the
Y
are
such
that
Var (X )
=
2
and
3Y
−
Y
in
linear
way
is
in
are
terms
have
to
such
same
terms
of
of
prove
have
distributed
E(X )
=
of
9
and
success, p,
p
Y
are
same
such
that
probability
E(X )
of
Theorem
we
have
of
this
explored
variables
Poisson
in
3.
so
far,
which
distribution)
two
X
+
b
X
− Y
c
4 X
a
E
(
are
will
Poisson
the
random
variance
of
variables
the
X
random
X
∼ Po ( μ
)
and
∼ Po ( μ
Y
+Y
)
=
E
(
X
)
+ E
(
Y
)
=
μ
+
(
X
+Y
)
=
Var
(
X
)
+
Var
Both
2
(
Y
)
=
μ
+
the
Poisson
μ
X
−Y
)
=
E
(
X
)
− E
(
Y
)
=
μ
−
1
expected
random
X
−Y
)
=
Var
(
X
)
+
Var
(
Y
)
=
μ
+
4 X
− 3
Y
)
=
4E
(
X
)
− 3E
(
Y
)
=
4 μ
= 16 μ
1
We
can
random
+
− 3
Y
)
=
4
Var
(
X
)
+
to
the
Y
of
the
equal
variance.
could
2.
to
Notice
random
Hence,
also
apply Theorem
(
−3
)
its
of
the
random
be
of
a
the
2.
that
variable
we
the
X +
might
Y
is
speculate
Poisson.
Notice
variable
variance. Thus,
Again
− 3μ
that
the
−
is
expected
X
apply Theorem
−Y
X
2
is
and
not
Y
not
equal
Poisson.
notice
that
2
2
2
4 X
variance
are
2
1
(
and
μ
to
Var
expected
2
value
(
value
variable
Apply Theorem
value
X +
Again
μ
1
(
the
2
that
E
nd
μ
equal
c
)
2
expected
Var
a
way
.
parameter.
(
give
variables:
1
E
all
3
Y
1
b
haven’t
Y
−
Var
we

and
a
8
success, p,
1
value
=
p
combination
all
that
probability
and
the
variables
(e.g.
also
Y
the
they
random
same
and
variables X
induction
a
)
7X
have
that
X
of
that
−
X
random
whether
in
2X
of
manipulation
discussed
they
Given
nd
the
and
(
11
Y
variables
that
variance
E(Y
X
Var
b
random
Negative
4
variables
Calculate:
+ 5
Y
Binomial
3
In
random
=
Var
(
Y
the
expected
value
of
the
4X
−
3Y
is
not
equal
4X
−
3Y
is
not
Poisson.
random
variable
)
to
its
variance. Thus,
9μ
2
conclude
variables
that,
is
in
not
a
general,
Poisson
a
linear
random
combination
of
two
Poisson
variable.
Chapter
2
53
In
a
two
similar
situation
random
we
variables
ask
have
ourselves
equal
what
expected
is
going
to
happen
=
values
and
1
2
equal
2
2
σ
variances
if
σ
=
1
.
In
this
case
we
conduct
a
slightly
dierent
2
calculation.
In
the
following
variance
can
of
a
Mia
to
the
Mia’s
uncle
Their
aunt
Z
expected
Given
+
X
use
to
the
expected
investigate
value
whether
says
box
says
we
of
Y
the
and
can
the
2
the
will
give
and
a
€2
the
write Y
=
variables
a
of
of
or
of
colour
black
0
of
X
=
x
marble,
marbles
and
then
retur ns
it
Mia
draws.
X.
for
of
each
white
money
marble
Mia
the
for
money
same
from
each
they
box
the
and
box,
white
she
receives
will
notes
notes
marble
get
from
and
and
Z
=
X
+
X ,
what
do
draws.
from
her
uncle.
the
the
colour.
colour
drawn.
their
The
aunt.
He
and
then
replaces
it.
random
Find
the
you
notice
about
the
Z ?
probability
distribution
table
f or
the

variable
(
the
marbles.
white
The
x
i
P
we
Y
marble
€
3
of
Mia
of
a
=
not
Z
2X
Y
and
amount
from
them
amount
variance
to
variance
draws
the
number
variance
marble
give
notes
white
and
Mia
will
She
the
denotes
draws
denotes
that
he
value
she
box.
contains
value
that
Theo,
a
represents
variable
value
parameters
X
and
X
from
Afterwards,
variable
d
X
X
expected
brother,
it.
The
expected
the
replaces
=
variable
random
Find
2X
will
variable
marble
box.
the
Mia’s
The
c
one
Random
Find
b
that
we

draws
back
a
random
conclude
Example
example,
3
2
5
5
X
is
the
same
as
that
in
Example
)
i
2
2
E
(
X
)
2
,
=
(
Var
X
)
=
⎛ 2 ⎞
−
5
5
⎜
6
Calculate
the
expected
Draw
probability
value
and
variance.
=
⎟
⎝ 5 ⎠
25
b
Y
=
y
0
i
P (Y
=
y
2
3
2
5
5
the
the
variable
Y
and
distribution
ll
in
the
table
f or
probabilities.
)
i
2
4
E
(
X
)
=
8
,
Var
(
X
5
54
Expectation
algebra
)
=
⎛
−
5
and
Central
4
⎜
⎝
⎟
5
24
⎞
⎠
Limit
Calculate
=
25
Theorem
the
expected
value
and
variance.
5.
c
Draw
Z
=
z
0
i

f or
P (Z
=
z
the
probability
distribution
table
2
9
12
4
25
25
25
the
variable
values.
We
are
Z
and
ll
actually
in
all
adding
the
the
)
i
3
(
P
Z
=
0
)
3
=
=
5
2
(
)
Z
= 1
3
(
Z
=
2
)
5
)
Z
= 0
( Z
=
0 )
↔
(0
+
0)
( Z
=
1 )
↔
(1
+
0)
( Z
=
2 )
↔
(1
+
1)
or
(0
+
1)
25
=
5
25
3
(
E
draw.
4
×
5
they
12
=
5
2
=
2
×
5
2
P
3
+
5
marbles
25
×
=
of
9
×
5
P
number
12
×
+
4
1 ×
+
2
20
×
4
=
=
Calculate
25
25
5
25
the
expected
value
and
variance.
5
2
12
⎛
(
Var
Z
)
=
⎜
0
E
E
(
Y
(
)
=
2X
)
E
=
(
Z
E
(
⎞
+
25
4
⎛
−
⎟
25
⎝
d
16
+
⎠
⎜
12
⎞
=
⎟
5
⎝
)
X
+
that
Var( Z )
Notice
that
although
expected
)
X
values,
(
2X
)
≠
Var
(
X
+
X
4
Theorem
will
generalize
Var( X ).
we
and
don’t
Z
have
have
conclude
equal
equal
that
f or
)
random
Theorem
+
Var( X )
X
they
variances. Thus,
Var
=
Notice
25
⎠
the
conclusion
we
reached
in
variables
Example
in
general
2 X
≠
X
+
X
7:
4:
+
Given
that
independent
random
variables
X
,
X
,
1
X
2
, ...,
X
3
,
n ∈ 
n
2
all
have
E (X
+
equal
X
1
expected
+ ... +
X
2
)
=
value
E( X
n
and
) +
equal
E (X
variance
E
) + ... + E ( X
1
2
)
(
)
X
=
n
μ
=
+
and
(
Var
+ ... +
X
= σ
,
then:
= n
;




n
)
terms
and
2
Var ( X
+
X
1
+ ... +
X
2
)
=
Var ( X
n
) +
Var ( X
1
) + ... +
Var ( X
2
)
n
2
n
When
adding
expectation
n
of
random
the
sum
variables
+
(X
X

the
expectation
E (X
+
X
1
variance
of
+ ... +
2
the
X
)
+
...
+
have
X
2
variable
=
which
E ( nX ) .
)
is
the
equal
expected
2
= nσ
terms
value,
the
to
n
multiplied
The
by n,
variance
(nX ),
of
the
i.e.
sum
is
not
equal
to
the
n
in
the
case
of
a
linear
transformation
of
2
Var ( X
+
1
Therefore,
same
2
= σ
+ σ
+ ... + σ





by
comparing
X
+ ... +
)
=
nσ
random
2
≠
n
variable,
2
σ
=
Var ( nX ) .
n
parameters,
X
+ X
+ ... + X





n
X
2
a
≠
we
conclude
that
for
a
random
variable X:
nX .
terms
Chapter
2
55
Example
Anna
and
Anna’s
,
,
,
drawn
a
8
Sarah
box
2,
2,
by
Draw
and
Anna
the
variables
Anna’s
amount
by
of
b
Draw
c
Suggest
independently
3.
the
Let
and
and
is
Sarah
the
3,
4,
variables X
counters
and
and
5.
Y
from
Sarah’s
two
box
represent
dierent
contains
the
value
boxes.
the
of
the
counters
counters
respectively
.
give
of
Anna
probability
a
2,
distribution
to
value
money
drawing
,
tables
and
nd
the
expected
values
of
the
Y
going
the
counters
random
probability
X
father
multiplied
are
contains
Anna
Sarah’s
will
ear n
as
in
distribution
relationship
between
much
drawn
this
as
Let
the
value
random
of
her
drawn
variable Z
counter
represent
the
game.
table
the
money
counter.
and
nd
variables X,
the
Y,
expected
and
value
of
the
variable Z
Z
a
Notice
X
=
x
i

2
3
4
5
1
1
1
1
1
5
5
5
5
5
since
P (X
=
x
that
we
X
f ollows
have
one
a
unif or m
counter
in
distribution
each
value.
)
i
1
E
(
)
X
1
×
=
(
1+ 2 + 3 + 4 + 5
)
=
5
Y
=
× 15
=
3
Calculate
the
expected
value.
5
y
i

2
3
1
1
1
There
are
only
three
dierent
values
counters and there are 3 out of
P (Y
=
y
)
value “1”, 2 out of
on
the
6 counters with the
6 counters with the value “2”
i
2
3
6
and
1
E
(
Y
)
1
=
× 1
2
+
1
×
2
+
3
3
=
6
z

i
Calculate
2
3
4
1
=
z
of
6
counters
with
the
value
“3”.
1
the
expected
value.
need
2
1
to
look
at
all
the
possible
pairs
and
5
their
P (Z
out
3
We
=
1
5
×
b
Z
only
1
products. There
outcomes
and
since
are
the
altogether
drawings
11
pairs
of
are
)
i
10
6
15
6
10
independent
multiply
6
8
9
0
2
5
1
1
1
1
1
1
give
the
f or
their
same
probabilities.
15
30
15
30
30
1 :
(1, 1)
product
For
5
56
Expectation
algebra
and
Central
Limit
Theorem
them
so
we
example:
1
×
→
of
probabilities.
1
10
each
1
=
2
10
we
have
Some
have
of
to
to
the
add
pairs
their
1
2
*Unsaved
1.1
A
B
C
:
(1,
2)
or
(2, 1)
1
→
×
5
D
1
1
1
1/10
2
2
1/6
3
3
2/15
4
4
1/6
5
5
1/10
3
5
: (1,
3)
or
(3, 1)
6
1
:
(2,
2)
or
(4, 1)
5
C1
3
15
1
×
5
2
=
2
1
+
6
1
×
5
1
×
→
1
=
2
1
+
×
1
×
5
1
→
5
4
1
+
3
1
=
2
6
=mean(a1:a11, b1:b11)
To
(
E
Z
)
simplif y
can
use
a
the
calculation
f or
expectation
we
GDC.
= 5
5
Z
c
=
XY
Notice
that
5
=
3
×
3
(
E
Z
)
=
Exercise
1
Six
(
is
Explain
X
of
Find
d
Use
and
2
An
3
a
why
+
X
the
the
the
=
X
Find
+
the
is
die
)
of
+
the
the
outcome
and
the
number
of
is
X
X
value
r ule
rolled
and
experiment
+
+
of
and
to
these
table
of
one
nd
such
the
coin
ip.
expected
Write
value
and
X
outcome
on
X
denote
X ,
can
where
ipping
six
variance
determine
be
Y
of
all
written
is
a
as
random
variable
coins.
the
random
possible
variable Y
outcomes
of
Y,
outcomes.
four
X
variable
the
+
die
the
times,
and
outcomes
distribution
the
why
the
ipped
the
number
of
multiples
of
recorded.
variable
rolling
c
(
Y
independently
whole
X
probability
Explain
Y
× E
variable
empirical
variance
b
)
distribution
the
+
comment
obtained
are
expected
unbiased
Let
X
denote
the
representing
c
(
recorded.
X
probability
=
E
coins
variable
variance
Y
=
2C
obtained
Let
the
b
)
XY
unbiased
heads
a
E
X
whole
+
four
expected
X ,
table
for
and
one
nd
roll
the
of
the
expected
die.
Write
value
and
X
experiment
where
Y
is
a
can
be
written
random
as
variable
representing
times.
value
and
variance
of
the
random
variable Y
2
3
Random
variable
X
has
the
parameters
= 3
a
Find
the
expected
value
and
variance
of
b
Find
the
expected
value
and
variance
of
c
Show
+
X
that
Var
(
X
X
+
+ 3X
)
≠
Var
(
and
X
+
X
=
+
4.
X.
3X.
6X
)
.
Chapter
2
57
Random
4
variables
X
and
Y
have
2
=
2,
= 1
X
a
X
b
Y
c
X
the
+
+
Y
+
4
and
1,
1,
2,
Shikma
+
X
+
X
the
a
+
Y
variance
+
regular
Y
+
of
3
respectively
.
values
of
the
Y
tetrahedral
independently
of
the
is
die
rolling
Let
a
with
outcomes
faces
biased
random
faces
of
distribution
variables
going
the
this
Draw
probability
of
number
in
is
and
the
die
1,
with
variables X
obtained
by
2,
3,
and
faces
and
Shariza
Y
and
to
X
and
give
concer t
tickets
and
nd
the
expected
Y
them
obtained.
tables
as
Let
many
concer t
random
Shariza
and
tickets
variable Z
Shikma
as
the
represent
are
going
to
game.
the
value
of
probability
the
Independent
6
X;
X
friend
product
b
+
and
respectively
.
values
ear n
X
rolling
2,
Draw
the
value
Y;
+
is
Y
+
Shikma
1,
Their
X
+
represent
a
+
X
parameters
= 3.
= 5,
Y
expected
X
Shariza
5
and
X
Find
the
2
variable
random
distribution
Z
and
table
comment
variables
X
and
Y
and
on
nd
your
have
the
expected
result.
the
expected
values
3
and
+
non-zero
A
linear
In
this
2
values
we
variable.
Theorem
to
If
two
of
independent
going
star t
to
look
with
a
∼
(
)
XY
=
at
the
normal
μ
nd
all
random
par ticular
corollar y
take
Theorem
possible
that
is
a
case
variables
of
a
normal
consequence
of
2:
Coefcients
independent
normal
2
X
E
that
2.
Corollar y
we
are
We
Given
μ
of
combination
section
random
respectively
.
N( μ
, σ
1
random
variables
cannot
both
a
and
be
equal
2
)
and
Y
N( μ
∼
1
),
, σ
2
then
the
to
linear
0
(we
write
this
2
2
condition
combination
The
aX
expected
+ bY ,
value
a,
and
=
aμ
+ bμ
1
and
σ
b ∈ 
2
=
a
2
+ bY
∼
+ b
N (a μ
+ bμ
Expectation
algebra
,
2
and
a
normally
of
aX
+
bY
,
σ
so
we
2
2
+ b
2
σ
1
Central
are
can
2
σ
distributed
2
1
1
58
a
2
σ
2
2
aX
is
variance
2
μ
b
)
2
Limit
Theorem
write
variable.
as
a
2
+
b
≠
0).
Example
Given
the
9
normal
probabilities
a
b
c
a
X
Y
+Y

3X
X
X
<
of
random
the
variables
X
∼
N
( 3,
0.25 )
and
Y
∼
N
b
Y
P
c
X
−
 0
∼
−
∼
X
3X
<
3X
− 2Y
P
N
(
nd
the
5;
N
(3
+
4,
0.25 + 0.64
)
=
N
( 7,
0.89
2
Y
> 8
N
(
)
4
=
)
Apply
the
theorem
− 3,
≤ 0.5
)
⇒ 3X
(1,
3X
∼ N
4.97
− 2
Y
by
above
using
a
and
calculate
the
GDC.
0.145
=
0.64
+ 0.25 )
=
N
(1,
0.89
− 2
Y
(3
)
0.298
<
0
2
=
,
> 8;
+ Y
X
(
Y
)
2
Y
+ Y
(
0.64
following:
probabilities
P
( 4,
× 3 − 2 × 4,
3
2
× 0.25 + 2
× 0.64
)
)
<
)
0
=
0
327
Scratchpad
normCdf(8, 100, 7,
√0.89)
0.144573
normCdf(–100, 0.5, 1, √0.89)
0.298056
normCdf(–100, 0, 1, √4.97)
0.326874
3/99
Notice
lower
+∞
that
in
part
boundaries
and
−∞
a
the
are
upper
−00
boundar y
which
are
is
00
sucient
and
in
values
parts b
that
and
c
the
represent
respectively
.
Chapter
2
59
Example
There
are
course,
nal
5.2
0
two
whilst
grades
and
of
of
Find
a
school.
class
classes
deviation
.85.
the
at
other
both
standard
deviation
class.
classes
the
The
HL
probability
One
studies
are
of
class
the
distributed
.37,
class
that
the
normally
,
whilst
claims
this
studies
Mathematics
the
that
SL
they
statement
is
Mathematics
Standard
with
class
have
the
has
a
HL
a
(SL)
class
mean
obtained
Higher
Level
of
Level
course.
having
4.8
better
a
and
result
(HL)
The
mean
of
standard
than
the
SL
tr ue.
2
X
∼
N(5.2, 1.37
)
Denote
random
2
Y
∼
N ( 4.8, 1.85
> Y
⇒
X
−Y
two
=
P
−Y
∼
(
X
variables
SL
X
classes
and
Y
by
the
parameters
of
the
di erence
random
variables.
+
1.85
)
5.2294 )
−Y
> 0
)
=
0
569
Scratchpad
normCdf(0, 100, 0.4, √5.2294)
Since
the
conclude
result
probability
that
than
the
the
is
HL
SL
greater
class
than
have
0.5
we
obtained
0.569428
can
a
better
class.
1/99
The
corollar y
normal
to
random
Corollar y
to
If
n
we
take
Theorem
3
can

N (μ
1
Theorem
be
generalized
for n
independent
3:
independent
, σ
1
now
variables.
normal
2
X
random
variables
2
),
X
1

N (μ
2
, σ
2
2
), ...,
X
N (μ

2
n
, σ
n
+
),
n ∈
n
n
n
2
then
the
linear
combination
of
a
X
1
+ a
1
X
2
+ ... + a
X
n
2
=
n
∑
a
X
k
,
k
a
∈ ,
k
k =1
a
normal
variable.
The
parameters
are
μ
=
∑
a
μ
k
and
σ
2
=
k
k =1
n
n
we
can
write
∑
a
X
k
∼
N
∑
k
k =1
The
It
60
is
proof
left
as
of
an
Expectation
this
k
,
k
k =1
theorem
exercise
algebra
μ
a
for
and
can
the
be
∑
k
k
k =1
done
by
mathematical
student.
Central
2
σ
a
Limit
Theorem
∑
k =1
n
2
so
a
n
2
also
∑
k =1
n
is
the
respectively.
2
N (5.2 − 4.8, 1.37
N ( 0.4,
and
> 0
2
X
HL
)
Find
X
the
induction.
2
σ
a
k
k
,
≠
k
0
of
Example
Given
Z
the
∼ N(
(
a
P
b
P
c
P
d
P
a
X
(
(
(

X
following
3,
0 .4 ) ,
−Y
−
2X
−
2X
+ Y
X
Z
+ 3
Y
−Y
−
Z
and W
Z
<
−3
W
<
2W
random
∼ N ( 0 .5,
variables
0 .64 )
X
calculate
∼ N ( 2,
the
∼ N ( 4 , 1.44 ),
2.25 ), Y
following:
)
;
> 0
− 4W
>
normal
)
0
−
Z
)
;
)
−Y
∼ N (2
;
− 4
+ 3,
2 .25 + 1 .44
+ 0 .4 )
Find
the
=
N (1,
b
(
X
2X
+ Y
− Z
+
Z
)
> 0
=
0
− 4W
c
P
(
(
2X
−
2X
+ Y
 Y
∼ N(4

d
P
4 × 2 .25 + 0 .4
+ 16 × 0 .64 )
Z
− 4W
>
−3 Z
 3Z
+
4
<
0
)
=
−W
)
0
(
(
2X
X
+ Y
+ 3
Y
+ 3
Y
the
the
the
a
of
GDC
probability.
new
nd
mean
and
variable
the
=
P
(
2X
+ Y
+ 3Z
+ W
>
0
variance
and
use
a
of
GDC
probability.
)
Rewrite
 W
−
9
+
0 .5,
4
×
2 .25
+
1 .44
+
9
×
0 .4
+
>
<
−3 Z
2W
− 2W
−W
−Y
)
)
=
=
P
0
(
the
have
one
then
nd
+
12
−
+
of
the
new
to
nd
the
2 .25
X
+
+ 3
Y
− 2W
9
× 1 .44
+ Y
<
0
X
+ 3
Y
<
2W
Notice
)
=
0
and
variable.
and
variance
Use
a
GDC
probability.
that
+
4
×
0 .64
+
we
expression
3Y
haven’t
because,
+ Y
≠
simplied
f or
4Y
random
since
the
1 .44 )
and
−Y
variable
mean
we
)
values
N (17, 19 .21)
(
that
448
the
4,
the
so
0 .64 )
+Y
1
inequality
random
variables,
∼ N (2
P
use
130
N ( −0 .5, 14 .68 )
X
=
variance
and
N (5, 19 .64 )
2X
P
nd
Find
to
P
and
690
∼ N ( 2 × 2 − ( −3 ) − 4 × 0 .5,
=
mean
variable
4 .09 )
to
P
the
new
of
4Y
the
are
variances
of
3Y +
Y
di erent.
0000525
Scratchpad
normCdf(0, 100, 1,
4.09)
0.689512
normCdf(–100, 0, 5, √19.64)
0.129611
normCdf(0, 100, –0.5, √14.68)
0.448086
normCdf(–100, 0, 17,
0.000053
√19.21)
4/99
Chapter
2
61
Example
Bill,
Jill,
order:
the

and
John
X-large,
are
Large,
parameters
are
going
and
given
to
a
steakhouse.
Small.
in
the
The
following
Average
X-Large
430 g
30 g
Large
35 g
22 g
Small
50 g
8 g
orders
Find
the
one
X-Large
probability
weight
steak,
that
Bill
2
X
P
∼ N ( 430,
(
X
>
L
+ S
30
)
orders
get
a
P
L
(
∼ N (315,
L
+ S
−
X
22
< 0
+ S
−
X
one
∼ N (315 + 150
− 430,
three
the
steak
steak
and
are
than
John
Jill’s
S
∼ N (150,
8
)
Denote
L
+ S
−
X
)
=
0
John’s
the
weights
Small
steaks
and
respectively.
2
22
+
they
can
normally
and
one
Small
steaks
put
steak.
together.
by
of
X-large,
random
Large,
variables
X,
and
S
L,
2
8
+
30
)
and
< 0
dishes
)
N (35, 1448 )
(
steak
orders
and
the
inequality
construct
parameters
P
of
distributed
2
),
Rewrite
=
types
steaks
deviation
Large
heavier
2
L
are
all
2
),
=
Standard
Jill
will
of
table.
Steak
Bill
There
weights
a
of
new
the
in
a
simpler
variable.
new
Find
f or m
the
variable.
179
Scratchpad
normCdf(–1000, 0, 35,
Example
Ever y
orders
more
The
1448)
0.178844|

day
two
one
time
wrestlers
Large
to
weights
pizza
nish
of
all
his
Kenji
and
pizza,
sizes
of
and
Kazuyoshi
Kazuyoshi
Kenji
pizza
orders
gets
are
go
hungr y
normally
for
one
a
snack
Jumbo
again
at
a
pizza.
and
distributed.
pizza
Since
orders
Large
one
place.
Kenji
Kazuyoshi
small
pizzas
always
needs
pizza.
have
a
mean
of
2
900 g
and
a
variance
of
25 g
,
and
Jumbo
pizzas
are
.5
times
the
weight
of
Large
pizzas.
2
Small
pizzas
have
a
mean
a
Find
the
mean
and
the
b
Find
the
probability
of
440 g
variance
that
on
a
that
Kenji
and
of
a
variance
the
given
weights
day
Kenji
of
0 g
of
Jumbo
will
eat
.
pizzas.
more
pizza
by
weight
than
Kazuyoshi.
c
Find
a
62
the
three
probability
day
Expectation
will
eat
period.
algebra
and
Central
Limit
Theorem
more
pizza
by
weight
than
Kazuyoshi
during
Let
a
L,
J,
and
S
represent
the
random
variables
of
Use
the
weights
of
Large,
Jumbo,
and
Small
2
J
=
1.5
×
L
⇒
J
∼ N (1.5
×
the
E( aX )
900, 1.5
×
f or mulas:
pizzas.
=
a E( X )
25)
2
=
D
b
N (1350,
=
L
+ S
Var( aX )
56 .25 )
− J
⇒
Let
2
D
=
P
c
∼ N (900 +
N ( −10,
(
D
>
0
D
+
440 − 1 .5 × 900,
)
=
∼ N (3 × −10,
>
0
the
the
daily
weight
of
di erence
pizza
that
Kenji
× 25 )
and
Kazuyoshi
nd
the
eat.
)
+
=
D
0
probability
Use
that
a
GDC
D
>
to
0.
148
X
X
D
0
=
(
Var( X )
represent
between
25 + 10 + 1 .5
a
91 .25 )
X
P
D
=
⇒
Let
3 × 91 .25 )
=
N ( −30,
the
273 .75 )
X
represent
weights
0349
the
during
di erence
a
three
between
day
period.
Scratchpad
normCdf(0, 1000, –10,
91.25)
0.147585
normCdf(0, 1000, –30, √273.75)
0.034901
Exercise
1
Given
X
2D
the
∼ N ( 0, 1), Y
calculate
2
(Y
a
P
c
P(3 X
e
P( X
A
the
−
Z
− 4Z
standard
Z
∼ N(
2,
random
0 .25 ),

< 0
Z

 3X
W
be
)
− 2Z )
sells
to
)
beetroot
normally
deviations
are
(
b
P
d
P( X
f
P(W
and
X
sweet
distributed
given
in
the
 Y
 Y
and
table
240 g
20 g
Sweet
730 g
50 g
the
Find
probability
potato
chosen
∼ N (3, 1 .21)
is
that
more
the
than
weight
three

of
times
 W
2Y
 2Y
 W
)
)
 3
W
The
the
 0
)
weights,
mean
in
values
grams,
and
below
.
Beetroot
Find
variables
and W
Z
potato.
Standard
potato

 3Z
Mean
sweet
b
0.16 ) ,
normal
following:
stall
assumed
independent
∼ N (1
,
− W
 Y
market
are
a
following
deviation
a
randomly
the
weight
of
chosen
a
randomly
beetroot.
the
potatoes
probability
and
four
that
the
randomly
weight
chosen
of
two
randomly
beetroots
will
chosen
exceed
sweet
2.5 kg.
Chapter
2
63
3
Lillian
the
and
same
hospital.
Veronica
to
the
Veronica
walks
hospital
inde pendent
values
and
of
in
Lillian
and
are
live
the
always
takes
the
assumed
each
standard
same
other.
takes
tube.
to
In
be
of
Lillian
35
5
Veronica
45
8
less
the
than
Given
b
probability
30
that
minutes
to
Veronica
probability
hours
that
that
travelling
get
a
work
a
the
ve
week
(not
work
at
and
a
but
bus
to
travel
distributed
are
times
in
the
and
mean
minutes.
deviation
given
to
works
within
to
on
and
taken
below
these
Standard
Find
tram
times
table
Mean
a
a
The
nor mally
the
deviations
building
day
Lillian
will
take
hospital.
days
she
a
week,
will
nd
spend
including
her
the
more
than
jour ney
4
home
again).
Find
c
the
the
time
than
probability
taken
three
by
times
that,
Lillian
the
on
to
time
a
given
get
to
taken
day
,
the
by
four
times
hospital
Veronica
to
is
more
get
to
the
Statistical
hospital.
ver y
impor tant
researchers
4
Dominic
grows
apples
on
his
farm.
It
may
be
assumed
of
the
apples
are
normally
distributed
with
a
in
255
grams
and
a
standard
deviation
of
12
in
the
probability
that
a
randomly
chosen
apple
than
250
have
of
these
apples
are
selected
at
random.
Find
in
to
that
than
1
the
total
weight
of
the
four
apples
also
kilogram.
grows
the
the
plums
are
plums.
normally
and
a
standard
It
may
be
assumed
that
Find
the
distributed
with
a
mean
of
deviation
of
2
probability
that
the
is
more
than
four
is
the
if
it
is
the
of
a
randomly
weight
of
a
not
probability
that
the
weight
of
a
randomly
then
is
more
than
the
weight
of
four
randomly
chosen
plums.
since
be
if
distribution
of
the
mean
core
course
we
studied
population
and
a
a
population
sample
64
is
In
Expectation
sample
includes
all
the
members
of
a
just
a
representative
selection
dened
or
order
algebra
to
be
and
able
Central
to
predict
subset
Limit
something
Theorem
were
We
to
it
the
be
technique,
type
must
other
about
to
of
take
factors
in
such
said
the
duration
of
the
group
of
tested.
comes
the
researchers
samples.
from
entire
when
concerns
and
the
resources
population.
results
the
sampling
ethical
and
be
selecting
as
that
the
population
However
,
Sampling
the
the
from
different
account
In
meaningful
cannot
chosen
results
2.2
the
plum.
the
apple
of
randomly
will
Find
d
the
chosen
obtained
chosen
of
grams.
weight
times
these
60
drawn,
apple
they
small
weights
conclusions
c
of
representativeness
population
grams
a
population.
concern
representative
of
study
is
sample:
Dominic
a
select
from
main
researchers
greater
draw
the
The
probability
to
entire
grams.
sample
Four
b
the
weighs
may
more
order
for
grams.
population
Find
a
for
many
mean
conclusions
of
is
that
disciplines;
weights
sampling
the
available.
study,
population
from
is
where
to
a
sample
be
We
are
The
in
selected
the
For
there
The
is
a
in
mean
Let’s
a
the
ideas
it
of
to
the
sample
to
be
is
in
of
of
sample.
has
the
taken
of
we
is
not
nails
an
A
equal
of
a
of
random
sample
oppor tunity
a
normal
have
of
check
simple
is
are
mean
so
in
far
role
for
one,
we
a
in
nails.
within
sample
of
the
a
the
standard
can
consider
variables.
the
it
and
generally
so
since
acceptable.
and
random
nails
learned
to
Taking
independently
,
weights
impor tant
manufactured
nails
the
population.
applications.
range
sample.
change
a
individually
,
weight
take
an
decides
weights
population
can
plays
nails
the
from
practical
nails
what
independent
what
sampling
lot
all
doesn’t
deviation
ourselves
a
weigh
decide
population
sample
of
checking
entire
a random
population
distribution
has
variation
to
have
the
sample.
manufacturer
The
to
of
manufacturer
needs
standard
remind
as
process
whether
large
the
and
the
some
impractical
population.
nails
of
sampling,
The
need
probability
considerable
range,
from
par t
suppose
ver y
consider
size
of
we
element
develop
manufacturer
weight
the
a
be
to
nails.
be
sample
ever y
normal
example,
irregular
To
of
theor y
would
to
going
study
a
correct
certain
deviation
the
However,
we
have
independent
random
X
variables
,
X
1
what
,
...,
the
study
of
expectation
X
N (μ , σ

algebra.
+
),
n ∈ 
,
then
n
n
n
⎛
E
is
of
sample?
the
2
of
weights
2
If
xed
⎞
⎜
∑
⎝
i =1
⎛
=
X
i
nμ
and
⎞
Var
⎟
⎜
⎠
⎝
∑
i
2
nσ
=
X
i
⎟
⎠
=1
−
Now
,
however,
random
We’ll
we’ll
variable
look
at
consider
within
the
a
sample
a
variable X
random
mean
sample
and
its
which
of
is
an
size k,
independent
where
k
≤
normal
n.
parameters.
n
⎛
⎞
∑
⎜
X
i
⎟
n
1
E
i
(
)
X
⎛
1
⎞
i =1
=
E
⎜
⎟
n
⎝
=
E
n
⎠
⎜
∑
⎝
i =1
X
⎟
i
×
=
nμ
μ
=
n
⎠
n
⎛
⎞
∑
⎜
X
i
⎟
(
X
)
2
n
1
Var
ii
⎛
σ
1
⎞
2
i =1
=
Var
⎜
⎟
=
Var
2
n
⎝
n
⎠
⎜
∑
⎝
i =1
=
X
i
×
⎟
n
σ
=
2
n
n
⎠
2
If
the
population
distribution
of
variance
the
sample
is
σ
and
mean
of
samples
these
of
size
samples
n
are
will
taken,
have
the
then
same
the
mean
2
σ
value
as
the
population
and
the
variance
n
2
⎛
σ
2
X
∼ N( μ , σ
) ⇒
X
∼ N
⎜
μ,
⎝
n
σ
The
term
standard
er ror
is
used
to
describe
the
quantity
,
which
is
the
average
n
distance
lie.
By
from
the
looking
smaller
the
at
mean
the
standard
that
each
formula
error
we
we
random
notice
variable
that
the
within
larger
the
the
sample
sample
we
of
size n
take
will
the
obtain.
Chapter
2
65
Example
The
a
weight
mean
of
Find
a
The
of
in
and
a
a
hand
20
sh
farm
standard
probability
than
catch
trout
340 g
the
more
b

that
a
may
be
deviation
catch
of
0
assumed
of
to
be
normally
distributed
with
30 g.
trout
will
have
a
mean
weight
per
sh
of
370 g.
net
can
trout
in
hold
the
up
to
hand
7 kg.
net
Find
without
the
probability
breaking
that
we
will
be
able
to
it.
2
X
a
∼ N (340,
30
) ⇒
Use
the
f or mula
f or
nding
the
mean
and
2
2
⎛
X
∼ N
⎜
⎛
⎞
30
340,
10
⎝
30
⎛
⎞
er ror
of
the
sample
distribution.
N ⎜ 340,
=
⎟
standard
⎠
⎜
⎜
⎝
⎝
⎟
10
⎠
Scratchpad
30
normCdf
(
)
370, 1000, 340,
√10
0.000783
P
(
X
> 370
)
=
0
000783
1/99
Method
b
I
2
⎛
30
⎛
Y
∼
N ⎜ 340,
⎜
⎜
⎝
⎝
⎜
⎟
20
⎠
7000 ⎞
⎛
P
⎞
Y
<
⎟
20
⎝
=
P
(
Y
< 350
)
= 0
Use
932
the
GDC
to
nd
the
solution.
⎠
*Unsaved
Method
II
30
normCdf
2
⎛
Y
⎟
⎜
⎝
⎝
0.931981
)
–100, 350, 340,
√20
⎞
∼ N ⎜ 340,
⎜
(
⎞
30
⎛
20
⎟
⎟
⎠
⎠
⇒
T
=
20Y
normCdf(–100, 7000, 6800, 30 • √20)
0.931981
2
⇒ T
∼ N ( 6800,
(30
20 )
)
2/99
⇒
P
(
T
Example
In
a
nail
mean
4
factor y
,
and
a
Find
b
By
the
=
0
932

value
nails
)
< 7000
of
8.2 g
the
the
the
weights
and
mean
the
samples
of
empirical
of
4
all
manufactured
standard
weight
parameters
using
a
of
is
the
deviation
of
nails
follow
0.02 g.
a
normal
Suppose
that
a
distribution
control
with
scale
weighs
recorded.
sample
r ule,
nd
mean
the
distribution.
inter val
of
weights
that
would
be
acceptable
for
all
nails.
2
a
X
∼ N (8 .2,
0 .02
) ⇒
Use
the
f or mula
f or
nding
the
mean
and
2
2
⎛
X
∼ N
⎜
0.02
8.2,
⎝
b
8.2
±
3
×
⎟
4
0.01
=
⎛
⎞
=
N ⎜ 8.2,
⎠
8. 2
⎛
0.02 ⎞
⎜
⎟
2
⎝
⎝
±
standard
er ror
a
66
accept
mean
all
the
Find
0.03
weight
Expectation
samples
between
algebra
and
of
4
8.7 g
Central
nails
and
Limit
of
the
sample
distribution.
⎠
the
values
deviations
We
a
that
have
8.23 g.
Theorem
which
above
and
are
3
below
standard
the
mean.
The
fall
empirical
within
Example
rule
the
in
range
statistics
of
three
states
standard
that
almost
deviations
all
(99.7%)
from
the
the
mean
obser vations
value.

Manufactured
screws
follow
a
normal
distribution
with
a
mean
length
of
3 cm
and
a
2
variance
the
of
sample
.
0.04 cm
will
lie
What
within
sample
0. cm
of
size
the
should
be
taken
population
to
mean
be
99%
cer tain
that
the
mean
of
length?
2
X
∼ N (3,
X
∼ N
0 .04
) ⇒
Use
2
⎞
3,
⎜
⎟
n
⎝
P( 2 .9
⎛
0.04
⎛
≤
X
Method
≤
⎛
=
f or mula
standard
er ror
f or
of
nding
the
the
sample
distribution.
=
⎜
⎜
⎝
⎝
⎟
n
⎠
0 .99
Transf or m
the
variable
into
the
2
9
3
3
≤
Z
1
3 ⎞
nor mal
=
≤
⎜
0
variable
Z
=
99
σ
⎟
0
2
0
2
⎜
⎟
⎜
⎟
n
⎝
n
Notice
that
the
proper ty
⎛
0
Z
⎜
1
n
≤
the
boundaries
are
symmetrical
⎠
about
P
by
standard
μ
X
⎛
and
0.2 ⎞
I
P
mean
N ⎜ 3,
⎠
3 .1)
the
mean
of
the
value
so
standard
we
can
apply
nor mal
the
cur ve.
α
⎞
⎟
=
0
P( −a
995
<
Z
<
a )
=
1 −α
⇒
P(
Z
<
a )
=
1
−
2
0
⎝
2
⎠
Apply
the
inverse
function.
n
1
=
(
Φ
0
995
)
2
n
=
2
×
2. 57583 ⇒
n
= 5. 15166
integer
2
n
=
5.15166
Notice
=
26.5396
≈
that
the
value
sample
theref ore
size
we
must
round
be
it
a
to
positive
27.
27
Find
the
solution
by
using
the
function
Scratchpad
f eatures
y
on
your
GDC.
1
0.2
f1(x)=normCdf
0.5
2.9, 3.1, 3,
(
√x
)
–0.99
(26.5, 0)
x
0
5
10
15
20
25
30
–1
n
=
27
Chapter
2
67
Exercise
2E
2
1
Given
that
∼ N( μ , σ
X
population,
and
)
the
sample
size
n
is
taken
from
the
nd:
2
(
a
P
1 ≤
b
P(|X
c
P(|X |
≥
≤ 3
X
)
= 1.5,
if
=
4,
n
= 10;
= 9,
n
= 7;
2
−
5
≤
3)
if
= 5,
2
2
Sleeping
0.8)
habits
distributed
standard
were
3
A
4
less
the
mean
(
Find
P
b
Find
the
of
the
The
Limit
Moivre.
extended
Bessel
the
35
mean
hours.
Find
5
is
and
15
are
value
of
students
the
found
5.5
to
hours
from
probability
taken
the
that
in
mean
a
the
from
the
standard
total
large
value
The
Almost
of
of
the
that
be
and
the
university
on
normal
deviation
sample
shopping
of
62.5 kg
led
12
states
that
an
12
a
value
centre
and
the
maximum
average
people
will
that
people
Limit
a
not
the
will
average
they
distribution
6.
is
are
more
than
180.
distributed
standard
deviation
weight
exceed
total
not
load
of
800 kg.
randomly
70 kg.
weight
exceed
a
of
of
the
the
randomly
maximum
load
to
Expectation
one
centur y
work.
due
Theorem
mathematicians
of
later
the
the
rst
Later
of
including
Central
of
the
Limit
worked
about
Laplace
on,
von
Mises,
by
CLT
in
in
was
Theorem
in
published
Central
all
of
Pólya
it
and
Poisson,
named
theor y.
Lindeberg,
and
general
Markov,
setting.
and
the
probability
Chebyshev,
a
the
Abraham
Dirichlet,
George
CLT
on
theorems
it
Cauchy,
the
contributions
publication
have
impor tant
write
period
impor tance
Finally,
and
to
this
contributions.
to
CLT
.
algebra
most
Pierre-Simon
During
mathematicians
the
the
mathematician
impor tant
on
many
(CLT),
rst
Moivre’
s
other
working
that
probability
histor y
“central”
Lyapunov
label
group
Theorem
de
Pólya,
a
Central
made
theorem
68
university
elevator.
mathematics.
were
a
= 11
)
the
group
Throughout
with
the
size
probability
Calculate
selected
de
of
buyers
has
the
selected
2.3
≥ 37
with
elevator
Find
b
1.2
at
n
hours.
value
of
= 8,
18.5 kg.
An
a
of
probability
weights
normally
of
X
with
random.
5
sample
a
The
at
than
−0.2,
students
deviation
random
with
of
normally
selected
sleep
=
if
Along
Lévy
and
So
far
let’s
we
take
have
an
been
spreadsheet,
we
calculated
the
of
four
a
set
mean
of
of
each
repeated
taking
example
the
have
mean
simulated
throws
set
of
4
B
samples
doesn’t
value
process
A
that
of
(shown
ve
D
four
in
times
E
A
We
(shown
in
die
have
below),
column
G
populations.
distribution.
tetrahedral
throws.
in
normal
normal
a
column
(shown
more
from
a
throwing
the
throws
only
follow
B
columns
H
times,
and
performed
and
have
J
D
−
0
have
then
repetitions
calculated
We
time
a
four
below).
This
Using
the
then
Q).
K
M
N
P
1
80
3
81
4
1
2
3
3
2
82
2
2
4
4
1
4
83
4
3
2
4
2
3
84
3
3
1
1
3
4
85
4
2
2
1
1
4
86
4
4
2
1
3
3
87
1
3
1
1
2
2
88
4
4
1
3
2
1
89
1
2
1
1
2
90
2
91
2
2
4
3
4
3
92
4
2
3
1
2
4
93
3
4
4
2
2
2
94
1
1
1
3
4
4
95
2
4
3
4
2
1
96
2
4
3
3
4
1
97
1
4
3
2
4
3
98
1
1
2
4
3
1
99
4
4
2
1
1
100
2
2,2
1
2,7
2
2,7
2
2,4
2
2,8
2
2,3
101
mean=
2,53
mean=
2,56
mean=
2,46
mean=
2,54
mean=
2,62
mean=
2,44
102
sd=
0,275076
sd=
0,359629
sd=
0,392145
sd=
0,440202
sd=
0,285968
sd=
0,295146
We
X
=
1
that
a
a

x
variable
die
2
has
3
X
the
3
3
1
2,7
random
tetrahedral
3
2,4
3
2,9
know
throwing
3
4
2,6
4
Q
1
1
79
2
1,7
that
describes
following
the
3
2,8
outcomes
3
2,8
2
4
2
probability
2,7
2,3
3
2,8
2
when
distribution.
4
i
P( X
=
x
1
1
1
1
4
4
4
4
)
i
1
We
can
see
that
E
(
X
)
5
(
1+ 2 + 3 + 4
=
)
=
=
4
2
and
5
2
2
1
Var
(
X
)
=
(1 + 4 + 9 + 16 ) −
4
In
each
are
in
in
the
group
the
of
repetition
the
range
range
four
[0.317324 ,
sample
mean
⎜
⎟
⎝
2 ⎠
5
of
X,
we
= 1
12
2
notice
the
is
that
standard
The
expected
which
standard
=
4
throws
the
5
⇒ σ
=
0 .643342 ].
around
deviation
5 ⎞
2 .675] and
[2.175,
themselves
standard
the
of
⎛
mean
value
.2,
sample
deviation
values
of
is
the
X,
not
of
even
also
notice
that
the
standard
error
is n
=
4
⇒
this
value
is
in
the
are
sample
2.5,
the
mean
while
range
obtained
range [0.317324 ,
1.118033989
=
n
and
in
is
values
deviations.
σ
We
values
the
which
mean
= 0. 559017
2
0 .643342 ].
Chapter
2
69
We
are
We
can
n
10:
=
going
see
to
the
A
repeat
results
B
in
simulation
the
D
two
E
for
the
sample
spreadsheets
G
of
n
= 10
and
n
= 50.
below
.
H
4
size
J
K
3
M
4
N
P
1
Q
1
79
1
80
3
81
4
1
2
3
3
2
82
2
2
4
4
1
4
83
4
3
2
1
2
3
84
3
3
1
1
3
4
85
4
2
2
1
1
4
86
4
4
2
1
3
3
87
1
3
1
3
2
2
88
4
4
1
1
2
1
89
1
2
1
2
2
90
2
91
2
2
4
1
4
3
92
4
2
3
2
2
4
93
3
4
4
3
2
2
94
1
1
1
4
4
4
95
2
4
3
3
2
1
96
2
4
3
2
4
1
97
1
4
3
4
4
3
98
1
1
2
1
3
1
99
4
4
2
2
1
100
2
2,2
1
2,7
2
2,7
2
2,4
2
2,8
2
2,3
101
mean=
2,53
mean=
2,56
mean=
2,46
mean=
2,54
mean=
2,62
mean=
2,44
102
s.d.=
0,275076
s.d.=
0,359629
s.d.=
0,392145
s.d.=
0,440202
s.d.=
0,285968
s.d.=
In
each
lies
repetition
a
= 10
range
mean
⇒
for
1
n
=
ten
of
standard
throws,
of
This
four
3
1,7
notice
2.62].
repetitions
3
1
2,7
[2.44,
that
is
the
sample
The
3
2
3
narrower
throws.
2,8
2,7
3
2,8
2
2,3
4
2,8
3
2
0,295146
mean
than
the
standard
range
error
of
the
is
118033989
=
n
the
of
3
2,4
3
2,9
σ
n
1
2,6
within
sample
=
0
35355,
which
is
in
the
range
of
10
deviations.
[0.275076, 0 .440202 ]
.
50:
A
70
the
B
D
E
G
H
J
K
M
N
P
Q
481
4
4
2
2
1
4
482
1
4
1
3
2
2
483
2
3
3
4
3
3
484
2
3
2
3
3
3
485
2
4
4
2
4
3
486
1
1
1
3
3
3
487
3
3
1
1
2
4
488
1
1
2
3
1
3
489
1
4
1
4
1
4
490
1
3
4
3
3
2
491
1
2
1
3
4
2
492
2
4
3
3
3
1
493
4
1
3
1
1
4
494
2
2
2
2
3
3
495
3
1
1
1
4
1
496
3
2
4
2
2
3
497
4
2
2
4
1
4
498
2
4
1
2
1
2
499
1
3
1
4
2
500
1
2,28
2
2,4
3
2,34
1
2,5
1
2,42
4
501
mean=
2,514
mean=
2,552
mean=
2,442
mean=
2,508
mean=
2,444
mean=
2,516
502
s.d.=
0,224707
s.d.=
0,100311
0,114095
s.d.=
0,154402
s.d.=
0,101893
s.d.=
0,116924
Expectation
algebra
and
Central
s.d.=
Limit
Theorem
1
2,46
In
each
lies
repetition
within
sample
an
of
even
mean
for
50
throws
narrower
repetitions
notice
that
of
four
throws
σ
error
for
n
=
50
is
by n
given
the
sample
range [2.442, 2 .552 ] than
= 50
or
1
⇒
ten
Our
the
is
in
the
range
conclusion
closer
the
population
is
of
that
the
the
parameters
mean
and
the
standard
larger
of
the
the
The
of
the
standard
118033989
=
0
15811
,
50
deviations [0.100311
, 0 .224707 ].
number
sample
standard
range
throws.
=
n
which
mean
the
of
mean
samples
are
we
getting
to
take,
the
error.
2
X
is
a
Then
random
n
as
variable
→ ∞
the
with
sample
the
parameters
mean
(
E
distribution
)
X
=
μ
and
approaches
Var
the
(
X
)
= σ
.
normal
2
⎛
distribution
X
∼ N
⎜
σ
μ,
n
⎝
This
any
be
is
one
of
the
probability
a
large
The
of
mean
samples
The
●
the
of
With
can
a
be
(discrete
taking
means
the
and
of
a
or
the
is
most
amazing
continuous)
large
all
samples
also
number
samples
ver y
close
of
and
to
take
those
taken.
the
facts
in
from
samples
W
e
should
mean
mathematics.
of
it
a
we
sample
look
notice
the
at
the
Suppose
of
the
size n
(n
take
must
statistical
following
population
we
that
results:
we
took
the
from.
population
●
After
standard
sample
important
distribution
number).
characteristics
●
most
deviation
divided
size
n,
large
the
by
the
the
of
samples
factor
standard
number
better
of
that
is
is
propor tional
close
to
the
to
value
the
of
standard
the
deviation
square
root
of
of
the
the
error.
samples
approximated
by
a
of
size n,
normal
the
distribution
of
the
means
of
each
sample
distribution.
The
simulation
of
the
y
3
f(x)
sampling
n
=
process
50
Central
2
can
be
distribution
and
the
Limit
Theorem
found
at
http://onlinestatbook.
1
n
=
10
com/stat_sim/
n
=
4
sampling_dist/
0
x
1
2
3
4
This
5
website
both
predened
uniform,
Theorem
5:
The
Central
Limit
sampling
from
any
normal,
and
Theorem
skewed
When
has
population
X
with
the
nite
parameters
in
distributions,
addition
to
the
2
and
,
the
distribution
of
the
sample
mean
X
is
approximately
capacity
to
Normal
if
the
sample
size
n
is
large
enough.
The
mean
of
for
create
students
their
distributions,
distribution
of
the
sample
mean
is
equal
to
the
population
the
variance
of
the
distribution
of
the
sample
mean
is
a
probability
the
distribution
variance
of
the
parent
population
divided
by
the
sample
by
mean
graphing
and
own
the
function
size,
and
then
applying
the
2
⎛
σ
process
X
∼
N
⎜
⎝
of
sampling
μ,
n
distribution
on
it.
Chapter
2
71
When
we
restrict
say
ourselves
population
such
Example
From
girls
a
age
Find
5
of
the
as
taking
a
samples
normal
mean
from
distributed
data
collected
who
live
6 cm.
in
There
probability
and
Find
65
c
to
we
that
a
we
cer tain
are
not
type
going
to
of
population.
a
in
a
school
large
are
town
two
over
have
spor ts
a
number
of
years,
a
mean
height
teams
made
up
of
of
5
we
can
66 cm
girls
and
say
and
8
a
that
standard
girls
within
that
group.
65
b
population”

medical
of
deviation
age
“any
the
probability
and
What
that
the
mean
height
of
the
team
with
5
girls
is
between
that
the
mean
height
of
the
team
with
8
girls
is
between
72 cm.
72 cm.
can
you
conclude
about
those
two
teams?
2
⎛
X
a
∼ N
⎜
6
the
parameters
of
the
sample
mean.
5
⎝
P
Find
166,
(
165 ≤
X
≤ 172
)
=
0
Use
633
a
GDC
to
calculate
the
probability.
2
⎛
X
b
∼ N
⎜
6
Find
166,
P(165 ≤
the
parameters
of
the
sample
mean.
8
⎝
X
≤ 172 )
=
0 .679
Scratchpad
6
normCdf
(
0.632632
)
165, 172, 166,
√5
6
normCdf
(
0.678985
)
165, 172, 166,
√8
c
With
more
higher
probability
height
is
overall
T
o
closer
help
weight
char ts
assess
in
72
to
the
children.
a
growth
other
consist
illustrate
to
of
a
the
series
This
a
child
whether
there
Expectation
valuable
is
might
be
algebra
team’s
to
is
at
of
same
percentile
can
an
child’
s
doctors
Limit
height
that
measurements
appropriate
Central
use
Growth
cur ves
body
help
a
doctors
age.
determine
rate
or
Theorem
Notice
team
the
problems.
and
a
mean
height
compare
the
selected
tool
growing
the
there
development,
of
of
of
team,
girls.
char ts
children
a
mean
of
child’
s
distribution
whether
in
that
population
percentile
and
members
the
that
then
same
if
we
we
get
inter val
have
a
more
larger
of
members
probability
heights.
of
a
over
Example
8
2
A
population
from
the
with
the
Find
the
probability
b
Find
the
minimum
and
= 1
is
given.
We
take
a
sample
of
a
size n
∼ N
⎜
P(1 .5 ≤
sample
X
≤
2 .5 )
size n
given
such
that
n
=
P(1 .5 ≤
that
20.
X
≤
2 .5 )
is
at
least
0.9.
1
⎛
X
2
population.
a
a
=
parameters
Find
the
parameters
of
the
sample
mean.
2,
20
⎝
Scratchpad
P(1 .5 ≤
X
≤
2 .5 )
=
0 .975
1
normCdf
(
0.974653
)
1.5, 2.5, 2,
√20
1
⎛
X
b
∼ N
⎜
2,
1.1
Scratchpad
n
⎝
y
P(1 .5 ≤
X
≤
2 .5 )
=
0 .9 ⇒
n
= 11
3
1
f1(x)=normCdf
1.5
1.5, 2.5, 2,
(
x
)
–0.9
(10.8, 0)
x
0
5
10
15
20
–1.5
Round
Exercise
the
value
of
n
to
the
rst
larger
integer.
2F
If
the
population
2
1
Given
size
n
the
population
and
parameters
and
the
nd:
sample
distribution
symmetrical,
smaller
is
then
sample
even
we
are
with
a
going
2
a
P(1 .5 ≤
X
≤
2 .5 )
=
if
2,
= 9,
n
= 30;
to
P(1 .25 ≤
X
≤ 1 .35 )
= 1.3,
if
a
better
approximation
2
b
achieve
=
0.04,
n
to
a
normal
= 50;
distribution.
When
the
2
c
P( X
≥
−0 .48 )
=
if
−0.5,
= 1
,
n
= 100;
population
distribution
symmetrical
then
to
is
not
achieve
2
d
P( X
< 397 ) if
=
400,
=
234,
n
= 85;
a
good
normal
1 ⎞
⎛
e
P
⎜
X
⎝
<
⎟
⎜
⎝
we
to
a
need
= 1
,
=
6,
n
=
a
40;
larger
sample.
Usually
we
say
2 ⎠
1
⎛
P
distribution
2
if
that
f
approximation
X
− 2
need
a
sample
size
2
if
≥
we
= 1.9,
= 36,
n
= 120 .
of
at
least
n
=
30
to
achieve
5
normality
of
the
sampling
distribution.
Chapter
2
73
2
A
in
real
a
estate
city
€8,000.
3
agent
have
a
Given
nd
the
The
gestation
a
a
average
than
b
the
64
the
2
60
the
average
has
days.
probability
gestation
a
mean
apar tments
standard
were
price
Three
studio
and
apar tments
dogs
2
€30,000
of
doesn’t
value
female
deviation
selected
at
exceed
of
dogs
63
€32,000.
days
are
of
random
and
selected
during
that:
period
of
selected
dogs
will
last
longer
gestation
period
of
selected
dogs
will
last
less
period
of
selected
dogs
will
last
within
days;
average
days
for
prices
days;
average
than
c
Find
the
of
the
of
studio
that
period
that
value
15
deviation
period.
the
mean
that
probability
standard
this
claims
of
gestation
the
expected
value.
2
4
A
population
We
take
a
sample
the
of
size
Find
the
probability
b
Find
the
minimum
Svetlana
jumps
and
normally
≤
sample
a
of
X
4
is
given.
population.
≤ 7 .6 )
size n
given
such
that
n
P(5 ≤
that
the
with
distributed
a
that
average
of
mean
jumps
made
probability
specializing
deviation
Their
Blanka
than
athletes
normally
standard
9 cm.
competition
the
are
are
distributed
deviation
greater
P(5 .4
the
=
=
X
10.
≤ 7)
is
at
QUESTIONS
and
Blanka’s
Find
from
and
exercise
EXAM-STYLE
2.02 m
n
6
0.95.
Review
Blanka
=
parameters
a
least
1
with
ve
the
average
of
a
the
Svetlana’s
of
and
high
mean
1.98 m
independent.
jumps
height
with
4 cm.
value
are
in
height
of
Svetlana’s
value
jumps
and
a
made
Blanka’s
of
are
standard
During
Svetlana
jump.
a
four
jumps.
jumps
was
jumps.
2
2
Let
X
∼ Po( m )
a
Show
b
Hence
Another
that
Find
Let
74
Find
e
State
=
such
P( X
the
≥
variable
Var
Z
and
reason
algebra
(2 X )
=
( E ( X ))
− 5
Y,
(
that
3
Y
)
is
independent
of
X,
has
a
Poisson
= 18
 5)
mean
a
Var
6)
that
variable
with
Expectation
 Y
that
5.
P( X
random
random
d
m
nd
distribution
c
such
and
be
such
that
variance
of
whether
or
Central
Limit
Z
= 3X

4Y .
Z
not
Z
has
Theorem
a
Poisson
distribution.
3
A
of
sh
shop
these
values
and
three
may
be
standard
types
Standard
Bass
320
2
Bream
400
20
Cod
350
5
a
Find
the
probability
b
Find
the
probability
c
Find
d
Calculate
and
a
A
the
b
A
that
that
bass
parameters
the
bream
random
Given
one
is
random
n
the
and
of
less
than
=
27
X
Z
in
bream,
grams,
weight
one
the
total
twice
and
cod.
distributed.
are
follows
a
one
of
two
and
weight
the
the
the
of
bream
that
all
bass,
normally
weight
follows
nd
variable
be
given
in
The
The
the
weights
mean
table
below
.
deviation
the
probability
variable
that
sh:
to
deviations,
Mean
buys
of
assumed
Fish
Nicholas
4
sells
sh
total
a
cod
is
Alex
of
sh
of
binomial
less
of
one
670 g.
cod.
has
Nicholas’
bought.
bass
cod.
distribution
of
450 g.
than
Nicholas
Alex’s
values
Poisson
exceeds
buys
weight
weight
possible
bream
E
(
3X
distribution
Var ( X )
and
7
=
6.
)
and
2
(
Var
(
Z
)
)
Prices
mean
a
A
of
value
car
is
Given
the
of
)
+ 12
2Z
family
€12,000
selected
at
)
and
a
cars
adver tised
standard
random.
Find
on
a
deviation
the
website
of
have
a
€3,200.
probability
that
the
price
€13,000.
that
30
cars
are
that
selected
the
at
average
random
price
of
from
those
the
30
website,
cars
will
nd
be
less
€11,500.
Vladimir
spend
is
buying
between
randomly
that
Z
5 +
probability
that
c
(
(
second-hand
exceeds
b
E
Var
Calculate
5
=
an
second
€10,500
select
average
a
in
and
order
price
to
falls
hand
family
€12,500.
have
within
the
the
car
How
and
many
probability
desirable
he
is
cars
of
at
willing
should
least
to
he
85%
range?
2
6
a
Let
X
be
a
random
variable.
2
show
b
that
Given
where
two
p
E (X
= 1,
expanding
the
expression
E ( X
E
(
X
))
2
) ≥
(E ( X ))
independent
+ q
By
show
random
that
Var
(
variables
X
+ Y
)
=
such
E
(
X
that
+ Y
X
∼ Geo( p )
) (
(
E
X
+ Y
)
and
Y
∼ Geo( q ),
3)
Chapter
2
75
Chapter
Linear
that
and
b
X
∈ R
E(aX
i)
summary
transformation
Given
a,

+
is
a
random
of
a
variable
single
with
variable
nite
parameters
then
b)
=
aE(X )
+
b
2
Var(aX
ii)
Linear
Given
+
that
E(aX
=
a
Var(X )
transformation
X
parameters
i)
b)
+
and
and
bY )
Y
a,
=
are
b
two
∈ R
of
Var(aX
+
bY )
Independent
random
variables
with
nite
then
aE(X )
=
variables
independent
+
bE (Y
)
2
ii)
two
2
a
Var(X )
random
+
b
Var (Y )
variables
+
Given
that
independent
random
variables X
,
X

,
X
2
,
…,
X
3
,
n
∈ Z
all
have
equal
n
2
expected
value
and
equal
variance,
E
(
X
)
μ
=
and
Var
(
X
)
=
,
σ
then
2
E( X
+
X
1
+ ... +
X
2
)
nμ
=
Var ( X
and
Independent
+
X
1
n
normal
random
+ ... +
X
2
)
nσ
=
n
variables
2
If
we
take
two
independent
normal
random
variables
X
∼
N (μ
, σ
1
then
the
linear
combination aX
+
bY ,
a,
b ∈ 
The
parameters
=
are
a
+
b
and
1
2
aX
+
bY
∼
N ( aμ
bμ
+
1
Sampling
,
=
2
2
+
b
1
to
be
a
and
Y
∼
N (μ
2
normal
variable
also.
2
b
1
2
+
going
2
a
2
σ
a
2
is
2
2
2
)
1
,
so
we
can
write
2
2
σ
)
2
Distribution
of
the
Mean
2
σ
E
(
X
)
=
μ
and
Var
(
X
)
=
,
Normal
population
sample
n
2
⎛
σ
2
X
∼ N (μ, σ
)
⇒
X
∼
N
⎜
μ,
⎝
n
σ
The
term
standard
deviation
of
the
mean,
is
n
76
Expectation
algebra
and
Central
Limit
Theorem
also
known
as
the
standard
er ror.
, σ
)
2
The
Central
Limit
Theorem
2
When
the
distribution
size
to
sampling
n
the
is
large
from
of
any
the
enough.
population
population
sample
The
mean
mean
mean
and
the
of
X
X
with
is
the
the
nite
approximately
distribution
variance
of
the
of
and σ
parameters
Normal
the
if
sample
distribution
of
the
the
,
sample
means
is
sample
equal
means
is
the
2
⎛
variance
of
the
parent
population
divided
by
the
sample
size,
X
∼ N
⎜
⎝
σ
μ,
.
n
Chapter
2
77
Exploring
statistical
3
analysis
methods
CHAPTER
7.3
OBJECTIVES:
Unbiased
estimators
based
variances;
on
and
estimates;
comparison
of
unbiased
estimators
n
X
i
X
as
unbiased
estimator
μ.
for
X
=
n
(
X
=
X
i
2
n
Condence
7.6
Null
and
critical
inter vals
alternative
for
the
mean
hypotheses
values,
p-values,
including
calculations
a
population.
normal
Before
the
you
variance
of
the
table,
of
H
one-tailed
of
nd
continuous
a
normal
and
H
their
.
population.
Signicance
and
two-tailed
probabilities.
tests.
T
esting
x
<
0
3
0
≤
x
<
20
2
20
≤
x
<
30
2
30
≤
x
<
40
8
40
≤
x
<
50
6
Answers
obtained
that
P (X
the
mean
1
variable X
Given
and
P (X
c
P (1
T
ype
I
and
the
following
II
for
regions,
errors,
the
variance
of
the
table,
mean
nd
variable
the
=
X
5)
~
=
≤
6)
=
X
by
the
B(10,
≤
x
<
2
22
2
≤
x
<
4
37
4
≤
x
<
6
46
6
≤
x
<
8
5
GDC.
0.3)
nd:
2
Given
that
X
~
B(n,
p)
nd:
1
0.103;
P (X
=
5)
if
n
=
5
and
p
=
;
2
0.989;
1
≤
X
≤
3)
=
0.62.
b
P (3
≤
X
<
8)
if
n
=
10
and
p
=
5
3
Given
that
Y
<
Po
(2)
nd:
3
78
a
P (Y
b
P (2
c
P (Y
=
≤
≥
Exploring
0)
Y
=
≤
4)
=
Given
that
Y
<
Po
(m)
nd:
0.135;
5)
=
a
P (Y
b
P (3
=
0)
if
m
=
0.4;
0.577;
0.143.
statistical
of
mean
Frequency
0
a
b
Critical
hypotheses
X
≤
a
level.
1
Frequency
0
Given
.
start
following
X
2
σ
for
)
0
and
estimator
1
7.5
Given
unbiased
∑
i =1
1
2
as
n
i =1
S
2
S
∑
analysis
methods
≤
Y
<
8)
if
m
=
7.
Biased
information:
Statistics
are
proposals,
television
often
and
regret
on.
later
control
Statistical
In
we
work
particular
calculated
that
be
from
which
to
an
the
quality
in
use;
from
used
the
into
to
sense
add
can
this
put
statistical
making
by
data?
to
theories,
paying
forward
analysis.
decisions
about
of
credibility
see
numbers
lear ning
age,
to
pull
this
is
calculators
what
is
of
two
is
only
Y
ou
balanced
you
data
in
that
statistics
and
that
some
sampling
tells
data
its
useful
is
attention
to
through
They
you
a
can
nd
good
how
be
cause
step
Roughly
,
us
a
then
to
towards
values?
the
do
decide
used
is
a
to
same
which
do
or
the
not
properties
population
there
a
quantity
the
about
Fortunately
,
to
amounts
whether
about
lear n
huge
to
be
statistic
from
accurately
is
decide
something
samples
to
we
of
possible
task
can
can
out
and
Our
computers
and
way
of
information
technology
.
relevant?
dierent
there
a
or
useful
extremely
statistics
balanced
So,
of
reasons,
be
sample
an
is
of
can
entire
a
way
,
and
it
Estimator
is
a
proper ties
obtain
is
make
attempt
advancements
However,
Estimator
reects
a
Many
digital
But
we
life.
can
to
us!
an
represent
these
to
as
can
arguments.
entice
your
due
dierent.
called
An
of
statistic
population
is
For
want
for
sample.
ver y
not
might
current
systematically
hard
do
methods
our
statistics
and
adver tisements.
misleading
data.
presented
ideologies,
adver tisements
taking
how
sample
some
way
of
of
calculating
both
was
drawn.
estimators
terms
of
the
how
of
data
In
a
special
sample
this
chapter
parameters
accurately
and
they
of
a
type
also
we
of
statistic.
the
are
entire
going
population,
model
the
to
This
statistic
population
study
and
in
analyse
population’s
from
detail
how
their
parameters.
Chapter
3
79
3.1
Estimators
Suppose
Y
ou
have
samples
Let’s
have
by
that
no
lots
of
say
that
a
specic
at
normal
have
data,
to
you
that
the
but
a
of
do
draw
exact
sample
you
a
statistical
you
the
from
you
S
θ,
of
that
T,
a
we
sample.
that
are
a
company
.
you
the
use
entire
want
A
The
to
Since
estimate
random
variable T
for
is
is
Consider
estimators
you
this
parameter
variable T
parameter θ
below
.
these
population?
parameter, θ.
the
shown
for
can
about
has
approximate
and
How
population.
distributions
variables,
start?
that
value
from
obtain
which
analysis
conclusions
population
the
normal
random
where
have
by
two
perform
and
random
value
to
estimates
infer
knowledge
taking
Look
of
data
information
A
you
and
called
called
describes
an
an
the
estimator.
estimate.
two
μ
S
T
μ
Notice
to
the
and
In
S
E
that
mean
is
(
S
μ
)
value.
obviously
general
a
population
μ ,
≠
In
whilst
this
biased
random
case
and
variable
parameter
θ
if
(
T
E
)
we
=
hence
T
(
E
T
is
)
since
say
that T
should
called
an
T
is
has
an
not
a
symmetrical
unbiased
be
used
unbiased
for
shape
estimator
estimation
estimator
for
with
for
of
the
the
respect
mean
parameter.
the
= θ
Denition:
An
estimator
of
a
population
a
random
parameter
(such
as
the
μ,
mean,
or
the
2
variance,
The
A
)
specic
Let
is
estimator
us
value
now
parameter
provides
of
consider
E (T
)
=
that
two
E (T
1
)
variable
an
random
depends
variable
random
=
that
approximation
μ ,
and
to
is
variables,
consider
this
on
called
both
the
unbiased
which
one
1
T
2
Exploring
statistical
μ
analysis
methods
estimators
might
T
80
data.
parameter.
an estimate
2
μ
sample
unknown
be
a
of
better
the
value
same
estimator.
μ,
By
looking
at
the
graphs
of
both
unbiased
estimators T
and T

that
T
has
a
smaller
spread
than
of
smaller
T

(or
.
Thus,
the
,
notice
2
standard
deviation
2
variance)
T
is
than
the
standard
deviation
(or
variance)

of
T
.
For
a
good
estimation,
it
is
essential
to
use
a
random
variable
2
with
a
small
standard
deviation.
Denition:
Given
two
estimators
and
T
T

T
is
a
more
ecient
of
the
population
estimator
than
T

if
Var (T
Given
that
that
) <
Var (T
)
2

two
the
say
1
2
Example
we
2
distributions
better
of
estimator
is
sample
the

mean
one
with
taken
the
from
larger
the
same
normal
population,
show
sample.
2
⎛
2
X
,
X
1
, ...,
σ
+
 N (μ,σ
X
2
n ∈ 
),
⇒
X
 N
⎜
n
Take
μ,
n
observations
and
nd
the
n
⎝
parameters
of
the
sample
mean. Take
m
2
⎛
2
Y
, Y
1
, ..., Y
2
(μ,σ
 N
σ
m ∈ 
⇒ Y
 N
⎜
that
one
is
a
better
2
(
X
)
<
Var
(
Y
)
⇒
the
better
same
population
and
m
nd
the
parameters
Use
the
denition
of
the
sample
mean.
estimator:
σ
1
<
n
Therefore
the
2
σ
Var
from
μ,
⎝
Assume
observations
+
),
m
⇒
m
1
n
estimator
⇒
<
n
>
m
a
larger
of
better
estimator.
m
comes
from
sample.
Unbiased
normal
Two
we
of
we
will
normal
set
random
well-known
scenarios
i.e.
estimators
will
need
random
normal
the
mean
and
variance
of
a
variable
statistics
need
an
for
to
are
mean
determine
unbiased
variables,
random
the
an
variance.
estimators
estimator
and
and
for
the
unbiased
for
When
the
mean
mean
value
estimator
for
creating
and
of
the
the
statistical
variance,
set
of
variance
of
the
variables.
Denition:
2
For
normal
random
variables
X
~
N(
,
σ
),

≤
i
≤
n,
i
n
X
i
1
X
is
an
unbiased
estimator
for
X
.
=
∑
i =1
n
2
n
(X
2
2
S
2
2
is
an
unbiased
estimator
for
.
S
X )
i
=
∑
i =1
n
1
Chapter
3
81
A
well-dened
We
must
show
denition?
that
the
above
denitions
for
unbiased
estimators
of
2
μ
1
and
are
σ
E(X )
well
dened.
To
do
so,
we
must
show
the
following:
=
Since all
of
the
samples
are
taken
from
the
E (X
population
)
μ, i
=
= 1
,
2,...,
n
i
n
n
X
⎛
E
(
X
=
E
2
2
E(S
⎞
n
⎛
1
⎞
1
1
i
)
⎜
∑
⎝
i =1
=
E
⎟
n
∑
⎜
n
⎠
X
⎝
=
n
⎠
i =1
E (X
∑
⎟
i
)
=
n μ
×
μ
=
i
n
i =1
2
)
=
2
σ
By
now
we
have
found
out
that
E
(
)
X
μ
=
and
(
Var
X
)
=
,
but
also
n
2
E (( X
−
μ )
2
)
= σ
, i
= 1
,
2,...,
therefore
n
we
can
calculate
the
following:
i
n
2
E (( X
−
μ )
−
μ )
2
)
=
)
=
nσ
i
i =1
n
n
2
E (( X
2
E ((( X
i
−
X )
+
(X
μ ))
−
)
i
i =1
i =1
n
2
=
∑
E (( X
−
2
X )
+
2 (X
i
−
X ) (X
−
μ)
+
(X
μ)
−
)
i
i =1
n
2
=
∑
(E ( X
−
2
X )
+
2E ( X
i
−
X ) E
(X
−
μ)
+
E (X
−
μ)
)
i
i =1
n
n
n
2
=
∑
E (X
−
2
X )
+
∑
i
i =1
2E ( X
μ )E ( X
−
−
X ) +
i
∑
i =1
E (X
μ)
−
i =1
n
n
n
2
=
∑
E (X
−
2
X )
+ 2E ( X
μ)
−
i
∑
i =1
E (X
−
X ) +
i
i =1
∑
E (X
−
μ)
i =1



0
n
2
n
σ
2
=
∑
E
(X
−
X )
2
+
nVar ( X )
=
i
∑
i =1
E (X
−
X )
+
n
i
n
i =1
n
2
=
∑
E (X
−
2
X )
σ
+
i
i =1
n
2
⇒
nσ
2
=
∑
E (X
−
X )
2
+
σ
−
σ
i
i =1
n
2
⇒
∑
E (X
−
X )
2
nσ
=
2
2
=
i
(n
−
1) σ
i =1
2
S
is
an
unbiased
estimator
the
population
2
n
⎛
(X
2
E (S
of
−
X )
=
E
⎜
⎝
∑
⎟
n
i =1
− 1
because
n
⎞
1
⎛
2
i
)
variance
⎠
=
E
⎜
n − 1
∑
⎝
(X
−
1
⎞
2
=
X )
(n
−
1) σ
2
=
σ
⎟
i
n
⎠
i =1
−
1
2
Thus,
In
82
the
above
examples
Exploring
2
denitions
and
3,
statistical
we
for
will
analysis
unbiased
look
at
methods
estimators
how
to
use
of
these
μ
and
σ
are
denitions.
well-dened.
Example
Show

that
the
unbiased
estimator
of
the
variance
can
be
found
by
using
the
formula:
n
2
2
x
∑
n( x )
i
n
2
s
2
i =1
=
.
n
Hence
2
=
σ
n
1
n
2
s
s
that
1
n
∑
2
show
(
x
−
x
2
2
)
i
(x
∑
i =1
− 2x
i
x
(
+
x
)
)
i
i =1
=
Expand.
=
n
−
1
n
n
n
−
1
n
2
2
∑
x
−
2x
x
∑
i
i =1
+
(
∑
i
i =1
)
x
i =1
Use
=
n
the
distributive
proper ty.
1
n
n
2
2
2
∑
2
x
−
2x
×
nx
+
n
(
)
x
∑
i
i =1
x
−
n
(
x
)
i
i =1
=
Simplif y
=
n
−
1
n
−
the
n
the
expression
by
using
1
denition
of
the
mean.
n
2
2
2
∑
2
x
−
(
n
x
)
∑
i
x
−
n
(
x
)
i
n
2
s
i =1
i =1
=
Simplif y
=
n
− 1
n − 1
by
using
the
denition
n
of
variance.
n
⎛
⎞
2
∑
⎜
x
⎟
i
n
n
2
2
i =1
⎜
=
n
−
1
the
(
x
)
n
⎝
Example
Given
−
⎟
σ
=
n
⎠
−
1

following
set
of
data:
22,
24,
23,
a
an
unbiased
estimation
of
the
mean;
b
an
unbiased
estimation
of
the
variance.
Method
x
26,
26,
27,
25,
24,
24,
26,
25,
26,
27,
28
nd:
I
∑
a
22,
x
375
i
=
⇒
x
=
=
25
Unbiased
estimate
of
the
mean
is
the
mean
15
n
value
of
the
sample
itself.
n
2
2
∑
x
n
(
x
)
i
2
9421
2
b
s
−
15 ×
25
2
i =1
⇒
=
n
s
=
1
Use
of
46
=
=
the
=
3
unbiased
estimate
29
Use
*Unsaved
B
C
the
GDC
∑x
22
∑x
26
SX
In
unbiased
standard
x
23
statistics
order
to
estimate
nd
of
the
the
variance,
square
population
One–Var...
T itle
24
One-variable
D
the
3
the
II
22
2
f or
variance.
calculation.
1
f or mula
7
1.1
A
shor t
23
14
Method
the
14
deviation
s.
25.
375.
2
4
5
9421.
:=
Sn–...
1.81265
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Chapter
3
83
Exercise
1
Calculate
the
following
sets
a
{2,
b
{21,
c
{1,
The
2
3A
4,
6,
8,
24,
4,
unbiased
of
10,
36,
7,
10,
contains
x
10
12,
28,
...,
distribution
estimate
of
the
mean
and
variance
for
the
data:
30,
16,
22,
18,
25,
20}
26,
38,
32,
34,
29,
37,
33,
31,
30}
133}
of
eggs)
14,
broken
is
given
0

2
3
22
2
4
2
eggs
in
the
in
40
boxes
following
(where
each
box
table:
i
f
i
Find
the
broken
The

unbiased
of
of
the
mean
and
standard
deviation
of
eggs.
following
group
estimates
70
table
displays
students
T
0
≤
to
t
Frequency
the
travel
< 10
times, T
to
(minutes),
taken
by
a
school.
10
≤
6
t
<
20
20
≤
3
t
< 30
30
26
≤
t
<
60
60
≤
t
7
< 120
8
Find:
a
An
unbiased
estimate
for
the
mean
b
An
unbiased
estimate
for
the
standard
3.2
Condence
Let’s
A
analyze
large
a
per
from
of
single
ever y
six-week
sample
whole
old
from
farm
year.
released
broiler
real-life
poultr y
chickens)
the
raises
After
farm.
broilers
population
of
the
,000,000
weeks,
is
deviation
of
T
mean
so
weighed.
is
to
found
a
estimate
six-week-old
to
are
be
measure
the
mean
the
to
sample
of
Using
value
be
weight
weight
2.0 kg.
mean
(young
ready
certain-size
The
to
broilers
broilers
dicult
broiler,
are
wish
over
six
It
sample
we
for
T ;
scenario:
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this
mean,
intervals
of
of
of
a
the
the
broilers.
If
However,
how
condent
can
we
be
that
the
population
data
satises
conditions
lies
within
cer tain
limits
(in
kilograms)
of
the
sample
of
2.0 kg?
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we
quantify
such
a
condence
of
that
Theorem,
mean
lies
within
the
prescribed
ability
to
information
answer
available
such
to
a
us,
question
as
the
depends
calculations
on
to
condence
inter val
depend
on
how
much
we
nd
know
the
data
such
population
in
question.
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assume
that
the
84
the
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statistical
of
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analysis
Central
methods
able
to
distribution
calculate
Limit
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Using
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about
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the
real-life
data
problem
satises
a
approximation,
events
the
the
by
distribution.
the
to
a
are
inter val?
of
Our
Central
we
the
approximate
population
the
mean
Limit
value
the
mean
we’re
studying.
There
are
two
calculating
types
situation
inter vals
that
for
you
the
may
mean
faced
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with
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population
standard
deviation
is
known;
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II:
The
population
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not
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estimate
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normal
and
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.
Let’s
take
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level
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[a,
in
the
condence
value
to
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n
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α ,
−
we
can
the
of
where
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Ronald
to
be
based
5%.
developed
the
in
biometrics,
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other
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3
85
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2
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take
the
=
5,
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x
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c
x
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the
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of
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sample
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elements
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70,
75,
from
a
normal
Determine
a
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taken
value
dier
clementines
are:
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0.1;
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and
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Chapter
3
89
The
5
measurement
assumed
to
of
follow
a
n
independent
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random
distribution
measurements
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μ
2
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mean
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mean
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a
have
the
population
2
with
standard
Find
a
a
95%
values
of
i
n
= 10 ;
n
=
iii
n
= 50 ;
iv
n
= 150
II:
can
real-life
so
it
is
of
you
64
inter val
n
given
conclude
unlikely
for
the
mean
for
the
following
that:
there
is
about
that
we
Even
hardly
the
sample
will
if
any
Justify
when
is
constantly
know
we
relationship
sizes?
for μ
populations
population.
distributed,
and
interval
situations,
ver y
obser ved
size
inter vals
Condence
In
=
25;
lengths
Case
σ
condence
sample
ii
What
b
deviation
the
assume
situation
between
your
unknown
grow
or
decay
,
parameters
that
it
when
the
answer.
is
of
the
A
normally
we
are
degree
is
cer tain
the
that
about
population
parameters.
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we
often
need
of
can
var y,
the
parameter
to
of
take
like
to
a
Before

illustrated
certainty
large
be
working
as
our
sample
accurate
with
we
to
small
estimate
for
as
or
that
estimates.
practical
possible
large
the
larger
in
samples
give
However,
reasons.
our
it
We
a
is
greater
dene
often
would,
conclusions,
dicult
however,
standard
deviation σ
,
we
a
concept
that
we
will
use
to
make
our
we
of
If
we
σ
is
have
unknown
a
sample
we
of
must
size
n,
rst
estimate
since
we
it
need
of
the
of
the
of
,
we
say
that
we
have
=
n
90
Exploring
statistical
analysis
methods
sample
and
we
an
have
estimation
standard
−
1
deviation
population,
data
can
then
var y
available
calculate
− 1 degrees
last
one
must
prescribed
by
whilst
the
n
pieces
be
previous
data.
−
1
data
in
order
an
obtain
the
standard
of
deviation
freedom.
a
example,
must
to
estimation
n
are
For
estimate:
from
to
have
size
calculated
when
samples.
population
if
the
When
without
population
we
estimating.
n
rst
v,
data
them.
Investigation
degree
of
to
changing
estimate
freedom,
number
we
require.
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estimate
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population
standard
deviation σ
we
use
n
estimate
of
the
population
standard
unbiased
1
= σ
s
deviation
an
.
n
The
Central
distribution
question
of
still
William
quality
titled
name
The
is
not
as
early
the
Despite
hard
of
work
the
called
size
exceed
doesn’t
t-distributed
deviation
X
Z
by
σ
of
his
new
a
small
with
At
the
end
he
by
his
formalized.
of
of
a
numbers;
1935,
brewer y
continued
paper
strictly
result,
he
the
a
samples
random
Guinness
venture,
employer
As
results
of
in
published
t-distribution
method.
the
this
the
work
of
smaller
t-distribution
variable
estimate
−
to
in
London.
publish
samples,
after
we
Gosset)
use t-distribution
when
sample
of
the
T,
we
approximate
standard
the
standard
deviation, s:
μ
,
=
s
n
n
the
variable T
freedom
and
Student’s
t-distribution
distribution
the
y-axis
the
y-axis.
normal
in
of
statistician
papers.
control
the
distributed?
he
statistical
but
the
30.
random
the
⇒ T
use
charge
distribution
σ
Let’s
form
empirical
a
because
quality
Monte-Carlo
involved
X
=
where
the
as
1908,
scientic
impor tant
and
Student’s
μ
−
the
take
any
In
approximate
distribution,
worked
‘Student’
can
samples
brewer y.
ensure
discovered
to
(commonly
a
the
to
normal
1937)
a
we
papers.
approximate
For
as
used
a
that
smaller
−
of
publishing
mathematical
Ireland
the
statistical
(1876
well-known
Gosset
by
are
pseudonym
from
application
left
How
depar tment
initially
of
suggests
sample
Gosset
under
beer
.
Gosset
To
Sealy
was
combination
an
large
employees
t-test
brewed
a
Theorem
remains:
control
‘t-test’
forbade
Limit
we
graph.
and
a
follows
write T
GDC
They
to
distribution
∼ t (n
graphs
achieving
look
and
a
are
t-distribution
are
ver y
similar
bell-shaped
maximum
at
the
some
with n
− 1
degrees
of
− 1) .
to
the
cur ves;
value
when
probability
normal
symmetrical
the
density
t-distributions
standard
with
cur ve
about
crosses
function
dierent
of
the
standard
degrees
of
freedom.
Chapter
3
91
The
for
graphs
the
show
larger
degrees
t-distribution
insignicant
that
better
as
this
of
distribution
freedom
(i.e.
approximates
under
the
normal
is
aligned
for
the
larger
data
with
the
standard
samples),
since
the
tail
but
areas
for
normal
smaller
are
a
bit
distribution
samples
larger
and
the
not
as
distribution.
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y
0.5
f1(x)
=
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x
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–3
–2
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can
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Before
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p
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technology
, z-distribution
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and
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0.99
“invt”
0.995
feature,
i.e.
a
variables
that
calculates
the
critical t
value;
t
=
invt
c
⎜
the
condence
inter val
can
be
calculated
x
−
⎣
statistical
analysis
methods
t
,
x
+
formula:
t
c
n
, ν
−
2
⎤
×
c
n
the
s
×
⎢
Exploring
using
s
⎡
92
by
with
α
1
⎝
So
0.9995
function
⎛
two
⎥
⎦
a t-distribution.
t-distribution
p
0.975
shape,
normal
t)
P
in
.
table.
Regardless
standard
give
us
book
critical
we
are
deviation
is
Example
In
x
a
=
the
to
fact
that
the
distribution
t-values
going
for
use
any
t-distribution
for
a
sample
t-distribution
large
size
is
ver y
sample
we
choose.
whenever
the
close
size,
to
the
calculators
Therefore,
population
in
can
this
standard
unknown.

sample
of
2540 kJ,
Calculate
of
normal
ve
00 g
whilst
the
99%
the
chocolates
estimate
condence
for
the
the
inter val
mean
value
population
for
the
of
the
energy
standard
mean
value
level
deviation
of
the
was
was
energy
found
s
level
to
be
= 120 kJ.
of
00 g
of
chocolates.
Method
n
= 5
I
⇒ ν
=
4
⇒
t
=
4
=
604
n
− 1
and
use
a
GDC
to
nd
the
critical
c
t-value.
120
s
2540
±
× 4
604
Use
the
f or mula
x
±
×
t
c
5
n
⇒
μ ∈
2787 ] kJ
[ 2293,
*Unsaved
1.1
invt(0.995, 4)
4.60409
120
2292.92
5
120
2540+
2787.08
·4.6040948673027
5
3/99
Method
II
Use
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1.1
“T itle”
“t
“CLower”
2292.92
“CUpper”
2787.08
the
input
Inter val”
GDC
the
t-inter val
given
stats
input
method
and
statistics.
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“x”
2540.
t
“ME”
“df”
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247.082
4.
x:
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:=
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“n”
2540
120.
Sx:
5.
n:
120
5
1/1
C
Level:
|
0.99
ok
⇒
μ ∈
[ 2293,
2787 ] kJ,
given
correctly
to
Cancel
the
0/99
nearest
Let’s
tr y
kJ.
to
see
how
to
solve
the
problem
of
the
mean
weight
of
broilers.
Chapter
3
93
Example
A
large
the

poultr y
sample
was
population
90%
farm
found
with
inter val
grows
for
to
be
broilers.
2.0 kg.
the
unbiased
the
mean
A
It
estimate
weight
of
a
sample
is
of
20
assumed
of
the
broilers
that
standard
is
broilers
taken.
come
deviation
of
The
from
mean
a
0.3 kg.
weight
of
normal
Calculate
the
broiler.
Working:
Method
n
=
20
I
⇒ν
= 19
⇒
t
= 1
729
=
n
− 1
and
then
use
a
calculator
to
nd
the
t-value.
c
0
2.10
s
3
±
×
Use
1.729
the
f or mula
x
±
×
t
c
n
20
⇒
μ ∈
[1.98,
2.22 ] kg
*Unsaved
1.1
invt(0.95, 19)
1.72913
0.3
1.98401
√20
0.3
2.21599
√20
3/99
Method
II
Use
the
GDC
t-inter val
stats
*Unsaved
1.1
“T itle”
“t
1.98401
“CUpper”
2.21599
“x”
input
Inter val”
“CLower”
the
given
statistics.
2.1
*Unsaved
“SX
“ME”
0.115994
“df”
19.
:=
Sn–1X”
t
Inter val
0.3
x:
Sx:
n:
2/99
C
⇒
μ ∈
[1.98,
2.10
20.
“n”
Level:
0.3
20
|
0.9
2.22 ] kg
ok
Cancel
0/99
When
data
94
the
actual
features
Exploring
on
data
the
statistical
is
given,
GDC,
analysis
as
we
need
shown
methods
in
to
use
the
the
list
following
and
statistic
example.
input
method
and
Example
In
a
66.
box
of
six
Calculate
Method
eggs,
the
the
95%
weights
(measured
condence
inter val
in
for
grams)
the
of
mean
each
egg
weight
were:
of
an
62,
63,
65,
62,
66,
egg.
I
x
∑
x
9
384
i
=
⇒
x
=
=
n
64
Find
the
mean
value
Find
the
unbiased
of
the
sample.
6
2
2
x
∑
s
n
(
)
x
24594
i
=
⇒
n
s
24576
=
=1
1
897
standard
n
=
6 ⇒
= 5 ⇒
t
=
2
of
the
population
deviation.
=
n
and
− 1
then
we
571
look
1
64
estimate
5
at
the
tables
or
use
a
calculator
to
nd
897
±
×
2
571
the
t-value.
6
s
Use
μ ∈
⇒
[ 62,
the
f or mula
±
x
×
t
c
66 ] g
n
Method
II
Use
*Unsaved
1.1
“T itle”
“t
where
Inter val”
“CLower”
62.0088
“CUpper”
65.9912
the
GDC
data
a
method
list.
64.
t
Inter val
5.
“df”
:=
in
input
1.99116
“ME”
“SX
stored
data
*Unsaved
1.1
“x”
is
t-inter val
List:
|
a
List:
1
1.89737
Sn–1X”
Frequency
6.
“n”
C
Level:
0.95
ok
Cancel
3/99
{62, 63, 65, 62, 66, 66}
μ ∈
⇒
[ 62,
66 ] g
2/99
Example
A
0
random
sample
of
ten
inde pendent
obser vations
are
taken
10
from
a
nor mal
10
2
population.
The
sample
gave
the
x
results
= 527
and
x
i
i =1
Find
a
90%
∑
x
condence
x
inter val
for
the
=
28,157 .
i
i =1
population
mean.
527
i
=
⇒
x
=
n
=
First
52.7
we
must
nd
the
unbiased
estimate
of
10
the
mean
of
the
sample
and
then
the
unbiased
2
2
∑
s
x
n
(
x
)
estimate
of
the
population
standard
deviation.
i
=
n
1
2
28157 − 10
⇒
s
=
×
52
7
=
6
5328
Now,
use
the
GDC
t-inter val
stats
input
method
9
and
input
the
given
statistics.
*Unsaved
1.1
“T itle”
“t
Inter val”
“CLower”
48.9131
“CUpper”
56.4869
“x”
*Unsaved
1.1
t
Inter val
x:
52.7
“ME”
3.78694
“df”
9.
Sx:
n:
“SX
:=
Sn–1X”
“n”
6.5328
10.
C
Level:
52.7
6.5328
10
0.9
ok
Cancel
1/99
0/99
μ ∈ [ 48.9,
56.5 ]
Chapter
3
95
Exercise
1
Given
the
inter val
2
3C
values
for
= 15,
the
a
x
b
x
=
− 23,
c
x
=
3478,
Given
the
s
the
mean
= 1.2,
s
=
s
set
x ,
=
A
n
= 15
5.8,
n
429,
of
s,
n,
α ,
and
nd
(1
–
α )%
condence
if:
=
n
data
α
and
32
=
α
and
= 310
nd
0.1;
and
the
(1
0.01;
=
α
=
0.05 .
α )%
–
condence
inter val
for
if:
=
{1,
b
A
=
{0.1, 0.12, 0.15, 0.18, 0.13, 0. 12, 0. 09, 0. 11, 0. 13, 0. 21, 0. 15}
=
3,
Nadim
W
e
5,
6,
7,
8,
9}
and
=
and α
=
.
0.1
buys
12
mandarins
given
in
grams
may
assume
Determine
a
that
99%
(g):
the
and
66,
weighs
70,
75,
mandarins
condence
them
84,
90,
came
interval
for
with
77,
from
the
the
71,
a
following
68,
80,
normal
mean
85,
72,
63.
population.
weight
of
the
mandarins.
4
Eight
independent
population
with
random
the
measurements
unbiased
estimate
of
are
taken
from
population
a
normal
variance
2
s
to
5
=
2.25.
be
Given
[12.345,
the
mean
the
condence
box
a
with
cod
market
at
are
standard
a
Find
bought
b
Find
a
bought
c
value
15
is
market
95%
at
on
condence
is
at
a
sh
market.
Weights
the
mean
is
found
and
the
of
The
cod
mean
bought
unbiased
weight
at
the
estimate
of
the
95 g.
inter val
for
the
mean
weight
of
cod
inter val
for
the
mean
weight
of
cod
market.
market.
the
signicance
inter val.
statistical
for
inter val.
536 g.
condence
the
the
distributed
is
inter val
sample;
of
condence
the
99%
at
the
bought
normally
Comment
Exploring
cod
the
of
condence
nd:
level
deviation
a
the
14.355 ]
b
of
96
that
a
A
=
0
321
, 325, 330, 324, 325, 326, 317, 318, 329, 310, 314, 318, 327, 322, 328
and
results
4,
0.01;
A
A
2,
α
a
c
3
of
mean
analysis
methods
level
and
the
length
of
the
05;
Condence
Matched
interval
pairs
population.
matched
are
In
of
product,
two
satisfy
the
Matched
when
be
in
from
in
one
samples
in
before
and
is
sor t
after
of
the
on
used
same
is
the
We
or
same
might
compare
the
compare
not
take
the
two
product
two
is
samples
means
dierent
of
a
group
of
the
samples
variable against
group
of
measure
treatment,
to
a
circumstances.
would
used
same
whether
whether
to
content
We
medical
the
and
the
production.
dierent
iron
analysis
and
would
determine
An
people
the
same
these
form
example
whether
suer
sets
not
both
of
matched
or
might
group
two
our
who
itself
a
pair.
new
dr ug
eective.
Alter natively
,
of
be
from
know
factories.
factor y
,
of
taken
suppose
determine
condition.
measurements
This
to
pairs
to
example,
each
two
the
want
dierent
also
in
samples
we
standards
can
measuring
blood
two
order
same
pairs
For
from
measured
a
study
,
dier.
manufactured
matched
dierent
our
pairs
for
people
might
they
same
we
who
follow
measure
have
would
Matched
experimental
of
of
a
group
one
are
and
group
(such
the
drug
control
the
group.
This
the
experimental
the
effects
of
the
used
on
a
as
a
lot
gender
,
on
to
the
in
age,
sets
we
We
match
of
weight,
without
the
a
of
study
weight.
diet
to
a
group
We
see
whether
results from
experiments.
might
the
etc.).
effect
control
monitor
and
other
experimental
minimize
which,
two
lose
the
to
the
pair.
elements
the
to
after
biological
group.
analysis
order
and
population,
the
had
and
pairs
in
matched
control
done
group,
a
diet
before
goal,
our
with
has
is
matched
cer tain
their
drug
characteristics
impact
a
use
weights
form
pairs
effect
elements
their
achieved
group
the
might
We
would
group,
of
determine
groups:
compare
group
all
based
then
we
the
on
cer tain
with
inuences
might
an
compare
compared
other
group,
T
o
two
the
on
ascribe
to
dr ug.
Chapter
3
97
Example
A
group
two
of
new
days
iron

patients
dr ugs:
when
tests
the
at
dr ug
a
A
hospital
and
inuence
were
of
conducted
Patient
dr ug
dr ug
and
suer
B.
A
the
from
First,
had
chronic
they
wor n
results
b
c
d
e
deciency
.
treated
they
g/dl )
(in
a
were
o,
iron
with
were
are
f
dr ug
treated
shown
in
g
h
Dr ug
A
60
58
47
80
35
55
53
40
Dr ug
B
63
55
42
76
29
6
5
4
a
Find
b
Calculate
the
dierences
the
90%
between
the
condence
results
inter val
obtained
of
the
by
mean
dr ug
of
A
They
treated
A
and,
after
with
dr ug
B.
the
and
are
following
dr ug
dierences
in
a
with
few
Ser um
table:
B.
par t a
a
d
=
−3
A − B
3
5
4
6
−6
2
−
Subtract
the
results
of
drug
B
from
those
i
of
Method
b
∑
d
d
10
i
⇒
d
=
= 1
n
Calculate
the
mean
of
di erences.
Calculate
the
unbiased
25
8
2
2
∑
d
n
(
estimate
of
the
)
d
136
i
=
⇒
s
d
12
=
5
=
4
200
standard
deviation.
d
n
n
A.
I
=
s
drug
= 8 ⇒
1
= 7
7
⇒
t
= 1
895
Find
the
degrees
of
freedom
and
use
a
c
GDC
to
nd
the
characteristic
value
of
t.
4.200
1.25 ±
s
× 1.895
d
Use
the
f or mula
d
±
×
8
t
c
n
⇒
μ
∈
d
Method
[ − 1.564,
4.064 ]
II
Store
data
Apply
1.1
the
into
lists
and
t-condence
nd
inter val
*Unsaved
1.2
the
“T itle”
“t
di erence
–1.56353
“CUpper”
4.06353
1.1
1.25
*Unsaved
1.2
A
“x”
list.
Inter val”
“CLower”
B
a
C
D
b
2.81353
“ME”
=a[]-b[]
7.
“df”
“SX
:=
Sn–1X”
4.20034
1
60
63
–3
58
55
3
47
42
5
80
76
4
35
29
6
8.
“n”
2
3
1/99
4
5
⇒
μ
∈
d
98
[ − 1.564,
Exploring
4.064 ]
statistical
analysis
C
methods
=a
b
the
di erences.
with
data
in
In
the
and
previous
positive
there
such
is
a
a
reasonable
Given
and
we
can
and
the
we
between
do
90%
condence
cannot
the
another
conclusive
say
at
eects
this
of
dr ugs
hypothesis
results.
You
inter val
level
test
will
of
A
that
lear n
contains
both
condence
and
can
B.
Sometimes
give
more
negative
that
us
about
in
more
this
in
6.
Exercise
1
so
dierence
case
Example
example,
values,
3D
two
nd
sets
the
(1
of
data
–
)%
in
the
following
condence
tables,
inter val
for
calculate
the
mean
the
of
dierences
the
dierences:
a
Set
A
5
7
23
7
8
20
9
6
6
2
Set
B
8
5
23
9
5
8
20
5
8
22
α
=
0
01;
b
Set
C
98
03
02
88
96
05
0
93
02
06
99
85
Set
D
2
5
04
96
0
2
23
02
08
09

03
α
=
0
;
1
c
Set
E
0.55
0.7
0.66
0.58
0.82
0.77
0.9
.02
Set
F
0.62
0.68
0.74
0.69
0.78
0.65
0.8
0.95
α
2
Bob
=
and
levels
of
0
Rick
(Hgb)
that
They
05
are
in
the
laborator y
same
oxygen-transpor t
obtained
Blood
two
sample
the
blood
technicians
samples
of
metalloprotein
following
results
in
is
12
and
they
male
measure
patients.
between
138
and
haemoglobin
The
normal
level
175 g/l.
g/l:

2
3
4
5
6
7
8
9
0

2
Bob
44
53
70
83
25
95
48
77
60
55
70
35
Rick
4
6
73
74
9
04
35
75
64
58
67
42
a
Find
b
Calculate
the
obtained
dierences
a
by
95%
Bob
between
condence
and
the
results
inter val
of
obtained
the
mean
by
of
Bob
and
Rick.
dierences
in
results
Rick.
Chapter
3
99
3.3
Hypothesis
Setting
In
up
order
not
yet
that
W
e
a
and
to
testing
formulate
been
new
wish
proved
dr ug
to
set
to
up
parameter.
The
only
for
and
such
our
In
general,
of
testing
that
the
hypothesis
The
stated
be
test,
ght
test
a
always
the
study
we
are
of
are
be
essential
usually
this
a
a
Higher
mean
topic
to
a
or
us
to
with
we
each
been
a
the
has
a
but
been
current
whether
about
studies.
proposed
claim
this
has
made
dr ug.
claim
is
tr ue.
population
includes
normal
oriented
hypotheses
statistical
than
syllabus
random
be
contradictor y
provides
better
determine
Level
will
discussing
has
stated claim
of
of
suppose
works
with
part
theor y
example,
hypothesis
accepted
is
an
infection
star ts
directly
hypothesis
is
For
population
hypotheses
can
a
tr ue.
Mathematics
when
statements
such
to
and
testing
testing
hypotheses
help
Hypothesis
as
testing
distribution,
thus.
consider
other.
arguments
as
two
The
to
process
why
a
cer tain
rejected.
called
the null
hypothesis
and
we
denote
it
by
H
0
and
the
alternative
hypothesis states
the
opposite
and
is
denoted
by H
1
Let’s
consider
again
the
example
of
broilers
y
from
f(x)
section
3.2.
Suppose
that
data
from
previous
years
x
suggests
that
population
weight
of
the
is
mean
2 kg.
broilers
We
this
weight
wish
of
to
year,
the
broiler
estimate
and
to
do
the
so
Acceptance
mean
we
take
Rejection
a
sample
found
to
of
be
a
cer tain
size.
The
sample
mean
region
null
Rejection
2.0 kg.
x
hypothesis
always
states
no
value
Critical
that
the
mean
weight
of
value
change,
Two-tail
i.e.
region
is
Critical
The
region
test
broilers
y
this
year
is
also
2 kg.
We
write
this
as
(
: μ
H
=
x
0
)
x
Acceptance
The
alter native
statements
to
hypothesis
depending
on
can
the
have
type
of
region
dierent
test
we
wish
perform.
Rejection
There
three
i
are
two
types
of
Two-tail
types
the
test
of
hypothesis
alter native
(H
:
μ
≠
x)
testing
Critical
and
One-tail
hypotheses:
The
mean
weight
is
upper
region
value
x
test
y
1
not
equal
to
x
2 kg.
Acceptance
ii
One-tail
upper
test
more
than
(H
:
μ
>
x)
The
region
mean
1
weight
is
2 kg.
Rejection
iii
One-tail
lower
test
(H
:
μ
<
x)
The
region
mean
1
weight
is
less
than
Critical
2 kg.
One-tail
100
Exploring
statistical
analysis
methods
value
lower
x
test
W
e
must
also
decide
which
level, α,
signicance
we
require
in
order
The
to
conclude
that
a
certain
hypothesis
is
valid,
with
− α )
(
%
most
common
certainty
.
signicant
are
Signicace
we
are
95%
inter val
of
the
level
sure
we
can
in
the
We
will
the
variance
There
the
use
are
two
of
the
otherwise
the
have
null
We
have
value
is
the
value
null
is
condence
in
the
hypothesis
section,
when
If
make
the
the
do
is
is
no
not
the
tr ue.
If
region,
reject
the
we
at
inter val,
calculated
the
5%
so
1%,
5%
and
10%.
if
condence
signicant
level
when
use
of
or
z-
or
we
known,
sample
based
t-value
found
the
of
and
condence
or
t-statistics.
t-statistics
when
size).
upon
t-value
reject
the
either z-statistics
is
the
calculating
the
on
the
null
calculated
the
test
values.
signicance
lies
outside
hypothesis,
it.
the
that
the
parameter
rejection
p-value
accept
sucient
will
decision
calculated
is
the
evidence
region,
g reater
we’re
given
than
the
null-hypothesis,
to
reject’,
or
investigating
that
the
null
signicance
but
simply
rather
‘fail
to
say
level
that
reject’,
hypothesis
four
steps
in
hypothesis
1
State
Step
2
Set
Step
3
Calculate
Step
4
Decide
the
we
in
the
the
testing
calculation
z-statistics
which
a
as
variance
z-value
within
that
we
the
probability
lies
just
(regardless
Step
in
to
statistics,
acceptance
say
Hypothesis
(see
to
mean)
the
As
the
unknown
test.
p-value
cannot
‘we
related
mean
testing
ways
we
hypothesis
we
accept
previous
is
so-called
(i.e.
the
z-statistics
Critical
The
that
calculating
inter vals
level
directly
used
test.
When
The
is
levels
Example
the
upon
for
of
already
and
criteria
μ
for
alter native
a
hypothesis;
decision;
statistics;
calculated
when
is
condence
hypothesis
have
null
testing:
testing.
the
and
decision
criteria.
known
inter vals,
Let’s
calculated
statistics
look
at
we
an
are
going
example
condence
to
use
for
inter val
4).
Chapter
3
101
Example
In
a

cer tain
and
the
countr y
standard
population
a
State
b
Use
a
H
and
the
a
is
believed
deviation
the
null
it
mean
and
two-tail
test
is
that
5 cm.
height
A
was
the
mean
random
found
to
alter native
hypotheses;
at
signicance
the
0%
:
“The
mean
height
is
82 cm”
:
“The
mean
height
is
not
( μ
=
height
sample
be
level
82)
of
of
the
male
00
men
population
was
taken
is
82 cm
from
the
83.6 cm.
to
decide
Step
1:
whether
State
the
or
null
not
the
claim
hypothesis
is
tr ue.
that
0
conr ms
H
the
claim.
State
alter native
82 cm”

hypothesis.
(μ
≠ 182 )
Method
b
I
5
μ
=
182, σ
=
=
0.5,
x
= 183.6
Step
2:
Calculate
Step
3:
Find
Step
4:
Compare
the
z-value.
100
x
z
−
μ
183.6
=
⇒
z
− 182
=
=
σ
3.2
0.5
Either
1
α
=
0.1 ⇒
z
=
Φ
( 0.05)
= 1.645
the
z-critical
value.
α
2
3.2
> 1.645 ⇒
z
>
z
the
z-critical
value
with
the
α
2
We
reject
the
null
calculated
value
and
make
a
decision.
hypothesis.
Or
P ((180.4
>
X )
or
(X
> 183.6 )
μ
Step
= 182 )
3:
For
the
cor responding
= 1 − P (180.4
 1  P ( 3.2
<

0.0037 < 0.
the
0%
Method
X
Z
< 183.6
 3.2 )
we

reject
signicant
μ
z
value
nd
the
= 182 )
0.00137
the
null
hypothesis
at
level.
Step
4:
Compare
signicant
Use
II
z-test
1.1
1.1
calculated
p-value.
level
the
and
statistics
p-value
make
f eature
a
with
the
decision.
on
a
GDC
*Unsaved
1.2
*Unsaved
1.2
z Test
zTest
182, 5, 183.6, 100, 0:
stat.results
µ0:
“T itle”
σ
“
Alternate
182
“z Test”
Hyp”
“µ
“z”
≠
“PVal”
5
µ0”
3.2
x:
183.6
0.001374
n:
“x”
|
100
183.6
“n”
100.
σ
5.
Alternate
Hyp:
Ha:
µ
≠
µ0
ok
1/99
0.0037
<
0.
we
reject
0/99
the
null
hypothesis.
Compare
level
102
Exploring
statistical
analysis
Cancel
methods
and
the
p-value
make
a
with
decision.
the
signicance
Due
to
the
symmetrical
acceptance
impor tant
We

region
found
inter val.
Y
our
is
it
the
to
We
we
null
0.0037
Let’s
be
of
test,
condence
[82.8,
therefore
gives
simpler
to
the
this
condence
example
inter val
84.4]
and
expected
you
make
compare
signicance
the
proper ties
two-tail
inter val
highlights
and
the
some
the
a
to
for
we
the
see
reject
mean
that
the
null
calculated z-value
decision
based
82
upon
of
is
the
not
in
this
hypothesis.
“z”
=
3.2,
but
the p-value
it
when
available.
When

90%
calculator
much
is
a
results:
population

in
use
levels
the
(%,
hypothesis
is
smaller
another
p-value
5%,
upon
than
example
and
all
each
0%),
three
each
for
with
and
we
the
we
common
conclude
signicance
ever y
which
of
one
have
of
that
levels,
we
reject
since
them.
already
calculated
the
We
condence
the
inter val
Example
After
a
worms
2.0,
and
a
State
b
Use
is
.,
from
the
the
a
in
Example
5
2
worms
0.5,
a
0.8,
null
and
deviation
test
at
0.4,
that
alter native
upper
surfaced
2.,
population
standard
one-tail
have
0.9,
follows
of
on
2 cm.
a
sand.
2.2,
Their
0.9,
normal
There
is
a
lengths,
.9,
.2,
distribution
belief
measured
that
.6.
with
the
It
the
in
is
cm,
known
mean
worms
were
are
as
that
value
the
of
growing
larger.
hypotheses;
the
5%
signicance
level
to
decide
whether
or
not
the
claim
tr ue.
H
a
night,
came
0.5 cm
condence

rainy
follows:
calculated
inter val.
:
“The
mean
length
is
0.5 cm”
:
“The
mean
length
is
larger
(
μ
= 10
5)
Step
1:
State
the
null
hypothesis
that
0
conr ms
H
the
claim.
State
the
alter native
than

hypothesis.
0.5 cm” ( μ
Method
b
> 10
5)
I
2
μ
= 10.5, σ
=
=
0.577,
x
=
11.3
12
μ
x
z
11.3
=
⇒
z
10.5
=
=
σ
1.38564
Step
2:
Calculate
Step
3:
Find
Step
4:
Compare
the
z-value.
0.577
Either
1
α
=
0.05 ⇒
z
= Φ
(
c
1.38564
< 1.645 ⇒
z
0.05 ) = 1.645
<
z
the
z-critical
the
value.
z-critical
value
with
c
the
We
have
no
sucient
evidence
to
reject
the
calculated
value
and
make
a
decision.
null
hypothesis.
Or
P (x
> 11.3
μ
=
10.5)
= P
(
z
> 1.38564
)
Step
3:
For
the
cor responding
=
calculated
z-value
nd
the
p-value.
0.0829
Chapter
3
103
Since
to
0.0829
reject
the
>
0.05
null
we
have
hypothesis
no
at
sucient
the
5%
Step
evidence
4:
Compare
signicance
signicance
level
the
p-value
and
make
with
a
the
decision.
level.
Use
Method
1.1
II
the
z-test
data
on
a
GDC
to
nd
*Unsaved
*Unsaved
1.2
f eature
p-value.
z Test
“T itle”
“z Test”
µ0:
“
Alternate
Hyp”
“µ
“z”
>
10.5
µ0”
σ
1.38564
“PVal”
0.082928
“x”
11.3
2
List:
Frequency
“SX
:=
a
List:
1
0.636753
Sn–1X”
12.
“n”
Alternate
Ha:
Hyp:
µ
>
µ0
2.
σ
ok
Cancel
1/99
0/99
0.0829
>
0.05
we
have
no
sucient
evidence
to
Compare
reject
the
null
hypothesis
at
the
5%
signicance
level
Condence
through
inter val
region
When
one-tail
is
for
we
two
upon
levels
the
Exercise
is
the
is
5%,
we
and
0.0829
>
or
to
the
lower),
whereas
Example
0%)
level
no
0.05
compared
(upper
p-value
an
and
make
results
since
a
a
believed
that
with
a
random
the
mean
obtained
condence
acce ptance
or
rejection
it’s
3
clear
(0.0829
sucient
and
with
0.0829
the
that
<
we
0.),
reject
whilst
evidence
>
three
to
common
the
upon
reject
null
the
the
hypothesis
remaining
null
0.0).
sample
μ
value
of
n
and
obser vations
the
standard
is
taken
from
deviation
σ
the
.
0
The
sample
: μ
H
=
μ
0
=
the
: μ
H
0
n
a
,
has
mean
μ
≠
1
μ
20,
value
test
the
of
−
x .
Given
claim
at
the
2
α
=
the
α
hypotheses
signicance
level
if:
0
=
10,
x
=
12, σ
=
and
0.1;
0
n
b
=
μ
25,
=
2,
= 1.9, σ
x
=
0.3
and
α
=
0.05;
0
n
c
=
μ
119,
=
−235,
x
=
σ
−238,
= 12.8
and
α
=
0.01
0
2
It
is
believed
population
that
with
a
random
the
mean
sample
μ
value
of
n
and
obser vations
the
standard
is
taken
from
deviation
σ
.
0
The
sample
: μ
H
=
μ
0
a
has
, H
0
n
=
20,
the
: μ
<
mean
μ
1
μ
value
test
the
of
−
x .
claim
Given
at
the
the
α
hypotheses
signicance
0
= 10,
x
=
9,
σ
=
4
and
α
=
0.1;
0
b
n
=
50,
μ
=
21.4,
x
=
21.2,
σ
=
0.75
and
α
=
0.01;
0
c
n
=
119,
μ
=
−235,
x
=
− 238 ,
0
104
Exploring
statistical
analysis
the
decision.
3E
population
with
symmetric.
in
have
be
mean,
not
signicance
levels
(since
the
p-value
(%,
0%
cannot
testing
about
testing
compare
signicance
It
hypothesis
one-tail
hypothesis
1
analysis
symmetric
signicance
only
inter val
the
level.
methods
σ
=
12.8
and
α
=
0.01.
level
if:
the
signicance
3
It
is
the
believed
that
population
a
random
with
the
sample
mean
of
n
μ
value
obser vations
and
the
is
taken
standard
from
deviation
0
.
The
: μ
H
sample
=
μ
0
a
,
=
the
μ
>
1
μ
20,
: μ
H
0
n
has
mean
test
value
the
−
x .
of
claim
at
Given
the
the
hypotheses
signicance
α
level
if:
0
= 10,
= 12, σ
x
= 5
α
and
=
0.05 ;
0
b
n
=
μ
40,
=
27.3,
x
−73,
x
28.0, σ
=
= 3.6
and
α
=
0.1;
0
c
n
=
μ
92,
=
− 71.6, σ
=
= 3.72
α
and
=
0.01
0
4
After
a
Their
25,
rainy
weights,
30,
came
35,
27,
from
a
mean
value
belief
that
a
State
b
Use
the
an
A
eight
1.35,
is
6
is
State
b
T
est
An
the
the
not
the
a
of
the
following
288,
293,
301,
298,
299,
289,
304,
is
a
State
b
Test
a
belief
the
at
Hypothesis
null
the
are
at
is
level
A
box
(in
302,
20
ml)
290,
a
the
alter native
for
μ
fat
with
25,
forest.
31,
snails
35,
in
of
that
3.5 g.
28,
the
forest
with
There
38,
is
the
a
population.
level
to
decide
in
bottles
100
of
a
of
0.3 g.
were
is
ml
a
In
lactose
a
sample
measured:
belief
more
that
free
1.43,
the
of
1.52,
company
fat.
hypotheses.
whether
apple
or
juice.
volume
bottles
288,
signicance
from
the
distribution
signicance
fat
were
and
testing
of
288,
that
10%
of
level
of
that
normal
not
1%
pure
285,
302,
a
22,
deviation
There
drink
have
known
deviation
levels
100%
follows:
in
tr ue.
alter native
claim
is
har vested
hypotheses.
the
1.55.
free
volumes
295,
There
1.45,
It
as
follows
signicance
7.3 ml.
28.
standard
lactose
they
were
snails
the
been
standard
claim
produces
that
that
the
following
5%
g,
33,
test
that
and
have
alter native
the
1.46,
null
the
deviation
and
and
in
34,
har vested
with
eco-farm
bottles
26 g
1.42,
at
30,
claims
producing
a
in
or
drinks
1.38,
29,
appropriate
1.4 g
snails
measured
null
company
drink
15
population
of
the
whether
5
period,
of
is
not
the
They
300 ml
taken
claim
package
and
for
is
a
true.
juice
standard
inspection
and
measured:
290,
300,
305,
300,
293,
contain
less
305.
volume
than
stated.
hypotheses.
level
when
whether
is
or
not
the
claim
is
tr ue.
unknown
We
As
the
with
calculations
standard
t-statistics,
of
deviation
irrespective
condence
when
of
the
inter vals,
hypothesis
sample
if
testing
size.
we
we
do
not
use
know
have
found
condence
this
inter val
instance
Example
already
the
for
in
7.
Chapter
3
105
Example
A

chocolate
In
a
sample
of
ve
2540 kJ,
be
x
a
State
b
Test
a
H
=
company
the
at
00 g
whilst
null
the
claims
and
%
that
the
chocolates
the
estimate
alter native
signicance
energy
the
for
level
mean
in
value
population
cer tain
of
the
00 g
energy
standard
chocolates
level
deviation
was
was
s
is
2500 kJ.
found
to
= 120 kJ.
hypotheses;
level
whether
:
“The
energy
level
is
2500 kJ”
:
“The
energy
level
is
not
(μ
or
not
the
= 2500 )
claim
Step
1:
is
tr ue.
State
the
null
hypothesis
that
0
conr ms
H
2500 kJ”
(μ
≠
the
claim.
State
the
alter native
2500 )

hypothesis
Method
b
x
t
a
two-tail
test.
I
μ
−
f or
2540
=
⇒
t
− 2500
=
=
s
0.7454
120
n
Step
2:
Calculate
Step
3:
Find
Step
4:
Compare
the
t-value.
5
Either
n
=
5
⇒

=
4, 
=
0. 01 ⇒
t
=
4. 604
the
t-critical
value.
c
0.7454
<
4.604
⇒ t
< t
c
the
We
have
no
hypothesis
sucient
at
the
X )
or
%
evidence
to
signicance
reject
the
calculated
the
value
t-critical
and
value
make
a
with
decision.
null
level.
Or
P (( 2460
>
(X
>
μ
2540 )
=
2500 )
Step
3:
For
the
cor responding
=
1 − P ( 2460
<
x
<
 1  P ( 0.74536 
0.497
>
reject
the
Method
0.0
we
null
2540
t

have
μ
=
hypothesis

the
evidence
%
to
signicant
Step
level.
4:
nd
the
p-value.
Compare
signicant
Use
II
t-test
level
stat
the
and
p-value
make
f eature
on
a
a
with
the
decision.
GDC.
*Unsaved
1.1
*Unsaved
1.1
value
0.497
sucient
at
t
2500 )
0.74536 )
no
calculated
t Test
“T itle”
“
Alternate
“t Test”
Hyp”
“t”
“µ
≠
µ0”
µ0:
2500
x:
2540
0.745356
0.497471
“PVal”
Sx:
“df”
4.
“x”
2540.
n:
“SX
:=
Sn–1X”
5.
“n”
120
|
5
120.
Alternate
Hyp:
Ha:
µ
≠
µ0
ok
Cancel
1/99
0/99
0.497
>
0.0
reject
the
we
null
have
no
hypothesis
sucient
at
the
evidence
%
to
signicant
level.
Compare
the
signicance
106
Exploring
statistical
analysis
methods
p-value
level
and
with
the
make
a
decision.
Due
to
the
symmetrical
acceptance
important
found
00 g
the
this
evidence
Y
our
but
a
properties
two-tail
90%
chocolates
within
●
in
of
test,
the
this
condence
example
interval
highlights
and
the
some
results:
We
●
region
is
to
be
inter val.
to
reject
calculator
it
condence
much
[2293,
inter val
2787]
Therefore,
the
null
gives
you
simpler
to
we
for
and
do
the
we
not
energy
see
have
that
level
2500
in
lies
enough
hypothesis.
the
calculated t-value
make
a
decision
based
“t”
=
upon
0.745,
the
p-value.
When
●
we
compare
signicance
levels
the
p-value
(%,
5%,
with
and
each
0%),
of
we
the
common
conclude
that
We
we
the
have
no
evidence
to
reject
the
null
hypothesis
upon
of
these
three
signicance
levels
since
0.497
is
for
we
studied
Example
Example
a
box
9.

of
six
66.
eggs,
65,
62,
66,
the
5%
signicance
H
problem
each.
when
In
the
larger
following
than
condence
ever y
inter val
one
calculated
The
the
weights
farmer
level
(measured
claims
whether
:
“The
mean
weight
is
66 g”
:
“The
mean
weight
is
less
(μ
that
the
=
the
in
grams)
mean
eggs
in
of
weight
the
boxes
of
1:
egg
one
have
Step
66 )
each
a
of
were
their
mean
State
the
as
follows:
eggs
weight
null
is
62,
66 g.
less
Test
than
hypothesis
63,
at
66 g.
that
0
H
than
66 g”
(μ
< 66 )
conr ms
the
claim.
State
alter native

hypothesis
Method
x
=
t
=
64,
x
−
f or
a
one-tail
lower
test.
I
s
= 1.897
μ
64
⇒
t
− 66
=
=
s
Step
2:
Calculate
Step
3:
Find
Step
4:
Compare
the
t-value.
− 2.582
1.897
n
6
Either
n
=
6 ⇒ν
= 5,α
=
0. 05 ⇒
t
=
−2. 015
c
−2.582
<
−2.015 ⇒ t
< t
c
We
reject
the
signicant
null
hypothesis
at
the
5%
calculated
the
level.
the
value.
acceptance
t-critical
the
value.
t-critical
Notice
region
so
that
we
value
t
lies
make
with
the
outside
a
decision.
Or
P (x
<
64
μ
=
Step
66 )
3:
For
the
cor responding
P ( −2.582
0.02466
<
<
hypothesis
t )
=
0.05
at
calculated
t-value
nd
the
p-value.
0.02466
therefore
the
5%
we
reject
signicant
the
level.
null
Step
4:
Compare
signicance
level
the
p-value
and
make
with
a
the
decision.
Chapter
3
107
Method
1.1
II
Use
*Unsaved
1.2
“T itle”
“
Alternate
the
“t Test”
Hyp”
“µ
“t”
<
t-test
stat
on
a
GDC
p-value.
µ0”
–2.58199
1.1
“PVal”
f eature
*Unsaved
1.2
0.024657
t Test
“SX
“df”
5.
“x”
64.
:=
µ0:
1.89737
Sn–1X”
List:
66
a
6.
“n”
Frequency
Alternate
List:
Hyp:
1
Ha:
µ
<
µ0
1/99
ok
Cancel
0/99
0.024657
<
hypothesis
5%
As
0.05
that
the
signicance
with
cannot
symmetric
about
1
A
(upper
the
one-tail
Exercise
is
the
null
66 g
inter val
compared
t-testing
for
reject
weight
condence
be
hypothesis
region
mean
we
at
the
level.
z-testing,
variable
therefore
to
or
mean,
t-testing
the
lower),
whereas
is
not
analysis
results
since
an
for
a t-distributed
obtained
a
through
condence
acceptance
or
one-tail
inter val
is
rejection
symmetric.
3F
random
sample
population
with
of
the
n
obser vations
mean
value
μ
.
is
taken
The
from
sample
the
has
the
mean
0
value
x
of
deviation
and
s.
an
unbiased
Given
the
estimate
of
: μ
H
hypotheses
the
population
μ
=
0
claim
n
a
at
=
the
μ
10,
signicance
α
= 5,
x
=
4.8,
level
s
,
H
0
: μ
≠
standard
μ
1
test
0
if:
= 1.3
α
and
=
0.01;
0
n
b
=
μ
25,
=
2,
x
= 1.9,
s
=
0.3
α
and
=
0.1;
0
n
c
μ
= 7,
=
−36,
x
=
−35.3,
s
=
0.523
α
and
=
0.05
0
2
It
is
believed
from
the
that
a
random
population
with
sample
the
of
mean
n
obser vations
value
μ
.
The
is
taken
sample
0
has
a
mean
population
: μ
H
μ
=
0
n
x
of
standard
,
H
0
level
a
value
: μ
the
deviation
μ
<
and
1
test
unbiased
is
the
s.
estimate
Given
claim
at
the
the
α
of
hypotheses
signicance
0
if:
= 30,
μ
= 15,
x
= 14.2,
s
=
2.2
and
α
=
0.05 ;
0
b
n
= 10,
μ
= 122,
x
= 119.8,
s
=
2.32
and
α
=
0.01;
0
c
n
=
6,
μ
=
627,
x
=
622.8,
s
= 12.6
0
108
Exploring
statistical
analysis
methods
and
α
the
=
0 .1
.
the
to
nd
It
3
is
the
believed
that
population
a
random
with
the
sample
mean
of
n
obser vations
value
.
The
estimate
of
sample
is
taken
has
the
from
mean
0
value
x
of
deviation
and
s.
the
Given
unbiased
the
: μ
H
hypotheses
the
=
population
μ
0
claim
n
a
at
=
the
μ
20,
signicance
α
= 1
,
level
,
: μ
H
0
standard
μ
>
1
test
the
0
if:
x
=
0.95,
s
=
0.335
x
=
26.4,
s
= 1.12
and
α
=
0. 1
;
0
n
b
μ
= 8,
=
25,
α
and
=
0.05
;
0
n
c
μ
= 15,
x
= 754,
= 758.6,
s
= 14.2
and
α
=
0.01
0
An
4
ice-cream
par ticular
in
a
factor y
ice-cream
box
and
119,
123,
121,
and
test
their
120,
claims
that
product
volumes
118,
116,
at
the
1%
adver tised
the
correct
is
in
ml
123,
signicance
the
average
120 ml.
are
122.
There
as
are
of
eight
a
ice-creams
follows:
State
level
volume
the
whether
hypotheses
or
not
the
factor y
volume.
In
A
5
manufacturer
claims
that
the
life
expectancy
of
some
used
LED
lamp
is
30,000
hours.
A
random
sample
of
lamps
29,500
is
tested
28,350
and
the
30,300
following
30,250
data
is
29,350
the
hypotheses
and
test
at
the
10%
ever y
making
whether
or
not
the
manufacturer
29,600
easier
to
read.
space
is
used
expectancy
Signicance
In
Example
the
illustrate
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a
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6
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some
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six
for
LED
countries
an
B.
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a
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ser um
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there
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dr ugs.
Chapter
3
109
Subtract
d
a
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−3
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3
5
4
6
−6
2
the
results
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drug
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−
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into
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level
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Exploring
statistical
analysis
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and
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the
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In
Example
contains
could
the
the
both
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say
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more
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The
saw
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at
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table
players
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D
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22
35
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and
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of
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whether
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F
the
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cubes.
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level
solve
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signicance
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after
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or
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Test
of
whether
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and
taking
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information
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1
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table
members
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level
whether
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3
111
3.4
In
Type
this
make
that
in
of
the
say
We
Type
are
II
we
discuss
we
can
statements:
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testing
hypothesis
as
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statistical
null
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them
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●
and
section
ever y
think
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by
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noting
values
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fail
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hypothesis
make
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Type
II
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reject
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tr ue
null
hypothesis
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probability
making
A
real-life
example
of
these
two
types
of
errors
can
be
seen
in
trial.
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other
times
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innocent
person
is
convicted,
the
area
guilty
person
walks
free.
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a
democracy
,
as
presumed
innocent
at
the
beginning
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trial
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region
by
we
say
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ype
II
denoted
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I
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Null
hypothesis
is
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based
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hypothesis
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hypothesis
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collected
data
decision
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In
medicine,
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health
a
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desirable
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of
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statistical
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I
the
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other
nothing
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to
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with
that
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II
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health,
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negative
their
is
means
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er ror
lack
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er ror
positive
means
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infection
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of
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ascer tain
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patient
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conclude
type
testing .
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but
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due
hand,
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negative
which
the
test
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have
body,
example
er ror.
analysis
false
whilst
a
vir us.
patient’
s
tests,
statistical
positive
does
eventually
thus
of
perfor m
wondering
the
it
can
we
cer tain
medical
that
lot
patient
different
health.
be
are
a
a
false
vir us.
in
wor r y,
assurance
patient.
the
the
this
to
results
with
that
by
as
example
presence
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perfor m
known
For
infected
patient
not
than
Exploring
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vir us.
their
false
is
infected
see
with
er ror
shown
viral
the
cause
patient
the
not
can
combined
is
has
shown
infected
I
negative.
patient
actually
hasn’t
researcher s
Type
whether
are
112
a
false
of
making
error
by
β.
is
nothing
give
and
for
a
since
the
is less
of
1
–
β
The
is
called
error.
the
on
is
The
error;
value
●
α.
that:
a
●
error
null
probability
hypothesis),
I
the
ever yone
denoted
is
of
whilst
rejection
a
Type
a
or
cour t
a
of
power
of
the
test.
Suppose
we
Our
hypothesis
die
null
roll
is
biased.
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=
the
0.026),
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●
it
Type
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the
A
(i.e.
II
the
There
are
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is
not
fair,
roll
and
obtaining
occur
is
conclude
might
when
tr ue),
that
occur
hypothesis
and
do
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a
six.
the
alter native
sixes
in
20
hypothesis
rolls
of
the
die
is
is
that
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fairly
small
possible.
hypothesis
and
and
the
might
error
hypothesis
Example
times
cer tainly
null
Type
is
error
hypothesis
●
20
probability
is
I
die
is
conclude
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but
the
that
we
die
when
false),
die
but
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the
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biased.
die
we
die
actually
reject
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is
is
indeed
accept
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biased
null
fair.

two
coins.
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coin
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coin
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2
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coin
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p
.
=
W
e
take
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it
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3
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probability
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hypothesis
number
the
biased
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is
H
of
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variable Y
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coin.
tables
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coin
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coins.
fair”
whilst
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alternative
hypothesis
H
0
selected
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coin
is
W
e
decide
that
we
are
going
to
reject
the
null
hypothesis
Find
the
probability
of
obtaining
a
Type
I
c
Find
the
probability
of
obtaining
a
Type
II
a
Fair
we
“the
obtain
1
⎛ 4
⎞
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x
0
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2
3
4
1
4
6
4
1
the
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P
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X
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x
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16
16
16
16
16
0
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2
3
4
1
8
24
32
16
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⎜
⎟⎜
x
⎝
⎛
⎝
⎠
1
2
=
y
the
=
binomial
81
81
81
P
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X
=
4
H
0
)
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⎠
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≠
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1
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)
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er ror
hypothesis
that
for
=
⎟
⎠
2
x
16
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we
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instance
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⎜
⎟⎜
y
is
⎠
⎛ 1
⎝ 3
y
⎞
y
⎛
⎟
⎜
⎠
⎝
when
we
2
y
⎞
=
⎟
3
⎠
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⎜
⎟
⎝
reject
the
y
⎠
4
3
null
have
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when
it
is
actually
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er ror
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when
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f ail
to
reject
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null
81
hypothesis
event,
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65
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81
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1
=
16
=
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16
β
y ) =
81
1
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=
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81
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x
⎛
4
P (Y
P(Y
x
⎞
coin:
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Y
c
if
“heads”.
b
b
is
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biased”.
decided
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to
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“heads”.
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hypothesis
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is
for
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a
dierent
probability
of
1
making
a
Type
I
error
will
remain
the
same
P (X
=
0
H
)
=
,
but
the
probability
0
16
1
of
making
a
Type
II
error
will
increase,
P (Y
≠
0
H
)
=
1−
80
=
65
,
>
as
expected.
1
81
81
81
Chapter
3
113
Since
than
test
the
it
is
probability
in
where
better
test
the
when
that
parameter
hypothesis
Find
b
It
c
If
a
we
of
α
=
II
β
P (X
=
II
no
hypothesis
the
null
error
is
“heads”,
when
we
~
the
Po(30)
P (X
and
distribution
all
the
for
≥ 36
the
of
the
x
a
we
is
values
Type
actual
hypothesis
I
test
this
greater
less
can
after
example
conclude
four
heads
obtaining
7
that
is
no
the
a
heads.
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than
than
or
hypothesis
30.
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against
acceptance
to
the
hypothesis
region
for
the
that
null
35.
of
the
parameter
was
40.
Find
the
probability
of
m
Type
I
region
and
II
to
all
errors.
m
=
values
What
: X
H
= 30 )
m
the x
can
~
less
you
than
or
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38,
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conclude?
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0
= 30 )
40 )
=
=
Use
0.157
H
0.242
a
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~
to
nd
the
probability.
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1
≥ 39
= 1 − P (X
β ′ =
are
in
error;
value
acceptance
both
≤ 35
≤ 35
α ′ = P (X
We
we
obtain
Use
c
smaller
error;
= 1 − P( X
b
X
that
expand
probabilities
a
Type
obtaining
reject
probability
found
Type
of
null
we
contains
the
was
the
a
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assume
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making
instance
reject
than
Example
Let’s
the
we
of
P (X
notice
m
≤ 38
≤ 38
that
connected,
m
m
=
Type
so
a
GDC
to
nd
all
the
probabilities.
= 30 )
= 30 )
40 )
I
=
and
when
=
Scratchpad
0.0648
0.416
Type
we
II
errors
decrease
1–poissCdf(30, 0, 35)
0.157383
poissCdf(40, 0, 35)
0.242414
1–poissCdf(30, 0, 38)
0.064844
poissCdf(40, 0, 38)
0.416024
a
4/99
Type
I
error
we
increase
a
Type
II
error.
α
Example
> α ′ ⇒
β
<
β ′
9
1
Let’s
assume
that
X
∼
B
and
100,
we
test
this
null
hypothesis
against
the
hypothesis
4
1
that
the
probability
p
≠
.
The
acceptance
region
for
the
null
hypothesis
contains
all
the x
4
values
that
a
Find
b
It
are
E(X )
within
and
the
8
of
the
expected
probability
of
a
value
Type
I
of
X
error;
2
was
found
that
the
actual
value
of
the
probability
was
p
=
.
Find
the
probability
of
5
a
c
If
Type
we
114
error;
expand
within
What
II
6
of
can
Exploring
the
the
you
acceptance
expected
region
value,
conclude?
statistical
analysis
methods
nd
of
the
the
null
hypothesis
probabilities
for
to
both
all
the x
Type
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values
and
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that
are
errors.
1
E
a
(
)
X
= 100
×
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= 25
the
f or mula
f or
binomial
4
distribution
E(X)
=
np
1
α
=
P
( X
≤ 16
or
)
(X
≥ 34
p
)
=
1
4
H
:
X
∼
B
100 ,
0
4
1
⎛
=
1 − P
17
≤
X
≤
33
p
=
⎜
=
P
Use
a
GDC
H
X
to
nd
17
≤
X
≤
33
p
probability.
2
=
=
⎜
the
⎠
2 ⎞
⎛
β
0.0487
⎟
4
⎝
b
⎞
=
0.0913
⎟
:
∼
B
100 ,
1
5
⎝
⎠
5
Use
1
c
α ′ =
P
(
X
≤
18 )
or
(
X
≥
32
p
)
a
GDC
to
nd
all
the
probabilities.
=
4
Scratchpad
1
⎛
= 1 − P
19
≤
X
≤
31
p
=
⎜
19
≤
X
≤
31
p
=
5
0.091253
0.13236
0.0398
⎟
⎝
0.048705
binomCdf(100, 0.4, 17, 33)
1–binomCdf(100, 0.25, 19, 31)
=
⎜
1–binomCdf(100, 0.25, 17, 33)
⎠
2 ⎞
⎛
P
0.132
⎟
4
⎝
β ′ =
⎞
=
binomCdf(100, 0.4, 19, 31)
0.039846
⎠
4/99
Again
errors
we
notice
are
that
Type
connected,
but
I
and
this
Type
time
II
when
we
α
increase
Type
The
II
a
Type
I
error
we
decrease
< α ′ ⇒
β
>
β ′
a
error.
calculations
we
performed
in
Examples
8
&
9
can
be
seen
Despite
visually
in
the
following
two
normal
distribution
the
not
being
normal
Examples
Normal
distribution
graph
for
a
one-tail
distributions
graphs.
can
test
still
both
18
&
in
19,
we
approximate
binomial
and
y
H
Poisson
H
0
distributions
by
1
a
Critical
normal
distribution.
value
x
0
1
2
3
β
The
●
diagram
to
the
demonstrates
right,
increase
●
to
the
α
Type
left,
decrease
we
we
that
decrease
II
II
moving
Type
I
error
the
critical
but
at
the
value
same
ver tical
time
line
we
error;
increase
Type
by
Type
I
error
but
at
the
same
time
we
error.
Chapter
3
115
Two-tail
test
y
H
H
0
1
Critical
value
Critical
value
0
x
1
α
α
β
2
Again,
this
vertical
critical
that
diagram
lines
value
the
2
demonstrates
simultaneously
line
critical
to
the
values
left,
are
that
(notice
the
by
that
other
moving
when
one
symmetric
is
about
the
we
move
moved
)
H
critical
the
to
one
the
value
vertical
right,
following
so
occurs:
0
●
When
Type
●
II
When
Type
Exercise
1
We
we
Type
I
error,
we
simultaneously
increase
error;
we
II
decrease
increase
Type
I
error,
we
simultaneously
decrease
error.
3H
believe
that
a
normal
random
variable, X,
is
distributed
2
such
that
X
hypothesis
null
~
N(5,
that
hypothesis
a
Find
b
It
the
was
Find
0.4
the
is
).
mean
{4.2
≤
probability
found
the
We
that
X
of
the
probability
μ
test
≠
≤
5.
The
null
hypothesis
acceptance
against
region
for
the
the
5.8}
Type
actual
of
this
I
error.
mean
Type
II
value
was
μ
=
4.5.
error.
1
2
Let’s
assume
that
X
∼
B
and
50,
we
test
this
null
2
1
hypothesis
against
the
hypothesis
that
the
probability
p
.
>
2
The
acceptance
a
Find
b
It
E(X )
was
region
and
found
the
that
for
the
null
probability
the
actual
hypothesis
of
Type
value
of
I
the
is
{X
≤
30}.
error.
probability
4
was
p
.
=
Find
the
probability
of
Type
II
error.
7
3
Let’s
assume
against
less
is
{
the
than
X
X
∼ Po
hypothesis
45.
The
(
that
45
)
the
acceptance
and
we
test
parameter
region
for
this
of
the
null
the
null
hypothesis
distribution
is
hypothesis
}
.
≤ 52
a
Find
b
It
the
was
Find
116
that
Exploring
probability
found
the
that
the
probability
statistical
of
actual
of
analysis
Type
I
error.
value
Type
methods
II
of
the
error.
parameter
was
40.
Review
exercise
EXAM-STYLE
1
A
box
of
a
of
20
salmon
salmon
standard
Find
a
QUESTIONS
was
deviation
the
95%
produced
Find
b
the
at
Comment
c
2
In
a
by
medical
two
level
The
in
table
mg/dl,
of
condence
he
a
sh
and
farm.
the
produces
The
farm
is
mean
owner
weight
states
that
the
114 g.
inter val
for
the
mean
weight
of
salmon
inter val
for
the
mean
weight
of
salmon
farm.
the
signicance
the
level
biochemical
blood
10
from
652 g
level
and
the
width
of
the
inter val.
below
for
be
salmon
laborator y
types
the
to
farm.
the
on
condence
of
the
at
bought
condence
99%
produced
is
found
of
a
lists
potassium
analyzers.
healthy
the
of
person
measured
The
is
in
range
between
levels
blood
of
of
is
the
270
potassium
and
potassium,
measured
390
given
mg/dl.
in
patients.
Patient
A
B
C
D
E
F
G
H
I
J
Analyzer
I
235
352
40
280
34
325
428
388
272
30
Analyzer
II
237
343
46
272
336
329
43
396
265
35
Find
a
the
dierences
between
the
results
obtained
by
the
two
analyzers.
State
b
or
the
not
hypothesis
there
is
biochemical
3
Fifteen
a
and
test
dierence
in
at
a
1%
signicance
measurement
of
level
the
whether
two
types
of
analyzers.
independent
obser vations
of
a
random
sample
are
15
taken
from
a
normal
15
2
population.
The
sample
gave
the
results
∑
x
=
80
and
∑
i
i =1
Calculate
a
given
4
Find
c
Inter pret
be
unbiased
estimates
of
the
=
488.
i
i =1
mean
and
variance
for
the
obser vations.
b
The
the
x
a
99%
condence
the
meaning
measurement
assumed
to
of
six
follow
a
inter val
of
the
for
independent
normal
the
population
condence
inter val
random
distribution
mean.
at
the
given
measurements
with
the
mean
level.
may
value
2
μ
and
for
the
the
variance
mean
is
σ
found
a
the
mean
value
b
the
condence
of
=
to
25.
be
the
level
of
Given
[47.2,
that
55.2],
the
condence
inter val
nd:
sample;
the
inter val.
Chapter
3
117
5
An
automotive
their
speedometer
driving.
tested
set
at
If
a
A
on
the
par t
1
claim
31.3,
auto
State
The
exact
of
straight
tr ue,
ten
1 km
speed
cars
company
at
was
racing
which
taken
track
claims
the
and
with
car
they
the
that
is
were
autopilot
to
seconds
31.4,
would
were
30.9,
unbiased
claims
up
the
hypothesis
measurement
assumed
long
it
take
each
car
to
travel
and
estimate
as
follows:
31.2,
30.8,
30.4,
30.5.
of
the
mean
and
variance
of
the
times.
sets
the
in
30.3,
the
how
track?
magazine
deliberately
6
is
32.1,
measured
c
a
the
sample
measured
Calculate
An
of
kilometer
times
30.8,
b
shows
random
manufacturing
120 km/h.
the
The
instr ument
follow
of
a
n
that
due
to
speedometers
and
test
the
to
reasons
show
claim
independent
normal
safety
at
random
distribution
a
the
higher
the
5%
company
speed.
signicance
measurements
with
the
unbiased
may
level.
be
estimate
2
of
population
for
7
the
mean
is
a
the
mean
b
the
sample
A
radar
bicycle
The
found
value
size
records
lane.
results
s
=
to
be
variance
[204,
216]
that
the
95%
condence
inter val
nd:
sample;
speed,
speed
150
Speed
the
Given
n
the
The
for
of
144.
of
v,
kilometres
these
bicycles
Number
in
are
bicycles
recorded
is
in
per
hour,
of
normally
the
bicycles
on
distributed.
following
table.
of
bicycles
0
≤
v
<0
9
0
≤
v
<20
56
20
≤
v
<30
47
30
≤
v
<40
25
40
≤
v
<50
3
a
For
i
ii
b
c
bicycles
on
the
bicycle
lane,
calculate
an
unbiased
estimate
of
the
mean
an
unbiased
estimate
of
the
standard
For
the
bicycles
on
the
bicycle
lane,
speed;
deviation
a
95%
condence
inter val
for
the
mean
speed;
ii
a
90%
condence
inter val
for
the
mean
speed.
Explain
the
Exploring
why
one
of
the
inter vals
other.
statistical
analysis
methods
of
the
speed.
calculate
i
of
118
the
found
in
par t b
is
a
subset
a
8
A
population
follows
a
normal
distribution
with
the
following
2
parameters
following
H
:
=
,
N (
=
2).
Let’s
assume
μ
that
=
10
to
test
the
hypotheses:
10
0
H
:
<
10,
1
using
a
the
Find
level
If
of
the
ii
actual
a
0.1;
Explain
the
a
sample
appropriate
making
i
c
the
of
of
size
critical
5.
regions
corresponding
to
a
signicance
of
0.1;
i
b
mean
population
Type
ii
the
change
0.05.
II
error
is
when
9.3,
the
calculate
level
of
the
probability
signicance
is
0.05.
change
in
mean
the
in
the
probability
probability
of
a
of
Type
a
II
Type
I
error
related
to
error.
Chapter
3
119
Chapter
Estimator
A
if
random
E(T
A
)
and
variable
summary
estimate
T
is
called
an
unbiased
estimator
for
the
population
θ
parameter
θ
=
specic
Given

value
two
of
that
estimators
random
T
and
T

estimator
than
if
T
variable
of
the
is
called
an estimate
population
we
say
that T
2
Var (T
2
)
<
is
a
more
ecient

Var(T

).
2
2
n
n
X
i
X
=
(
X
∑
i =1
is
unbiased
estimator
for
μ.
S
X
)
i
2
=
2
is
∑
n
unbiased
estimator
for
σ
1
n
i =1
n
2
∑
n( x )
i
n
2
S
2
x
2
2
i =1
=
or
n
s
σ
=
1
n
Condence
interval
1
for
mean μ
of
population
⎡
i)
When
the
population
standard
deviation σ
is
known
⎢
σ
x
−
the
population
deviation σ
standard
is
where
t
=
invt
Calculate
for
the
the
mean
2
dierences
Hypothesis
There
i)
are
two
are
four
test
of
upper
lower
for
matched
between
: μ
≠
State
2
Set
Step
3
Calculate
Step
4
Decide
use
We
use
the
the
test ( H
the
obser vations
testing
: μ
>
x )
: μ
<
x )
null
statistical
and
the
upon
in
in
for
and
then
nd
the
and
three
types
of
the
⎤
×
t
c
n
condence
alter native
testing:
alter native
a
hypothesis;
decision;
statistics;
calculated
for
μ
when
hypothesis
testing
t-statistics
hypothesis
criteria
testing
z-statistics
Exploring
+
⎥
⎦
pairs
x )
test ( H
in
1
Hypothesis
120
steps
Step
We
x
⎠
hypothesis
(H
Step
Hypothesis
,
n
1
There
t
inter val
testing
Two-tail
One-tail
s
×
c
1
iii)
⎦
dierences.
types
One-tail
⎥
2
n
−
1
ii)
α
⎟
interval
of
z
,ν
1 −
⎝
Condence
×
⎞
⎜
c
+
s
x
unknown
⎣
α
x
2
⎢
⎛
,
n
⎡
When
z
α
⎣
ii)
⎤
σ
×
for
μ
analysis
is
and
decision
criteria.
known
testing.
when
hypothesis
statistics
is
testing
methods
unknown
regardless
the
sample
size.
hypotheses:
Signicance
We
calculate
hypothesis
Type
I
testing
the
Type
●
We
say
●
We
make
matched
dierences
testing
and
for
that
a
between
regardless
II
the
pairs
the
sample
obser vations
and
then
use t-statistics
in
size.
errors
we
make
Type
II
a
Type
er ror
I
er ror
when
we
when
reject
we
a
fail
tr ue
to
null
reject
a
false
null
hypothesis
hypothesis
Reality
Null
is
Decision
Fail
to
based
null
hypothesis
on
reject
hypothesis
tr ue
Good

Null
−
is
hypothesis
false
Type
decision
α
I
error
(Failing
to
reject
false)
collected
data
Reject
null
Type
hypothesis
II
error
(Rejecting
Good
tr ue)

−
decision
β
β
One-tail
test
y
H
H
0
1
Critical
value
x
0
1
2
3
β
Two-tail
α
test
y
H
H
0
1
Critical
value
Critical
value
x
0
1
α
α
β
2
2
Chapter
3
121
Statistical
4
CHAPTER
OBJECTIVES:
Introduction
7.7
moment
ρ
and
=
and
the
Y.
Its
=
y)
Expectation
two
−
value
Y )
ρ
=
and
of
of
a
Y
given
ρ;
linear
the
use
the
of
scatter
=
of
the
x)
are
of
ρ;
facts
2
Find
of
a
e.g.
of
−
the
random
2E(X )
=
2
=
4Var(X )
data
nd
the
mean
set
−
the
and
following
use
that
of
to
denition
the
t-statistic
regression
of
to
of
X
the
on
r,
test
the
on
estimates
predict
of
obser vations
interpretation
least-squares
lines
Y;
and
paired
the
independence
value
of
of
these
one
e.g.
If
X
and
variables,
Y
are
1
+
+
If
nd:
of
and
Z
E(2Z
−
c
E(XYZ )
a
Using
are
three
normal
3Y
+
standard
random
2X )
variables,
Var(2Z
b
3Y
+
2X )
Var(Y ).
X
Y
5.7
−2.
2
7.
−3.3
2.3
0.9
3.9
.2
variance
X
Y
33
−2
49
−44
table.
50
−23
Using
42
−39
X
2
∑
−
nd:
Var(Y )
n
i
Y,
a
variables:
x
X,
independent
the
formulae,
nd
formulae,
and
n
∑
n
product
of
other
.
variance
the
mean
of
case
X
informal
linear ;
E(Y );
Var(X )
using
(population)
the
start
=
Y )
in
between
terms
regression
the
and
0
in
2
Var(2X
=
diagrams;
of
these
value
R
estimation
(E(Y )|X
ρ
that
relationship
coefcient
and
X
covariance
proof
knowledge
on
the
to
R
0;
Algebra,
independent
E(2X
case
correlation
lines;
you
distributions;
coefcient
application
variables
Before
Use
the
hypothesis
(E(X )|Y
the
bivariate
moment
regression
1
in
obser ved
null
Y
to
correlation
±1
product
X
modeling
(x
b
and
Y,
a
as
shown
GDC,
in
the
conrm
x )
i
2
n
x
i =1
x
i =1
=
;
Var ( X )
2
i
=
=
n
n
−
∑
i =1
your
On
5.7 + 7.1 + 2 .3 + 3.9
x
4
=
=
∑
x
5 .7
a
lists,
2
2
+
7 .1
2
+
par t
a
GDC,
you
can
enter
the
data
in
two
different
and
then
select
from
the
Stats
calculation
2
2 .3
+
menu,
3 .9
=
=
25
‘2-Variable
Statistics’,
and
see
the
values
85
for
n
for
75
4
2
values
x
n
both
variables
on
the
same
screen.
4
2
Var ( X )
=
25 .85  4 .75
= 1 .81
3
Similarly
,
y
=
−0.825,
Var(Y
)
=
Given
GDC
3
Use
e.g.
the
a
c
122
the
t-statistic
given
that
following
P( X
≤ 1.2 )
P( 0.3
Statistical
≤
X
X
to
~
calculate
t (v
=
6),
that
X
~
t (v)
use
the
.93
to
nd
the
following
probabilities
probabilities,
use
the
GDC
to
a
=
2,
P( −0.4
b
= 10,
≤
X
≤
0.8) ;
nd
P( X
≤
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sample
population
y ,
calculated
and Y
will
correlation
correlation
is
coecient R,
for
n
.
In
form
(rho).
obser vations
mean
linear
gdc
A
A6
most
points
lie
point.
quadrants.
c
1.1
the
paired
a
cor relation
If
we
take
the
numerator
of
R,
n
obtain
(X
−
X )(Y
i
−Y
)
=
and Y
X
i
are
constants,
n
(X Y
i
) −
X
Y
−Y
X
brackets
to
XY
−
X Y
i
+
XY
).
i
expression
is
equal
to
∑
XY
i =1
Y
=
, Y
n
,
=
n
∑
i
i =1
therefore,
n
n
X
i
i =1
n
i
+
i
∑
i
i =1
∑
the
n
∑
) −
X
i =1
n
(X Y
out
n
∑
i
i =1
X
−
i
this
n
∑
i
i =1
Since
multiple
i =1
n
∑
can
(X Y
i
i =1
Since
we
n
Y
−Y
∑
i
i =1
n
X
+
i
i =1
n
∑
XY
=
(X Y
∑
i =1
i
i
) − 2 nXY
+ nXY
i
=1
n
=
(X Y
∑
i
) − nXY .
i
i =1
n
n
(X
Hence,
−
X )(Y
i
−Y
)
=
(X Y
i
i
i =1
) − nXY ,
and
an
alter native
i
i =1
n
X
∑
Y
i
nXY
i
i =1
formula
for
R
is
therefore
R
=
n
2
∑
(X
2
− nX
2
2
)(Y
i
− nY
)
i
i =1
We
can
classify
the
strength
of
the
correlation
between
the
random
As
variables
X
and
Y
by
using
the
following
general
for
the
●
± 
indicates
a
perfect
positive/negative
0.5
≤
●
− ≤
R
R
<
≤
R

indicates
−0.5
<
0.5
a
strong
indicates
●
0. ≤
indicates
●
−0.5 <
R
≤
−0.
●
−0. <
R
<
0.
a
a
positive
strong
weak
values
indicates
a
a
negative
positive
weak
highly
earlier
of
of
R
r,
a
near
0
weak
correlation.
correlation
between
and
values
X
correlation.
Y,
and
of
R
correlation.
nearer
indicates
values
correlation.
indicate
●
stated
classication:
negative
weak
correlation.
correlation,
or
strong
±1
negative
no
indicate
positive
a
or
correlation.
correlation.
Of
course,
they
could
if
=
have
sinusoidal,
this
R
etc.
a
0,
X
and
dierent
These
Y
may
not
have
correlation,
kinds
of
e.g.
correlation,
a
linear
correlation,
quadratic,
however,
but
exponential,
are
not
par t
of
course.
Chapter
4
131
Example
Using
the
calculate
X

table
the
(percentage
retur ns
rate).
The
below
,
product
of
Check
of
of
this
and
stock
percentage
X %
economic
Standard
shares
draw
scatter
500
would
economic
growth
by
500
rate)
using
index
publicly
have
diagram,
correlation
result
Poor’
s
of
you
a
moment
is
earned
use
one
GDC
stock
to
you
Y %
nd r,
mar ket
invested
of
S&P
500
.9
7.7
2.6
2.6
3.9
3.7
3.2
9.8
the
formulae
the
two
Standard
inter pret
based
S&P
your
of
and
index
The
of
between
(percentage
companies.
had
growth
and Y
a
a
traded
and
coecient, R,
500
money
on
the
retur ns
in
all
and
your
total
rate
500
above
random
Poor’s
500
result.
value
is
to
variables
of
issued
the
companies.
retur ns
Make
sure
that
*Unsaved
1.1
the
y
mean
point
is
included.
20
label
(3.9, 13.7)
15
(2.6, 12.6)
Economic
Growth
10
(2.9, 10.95)
(3.2, 9.8)
(1.9, 7.7)
5
S&P
Returns
x
0.5
Method
1
1.5
2
2.5
3
3.5
4
I
n
[( x
−
x )( y
−
y )]
=
(.9
–
2.9)(7.7
–
0.95)
+
(2.6
–
2.9)(2.6
–
0.95)
i =1
+
f or
(3.9
–
2.9)(3.7
–
0.95)
+
(3.2
–
2.9)(9.8
–
0.95)
=
5.6
n
2
(x
x )
( y
y )
2
=
(.9
–
2.9)
2
+
(2.6
–
2.9)
2
+
(3.9
–
2.9)
2
+
i =1
n
2
2
=
(7.7
–
+
(3.7
2
0.95)
+
(2.6
–
0.95)
–
0.95)
i =1
2
–
0.95)
5.16
r
≈
2.18 × 22.17
132
Using
Statistical
modeling
0.742
2
+
(9.8
=
22.7
(3.2
–
2.9)
=
2.8
r,
r
the
=
f or mula
0.742.
Method
II
(You
A
B
C
D
will
notice
E
that
the
gives
= TwoVar(
GDC
you
all
the
=
values
necessar y
2.9
2
2.6
12.6
x
3
3.9
13.7
Σx
4
3.2
9.8
Σx
to
evaluate
r
11.6
manually.)
2
5
sx
6
σx
35.82
:
=
:
sn–...
0.852447
σn...
0.738241
=
n
7
4.
8
10.95
y
9
Σy
10
Σy
11
sy
12
σy
13
Σxy
14
r
43.8
2
D26
r
=
=
501.78
:
=
:
=
sn–...
2.71846
σn...
2.35425
132.18
0.74223
22.17
0.742
There
is
a
strong
positive
correlation
between
the
two
random
variables.
Inter pret
the
r
value.
Although
the
it
mathematician
He
borrowed
‘Pearson
method
linear ly
the
Any
For
related.
at
assume
other
that
the
The
both
is
Correlation
founded
College
UK
the
data
in
in
Y
the
wor ld’
s
be
mental
work
be
are
rst
other;
here,
and
by
For
is
there
an
we
a third
concept.
the
name
most
the
common
variables
that
are
statistics
or
R
great
was
can
be
used
care.
seen
increase
would
there
so
the
developed
coefcient.
with
It
this
two
Either r
correlation
between
aected
It
rst
English
hence
university
1911.
930s
who
the
formalized
between
illness.
the
related.
in
was
Physics,
treated
early
of
Bravais
it
Coefcient’.
moment
the
who
from
London
must
cause
may
and
1800s,
coefcient
relationship
at
Auguste
mid
(1857–1936)
product
factors
X
the
cor relation
increase
one
physicist
in
‘moments’
coecient
in
unrelated
which
a
sample
an
of
Moment
signicant
and
cor relation.
Pear son
Pear son
correlation
example,
French
concept
Univer sity
Pearson
licenses
the
coefcient
computing
statistically
to
Kar l
the
Product
of
depar tment
for
was
correlation
are
call
be
in
this
or
there
be
a
radio
absurd
probably
factor,
example,
to
to
many
a spurious
more
f actors,
could
be
a
Chapter
4
133
correlation
could
both
actually
between
be
be
smoking
related
causing
to
a
both
and
third
the
high
blood
variable
high
blood
pressure,
(stress
level)
pressure
however
which
and
the
they
could
desire
to
smoke.
Hence,
great
change
in
caution
one
factors
should
This
why
,
is
statistician,
their
For
the
point
be
as
The
of
sets
repor t
5
change
and
in
other
nding
causation
a
before
the
other.
statistical
be
left
Many
tests
correlation
should
asser ting
is
to
that
a
other
performed.
the
the
job
of
the
specialists
below
,
Y
being
by
of
using
ten
with
(Y
),
are
the
draw
the
a
a
how
a
in
highest
and
were
daily
informed
inter pret
sought
of
table
including
for
the
this
world
below
,
(X ),
with
1
mean
value.
their
and
aairs.
the
random
regarding
newspaper
them
the
diagram,
Calculate R
GDC,
people
shown
scatter
point.
a
television
Their
being
levels
levels
the
of
lowest
satisfaction.

3
2

2
5
4
4
2
Y


2
4
3
4
2



The
lengths
given
Y
The
are
be
in
in
mm
the
(X )
table
and
widths
in
mm
(Y
)
of
cer tain
leaves
below
.
00
5
40
49
88
32
52
44
2
28
33
38
40
5
36
40
5
43
32
42
percentages
shown
the
in
study
.
4
X
134
a
taken
X
are
3
of
through
and
satisfaction
2
elds
satisfaction
and
earlier,
asser ting
opinions
news
causes
be
4A
axis
X
always
considered,
stated
but
data
and
variables
1
variable
respective
Exercise
should
scored
below
.
percentage
Let
X
in
the
be
scored
in
same
the
subject
percentage
Test
on
two
scored
dierent
in
Test
1,
tests
and Y
2.
X
55
35
66
82
9
79
48
52
7
88
Y
58
65
52
35
36
42
60
55
50
38
Statistical
modeling
4.2
Covariance
Another
statistical
between
two
implies,
the
measure
random
used
variables
covariance
tells
us
to
is
determine
called
whether
the
or
if
there
is
covariance.
not
the
a
relationship
As
the
name
variables var y
together.
A
●
positive
one
the
A
●
variable
other
the
Zero
In
other
of
this
formula
tend
covariance
words,
for
paired
to
with
indicates
be
paired
by
each
higher
than
than
that higher
with
lower
the
and
that
sum
i.e.
of
the
is
thus
the
no
the
of
the
of
of
dividing
obtain
−
X )(Y
i
r,
the
sample
of
the
dierent
two
in
unit
random
this
issue
by
deviations
that
is
The
data.
−Y
used.
)]
For
X
the
and
individual
take
the
values’
average
of
the
Cov (X,
Y )
XY
=
XY
dimensionless,
this
and
value
the
when
covariance
This
a
to
is
same
we
the
quantity
facilitates
it
dependent
represents
the
correlation
creating
units.
that
indicate
reason,
Y,
the
a
not
i.e.
have
a
if
distribution
coecient
product
of
between
the
strong
relationship
a
on
r
of
addresses
the
−
a
standard
and
comparison
+
of
sets.
covariance
relationship,
between
numerator
∑
means
variables,
of
not
might
normalizing
the
is
This
unit,
variables
data
of
of
n
however,
independent
dierent
values
values
i =1
one
is
of
of
covariance:
i
or
covariance,
units
of
n
[( X
=
relationship
average
average
n
The
values
values
correlation
products
numerator
equivalent
we
there
i =1
Y )
average
than
than
n
∑
Cov(X,
average
variable.
the
mean,
is
n
indicates
nd
the
This
r
of
we
from
sum.
be
that higher
variable.
variance
deviations
to
covariance
variable
other
the
tend
indicates
variable.
negative
one
●
covariance
indicates
but
it
does
whether
not
tell
there
us
is
a
anything
positive
about
or
the
negative
strength
of
the
relationship.
If
we
use
the
rst
denition
given
for R,
n
∑
(X
−
X )(Y
i
−Y
)
i
i =1
i.e.
R
=
n
n
⎛
and
E(X
∑
⎝
i =1
divide
)
2
⎜
and
the
2
⎛
⎜
∑
⎠
⎝
i =1
2
numerator
these
E[( X
⎞
⎟
− nX
i
replace Y
Replacing
ρ
X
by
by
2
⎞
− nY
⎟
i
⎠
and
E(Y
averages
− E[ X ])(Y
Y
),
denominator
the
the
sums
by n
become
expected
and
replace
X
by
averages.
values,
we
obtain,
− E[Y ])]
=
.
2
E[( X
− E[ X ])
2
]E[(Y
− E[
Y ])
]
Chapter
4
135
The
like
denominator
a
variance,
rather
than
a
covariance
equal
except
square
of
population
is
X
of
and
product
it
ρ
Cov(X,
Denition:
is
the
a
For
random
measure
deviations
Cov(X,
product
Y ),
and
and
in
The
of
deviation
fact
the
numerator
is
called
greek
letter
of
X
Y
each
of
the
rho,
ρ,
=
set
of
their
E[(X
n
paired
and
Y
value
−
E(X
))(Y
of
of
respective
=
E(X
μ
),
with
Y
)
=
=
observations,
a
joint
the
E(Y
the
covariance
distribution,
product
means,
−
=
E(Y
of
the
Cov(X,
two
=
this
denition
E[(X
−
)(Y
we
−
E[XY
−
E(XY
)
=
−
variables’
i.e.
))]
=
E(XY
)
E[(X
−
μ
)(Y
μ
−
=
E(XY
)
μ
Y μ
)
−
μ
+
μ
Properties
of
Symmetr y:
proof
two
If
E(XY
)
X
−
=
this
E(X
Y
E(Y
=
Y
is
are
)E(Y
μ
x
The
it
a
the
)
)
=
left
=
second
E(XY
)
=
throw
and
Students
should
Y
above
Theorem
proof
:
is
Statistical
If
left
μ
μ
y
μ
y
x
y
the
a
);
as
are
X
),
and
Var(X
)
=
Cov(X,
)
=
0.
X
)
exercise.
X
and
Y,
Cov(X,
variables,
Cov (X,
Y
Y
then
)
=
E(XY
)
μ
−
y
x
y
0.
is
X
not
and
‘the
die’.
not
and
an
modeling
necessarily
Y
are
If
=
if
U+V
Cov (X,
Y
‘the
and
)
=
Cov(X,
For
on
Y
0,
=
)
=
example,
the
number
Y
rst
0,
let
throw
appearing
U−V,
however
U
of
on
then
the
random
independent.
the
Y
appearing
variable
X
tr ue:
independent.
number
random
above
proper ties,
as
an
and
μ
hence,
verify
X
Cov(Y,
y
of
E(X )E(Y
the
=
that
be
X
The
μ
variable
V
variables
From
136
let
+
x
+
μ
x
events
μ
however,
follow
and
μ
independent
x
random
die’,
the
not
μ
y
converse,
does
be
−
)
y
x
μ
y
μ
independent
and
one.
covariance
Cov(X,
of
alter native
]
μ
x
x
μ
−
an
x
x
Proof:
)],
y
)]
μ
E(X
μ
−
derive
x
y
For
),
y
X μ
−
can
y

Y
)
y
The
the
y
x

is
coecient.
)
mean
x
Cov(X,
Y,
)
variables X
from
)
μ
Star ting
and
the
x
where
looks
=
Var ( X ) Var(Y
of
a
deviation,
moment
Cov ( X ,Y
Hence,
.
Var[ X ] Var[
Y ]
contains
a
Y,
to
we
are
result
can
as
establish
independent
exercise.
an
exercise.
the
following
variables,
then
ρ
theorems:
=
0.
Theorem
2:
If
X
and
Y
have
a
linear
relationship,
i.e.
Y
=
mX
+
c,
then
ρ
=
±.
Proof:
Y
=
mX
+
c
⇒
E(Y
)
=
mE(X )
+
c.
Hence,
2
=
m μ
y
+
c ;
Var(Y )
=
m
Var(X ).
x
Therefore,
2
E(XY
)
=
2
E(mX
+
cX )
=
mE(X
)
c μ
+
x
and,
using
the
formula
for
ρ
and
making
2
substitutions
2
) + c μ
m E( X
− μ
x
ρ
the
(m μ
x
+ c )
2
) + c μ
m E( X
x
− (m μ
x
=
above,
− u
x
c )
x
=
2
Var ( X )m
2
2
Var ( X )
m
2
( Var ( X ))
2
m ( E( X
μ
)
)
mVar ( X )
m
x
=
=
m
Var ( X )
Practically
ρ
is
R,
the
Y,
sample
=
Var ( X )
we
coecient
Example
Given
m
speaking,
correlation
for
=
can
of
a
± 1
m
estimate ρ
sample
product
for
drawn
moment
the
population
from
the
only
population.
by
using
Hence,
the
an
estimate
coecient.

that
Cov(X,
X
Y
~
),
N(0,
is
)
and
Y
=
2X,
show
that
the
covariance
of
the
joint
distribution X
and
2.
2
Cov (X,
Y )
=
E(XY )
2
E(2X
–
E(X )E(Y )
=
E
(2X
)
–
E(X )E(2X )
Use
denition.
Use
proper ties
2
)
=
2E(X
)
2
Var(X )
=
E(X
2
)
–
[E(X )]
2
⇒
E(X
2
)
=
Var(X )
+
of
expectation
algebra.
[E(X )]
Hence,
2
E(X
)
=
+
0
=

Cov(X,
Y
)
Substitute
2
Hence,
=
E(2X
=
2E(X
)
–
E(X )E(2X )
)
–
E(X )2E(X )
2
Exercise
Prove
the
Y
)
=
Cov (Y,
2
Cov (X,
X )
=
Var (X )
3
Cov (aX,
4
Cov ( X,
5
Cov ( X
Y
)
bY
+
=
)
, Y
+
If
X
=
2
Var(X ) = 1.
Evaluate.
+
and
Y
Y
)
)
=
)
Cov ( X
, Y
)
=
=
are
+
Cov ( X
Var(X )
, Y
)
+
Cov ( X,
Y
1
+
Var(Y
independent
)
2
Cov ( X , Y
2
)
)
1
Y
1
Y
b Cov ( X , Y
2
Cov ( X , Y
Var(X
0
and
X )
aCov (X,
=
X
1
8
–
0
following:
Cov (X,
7
2
=
4B
1
6
=
E(X )
)
2
)
+
2Cov (X,
variables,
then
Y )
ρ
=
0.
Chapter
4
137
4.3
Hypothesis
testing
Introduction
This
introduction
distribution,
Before
be
in
two
we
can
order
to
random
sample
Y
ou
are
single
not
We
already
be
understanding
f or mally
how
that
X
large
Y,
rst,
∼
N
( μ
Then,
the
ver y
familiar
signicant
will
random
two
for
with
the
variable.
two
perform
dene
nor mal
coecient
correlation
an
must
between
appropriate
what
normally
normal
We
random
will
we
mean
distribution
now
variables X
distributed
) and Y
∼
( μ
N
, σ
Y
bivariate
is
normal
dened
by
by
consider
and
variables,
for
the
a
joint
Y
i.e.
)
Y
distribution,
the
⎡
1
⎢
−
x −μ
x
2
)
⎣
σ
⎝
y −μ
y
+ ⎜
normal
function:
⎠
σ
⎝
⎢
⎣
⎤
⎞
⎛
− 2 ρ
⎟
⎜
x
joint
2
⎛
⎞
⎟
⎢
ρ
2 (1
⎢
the
density
2
⎛
⎜
⎢
1
or
probability
⎢
(x, y)
a
correlation
however,
⎡
f
bivariate
distribution
X
distribution,
the
2
, σ
X
the
is
we
2
X
of
assessed.
there
and
will
distribution
Y
aid
be
conclude
variables
nor mal
and
to
determine
continuous
normal
X
will
statistic.
bivariate
Let
but
ser ves
x −μ
⎟
⎟
y
σ
⎝
⎠
⎞
x
⎜
⎛
y − μ
⎜
x
⎠
y
⎜
⎤
⎞
⎥ ⎥
⎟
σ
⎝
⎟
y
⎥
⎥
⎠ ⎥ ⎥
⎦
⎦
e
=
2
2πσ
σ
X
ρ
1
Y
μ
y
μ
x
y
x
For
standardized
variables,
i.e.
z
; z
=
x
,
=
σ
y
2
z
2
becomes
f
(z
z
x
)
−2 ρ z
+ z
x
y
1
PDF
bivariate
σ
x
normal
the
y
z
x
y
2
2 (1
ρ )
e
=
y
2
2π
Y
ou
know
random
normal
that
the
variable
is
shape
that
distribution
has
of
of
a
a
ρ
1
the
normal
2D
bell-shaped
3D
distribution
bell-shaped
surface.
of
a
The
single
bivariate
surface.
0.15
0.1
0.05
0
–3
–3
–2
–2
–1
–1
0
0
1
1
x
y
2
2
3
Y
ou
have
variables,
a
138
already
then
bivariate
3
seen
Cov(X,
normal
Statistical
that
Y )
if
=
0
X
and
distribution,
modeling
and
Y
are
hence
the
two
ρ
=
converse
independent
0.
is
For
also
X
and
tr ue.
random
Y
having
Fur thermore,
Theorem
For
X
we
and
For
rst
⇐,
Y
with
par t,
we
now
state
the
following.
3:
independent
The
can
if
a
bivariate
and
i.e.
have
only
⇒,
the
ρ
if
was
normal
=
of
seen
a
in
joint
Theorem
1
⎢
−
2
ρ
2 (1
⎢
1
)
i.e.
f
(x,
y)
x − μ
2
⎛
⎞
⎟
x
⎛
⎟
⎜
⎠
⎤ ⎤
⎞
y
⎜
+
⎟
⎝
y − μ
x
σ
⎝
y
x − μ
⎟+
⎜
⎟
σ
⎝
⎠
⎛
⎞
y −μ
x
− 2 ρ ⎜
⎟
x
⎜
⎠
⎞
y
⎥ ⎥
⎟
⎥ ⎥
⎜
σ
⎝
⎟
y
⎠⎥
⎣
⎣
Y,
⎛
σ
⎢
function
2
⎜
⎢⎜
⎢
and
are
density
⎡
⎢
X
Y
.
probability
⎡
of
and
0.
already
denition
distribution, X
⎥
⎦ ⎦
e
=
2
2πσ
σ
X
Using
ρ
=
0
and
the
laws
of
ρ
1
Y
exponents,
2
2
1 ⎛
μ
x
x
⎜
1
(x,
y)
⎜
2πσ
will
x
any
result
2πσ
≤
X
≤
and
therefore
X
t-Statistic
For
be
a
used
linear
to
and
for
bivariate
,
y
2
Y
are
determine,
correlation
≤ Y
≤
by
a
y
1
)
the
f
=
joint
P( x
2
(x)
×
f
( y)
≤
distribution
X
≤
x
1
)
×
of
P( y
2
X
and
≤ Y
≤
Y
y
1
),
2
independent.
distribution
at
⎠
=
from
dependence
normal
⎟
y
Y
calculated
x
σ
σ
X
1
⎜
e
Y
P( x
⎞
y
⎟
⎝
×
probabilities
in
2
⎠
σ
X
Hence,
1
⎟
σ
2
e
=
μ
y
⎜
⎟
⎝
f
⎛
1
⎞
given
of X
of
level
determining
X
and
of
ρ
if
=
and
Y
,
Y
the
correlation
signicance,
0.
We
will
coecient
whether X
use
and
Theorem
4
Y
to
can
have
make
a
this
determination.
Theorem
If
X
and
4:
Y
have
a
bivariate
normal
n
sampling
distribution
distribution
that ρ
such
=
0,
then
the
2
has
R
the
student’s
t-distribution
with
(n
−
2)
2
1
degrees
This
of
means
joint
R
freedom.
that
normal
for
any
random
distribution
(X,
Y
)
sample
with
ρ
of
=
n
0,
independent
the
obser ved
paired
value
n
product
moment
coecient
R
has
the
proper ty
that t
=
r
data
of
from
the
the
sample
2
,
r
2
1
i.e.
To
it
is
distributed
perform
a
as
student’s
hypothesis
test
t-statistic
with
H
:
with
ρ
=
degrees
0
and
H
0
of
n
independent
pairs
(x
,
1
distribution
●
(X,
Y
)
Calculate r,
correlation
y
),
(x
1
,
y
2
:
of
ρ
r
freedom
≠
0
on
a
v
=
n
−
random
2.
sample

),
…
(x
2
,
n
y
)
from
a
bivariate
normal
n
we:
the
obser ved
coecient
value
of
the
sample
product
moment
R;
n − 2
●
Calculate the
t-statistic,
t
=
;
r
2
1 − r
●
Calculate the
or
calculate
critical
the
values
for
the
indicated
level
of
signicance,
p-value.
Chapter
4
139
Example
Test,
at
the
evidence
growth
ρ
:
H

0%
of
rate)
=
0;
signicance
correlation
and
H
0
ρ
:
Y
≠
(percentage
=
of
the
random
Standard
data
from
variables X
and
0
Poor’s
Example
2
(percentage
500
retur ns
Step
1:
State
the
Step
2:
Find
Step
3:
Using
null
shows
of
signicant
economic
rate).
and
alter native
hypotheses.
4 − 2
r
t
⇒
=
≈
0.742
2
For
H
,
T
value
of
the
test
statistic.
:
t
(2)
P(T
1.1
the
1 − 0.742
0
=
1.565
2
1 − r
p
whether
the

n − 2
t
level,
between
≤
–.565)
+
P(T
≥
.565)
=
0.258
*Unsaved
2.1
a
GDC,
calculate
the
p-value.
tCdf(–∞,–1.565,2)+tCdf(1.565,∞,2)
0.258054
|
Alter natively
,
t
test’
we
in
the
obtain
using
Stats
the
the
tests
‘Linear
menu
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Regression
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1.9
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2.6
12.6
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ρ
R...
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a+b*x
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1.56634
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0.25777
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evidence
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we
H
do
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4:
Reading
compare
with
the
the
p-value
signicance
0
Hence,
there
correlation
X
and
Note:
not
evidence
the
of
signicant
random
conclusion.
variables
Y.
Although
cor relation
140
is
between
is
Statistical
the
not
r
value
seems
signicant.
modeling
relatively
high,
with
so
f ew
data
from
values
the
the
level
results,
and
state
we
Exercise
Use
1
the
level
4C
data
in
whether
Example
the
1
random
to
determine
variables
at
show
the
a
1%
signicance
signicant
level
of
correlation.
Below
2
is
a
table
Determine,
evidence
at
of
Using
3
there
X
supermodel,
Weight
At
Linear
the
star t
determined
in
The
68
69
70
72
74
69
64
67
64
62
73
7
determine,
a
of
known,
price.
but
random
(weight
of
04
6
2
26
7
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this
chapter
the
we
graphed
scatter
the
two
positive
between
values
Y
rather
of
on
X
than
can
be
or
the
X
of
negative.
two
on
X
are
the
sold.
the
of
E(Y
)|X
the
of
a
if
the
accurate.
on
joint
In
If
x,
In
is
other
shares
case,
the
bought
of
there
was
then
to
a
the
diagrams
showed
draw
Y
and
the
var y
and
X.
and Y
of
Y
a
estimate
words,
value
will
and
the
of
might
stock
the
Y
Y
For
according
is
line
linear
distribution
stock,
variable,
sets
whether
used
conditional
par ticular
independent
so,
could
distribution
this
data
scatter
we
=
of
whether
and
variables,
X,
price
number
is
on
pairs
diagrams
variables,
focuses
the
stock
the
Hence,
and Y
if
regression
the
amount
inches)
the
73
Y
X
between
level,
72
of
example,
in
signicance
72
line
X,
10%
Y
72
regression
of
the
supermodel,
regression.
that
at
correlation
of
given
the
66
line
regression
and
64
a
given
variables X
pounds).
through
or
signicant
7
was
t,
random
inches.
is
70
correlation
best
in
there
70
between
of
the
if
70
of
correlation
heights,
level,
67
correlation
a
of
sons’
between
evidence
(height
Height
4.4
below
,
signicant
variables
a
height
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and
signicance
height
the
is
fathers’
5%
correlation
X-Father’s
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of
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be
is
to
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dependent
variable.
Chapter
4
141
One
method
of
the
regression
the
least
calculating
line
is
to
y
nd
y
of
the
the
sum
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ver tical
points,
squares
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the
distances
i.e.
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the
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by
squares
and
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principle
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of
line
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following
ver tical
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the
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regression
regression
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formulae:
n
n
⎛
⎞
⎛
⎟
⎜
⎞
The
∑
⎜
[( x
−
x )( y
i
−
y )]
i
∑
i =1
y
−
y
⎟
(x
− x )
or
y
−
y
=
∑
⎝
that
Hence,
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variables,
of
Y
on
values
The
−
the
of
we
E(Y
X
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the
table
mean
includes
⎜
⎟
⎜
⎠
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of
formulae
⎟
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beyond
2
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x
⎟
2
regression
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− nx
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i
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line
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show
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(inter polation).
there
dierent
and
given
ways
Var(Y ).
to
work
to
write
Depending
with,
some
the
on
the
formulae
will
others.
the
data
nd
formula
the
estimate
valid
than
to
to
data
seen,
given
asked
that
used
inter val
Cov(X,
and
are
of
already
statistics
here
be
information
Instead
we
state
only
the
ts
following
directly
,
line
the
of
given
summar y
we
may
be
regression.
statistics
given
summar y
In
this
case,
we
best.
For
example,
if
statistics:
2
n
= 10;
and
∑
xy
asked
= 906.82;
to
nd
the
∑
x
= 103.8;
∑
regression
line
y
= 100.9;
of
Y
on
∑
X
x
for
2
= 1983.08;
these
∑
values,
y
= 1050 .67
the
n
⎛
⎜
⎞
∑
x
y
i
nx
y
⎟
i
i =1
formula
best
suited
is
y
−
y
=
⎜
⎟
(x
x )
n
⎜
⎜
⎝
2
∑
i =1
x
2
⎟
nx
i
⎟
⎠
Chapter
4
143
Using
the
given
values,
the
gradient
of
the
line
of
regression
is:
n
⎛
⎜
⎞
∑
x
y
i
nx
103.8
y
906.82 − 10 ×
⎟
i
m
=
100.9
×
10
i =1
⎜
⎟
10
=
=
n
2
directly
,
− 10.09
the
as
data
Y
a
car
Use
c
Use
on
a
the
it
in
− 10.38) ⇒
we
should
example
the
fuel
driven
in
y
⎠
=
have
−0.155 x
the
data
+ 11.7
to
work
with
shows.
on
consumption
a
test
in
litres/00 km)
sense
to
and
average
speed
of
a
typical
Euro
nd
the
0
5
20
30
35
40
45
50
3
0
9
7
6.6
6.2
6
6
line
of
best
t,
or
linear
regression,
for
table.
to
nd
the
equation
of
the
linear
equation
it
at
an
make
Inter pret
at
to
average
sense
80
your
estimate
to
the
speed
use
the
of
fuel
regression,
25
table
consumption
and
check
your
answer
of
a
typical
Euro
4
above
to
nd
the
fuel
consumption
car
of
a
car
km/h?
answers
for
this
table
of
values.
Since
A
passenger
km/h.
a
B
C
D
r
≈
−0.921,
this
E
indicates
a
strong
=TwoVar(
=
negative
3
20
9
Σx
4
30
7
Σx
5
35
6.6
sx
6
40
6.2
σx
7
45
6
n
8
50
6
y
cor relation,
245.
hence
we
can
2
8975.
of
:=
:=
sn–...
14.5006
σn...
13.5641
8.
7.975
9
Σy
63.8
10
Σy
11
sy
12
σy
13
Σxy
14
r
2
144
the
GDC.
the
Does
E1
4
circuit.
km/h)
makes
the
data
traveling
e
shows
when
that
travelling
d
(x
however,
speed
variables
b
−0.155
consumption
Justify
⎟
10
⎝
following
below
(Average
(Fuel
⎜

passenger
X
=
time,
the
Example
The
1983.08 − 10 ×
⎟
⎠
y
of
⎛ 103.8 ⎞
nx
i
i =1
Hence,
Most
⎟
2
x
∑
⎜
⎝
− 0.1551...
2
⎜
=“Two–Variable
Statistical
Statistics”
modeling
553.
:=
:=
sn–...
2.51268
σn...
2.3504
1719.
–0.9209...
best
t.
draw
a
line
b
From
∑
xy
the
GDC
= 1719; x
above,
we
= 30.625; y
obtain
the
following
values:
= 7.975
2
n
= 8;
∑
m
= 8975.
x
∑
x
y
i
nx
y
1719 − 8 × 30.625 × 7.975
i
=
∑
y
2
x
8975 − 8 × 30.625
x
y
i
nx
y
i
(x
=
2
∑
Hence,
A
− 0.1595 …
2
− nx
i
∑
−
≈
=
2
y
Hence,
i
y
=
B
−
x )
⇒
y
− 7.975
=
− 0.1595( x
− 30.625)
2
x
nx
i
−0.60x
C
D
+
2.9
(to
3
s.f.)
E
From
the
GDC,
we
see
=LinRegMx(‘xvalues. ‘yvalues.1
1
10
13
T itle
2
15
10
RegEqn
3
20
9
m
4
30
7
b
5
35
6.6
6
40
6.2
7
45
6
8
50
6
Linear
Regression
(mx+b)
that
the
regression
line
is
m*x+b
y
–0.159575
=
+
−0.160x
12.9.
12.862
0.848066
–0.920905
Resid
{1.7337579617839, –0.46836...
9
10
11
12
B12
c
y
=
at
a
−0.596(25)
constant
+
2.86
speed
of
=
8.87
25 km/h
,
hence
would
the
be
fuel
about
consumption
Use
8.9
the
equation
substitute
f or
x
to
and
nd
y.
litres/00 km.
d
No,
since
80
lies
outside
the
data
Check
domain.
if
within
e
Since
these
urban
lights,
or
and
averages,
areas
congestion
‘stop-star t’
in
where
they
there
general.
mode,
indicate
The
which
is
driving
more
car
causes
is
trac,
Attempt
in
trac
therefore
in
a
value
data
reasonable
explanation,
that
a
the
lies
domain.
assuming
data
is
cor rect.
increased
usage.
Example
The
are
congested
continual
fuel
speeds
the
the
table

below
continues
for
the
following
speeds
of
the
same
test
cars
on
the
same
test
circuits.
X-Average
Y-Fuel
line
Driving
in
km/h
consumption
Describe
The
speed
and
inter pret
joining
at
the
speeds
consumption
in
is
litres/00
the
points
allowed
almost
km
relationship
is
on
just
about
priority
constant,
6
50
55
60
6
6
6
between
parallel
roads,
i.e.
the
to
two
65
70
6
6
75
80
6.
6.
variables.
the x-axis,
between
50
hence
and
its
gradient
80 km/h,
the
is
0.
fuel
litres/00 km.
Chapter
4
145
Example
The
table

below
continues
for
the
following
speeds
of
the
same
test
cars
on
the
same
test
circuits.
X-Average
Y-Fuel
a
speed
that
variables
b
Use
a
c
Use
the
speed
it
in
e
Determine,
and
sense
to
nd
the
85
90
95
05
0
5
20
6.
6.3
6.4
6.4
6.6
6.7
6.8
6.9
line
of
best
t,
or
linear
regression,
for
the
nd
the
to
equation
estimate
of
the
the
fuel
linear
regression
consumption
of
a
line.
car
travelling
at
an
average
00 km/h.
Inter pret
average
to
litres/00 km
80
table.
equation
of
in
makes
the
GDC
d
a
km/h
consumption
Justify
0
in
your
answers
at
the
speeds
0%
and
50 km/h,
for
H
table
signicance
fuel
and
this
of
level,
consumption
between
80
values.
and
of
if
there
the
test
is
a
cars
correlation
for
average
between
speeds
the
between
20 km/h.
L
Use
a
GDC
Use
the
to
nd
r.
=TwoVar(
4
95
6.4
Σx
5
105
6.6
sx
6
110
6.7
σx
7
115
6.8
n
8
120
6.9
y
81500.
:=
sn–...
:=
14.6385
σn...
13.6931
8.
6.525
9
Σy
10
Σy
11
sy
12
σy
13
Σxy
52.2
341.12
:=
sn–...
:=
0.271241
σn...
0.253722
5247.5
14
0.989426
L
Since
r
≈
0.989,
correlation,
b
this
hence
H
indicates
we
can
a
nd
strong
the
positive
regression
line.
L
GDC
to
nd
the
line
of
best
interval
of
the
t.
=LinRegMx(‘xvalues, ‘yvalues,1):
1
80
6.1
T itle
2
85
6.3
RegEqn
3
90
6.4
m
4
95
6.4
b
5
105
6.6
r
6
110
6.7
r
7
115
6.8
Resid
8
120
6.9
Linear
Regression
(mx+b)
m*x+b
0.018333
4.69167
2
0.978964
0.989426
{–0.05833333333334, 0.049..
9
10
11
12
I9
The
c
equation
+
=
0.083x
y
=
0.0833(00)
6.5
The
as
4.69
at
+
line
(to
3
of
best
t
is:
s.f.)
4.692
=
00 km/h
6.53,
is
hence
the
fuel
speeds
little
speeds
Statistical
in
the
modeling
the
value
table
so
indicate
and
will
so
fuel
highway
it
makes
trac
sense
consumption.
that
is
in
substitute
Attempt
interr uption,
increase
If
then
about
litres/00 km
with
146
the
y
consumption
d
of
a
assuming
the
f or
x
and
reasonable
that
the
evaluate
y.
explanation,
data
is
cor rect.
data,
e
For
H
the
ρ
:
data
=
0;
H
0
:
≤
ρ
X
≠
≤
50:
Step
0
1:
State
the
null
and
alter native
hypothesis.

n
t
0
=
2
r
8
⇒
t
=
=
2
1
Step
2:
Find
t.
Step
3:
Find
the
2
− 0.921
− 5.79106
2
r
1
0
921
p-value
and
conr m
*Unsaved
1.1
using
the
stats
test
menu
of
the
GDC.
tCdf(–9.E999, –5.791, 6)+tCdf(5.791, 9.E999, 6)
0.001161
|
1/99
p
=
P (T
≤
0.006
<
−5.791) + P (T
0.,
hence
we
≥ 5.791)
reject
=
H
0.00116
,
since
there
is
Step
4:
State
conclusion.
Step
1:
State
the
0
signicant
random
For
ρ
:
H
the
evidence
variables
data
=
0;
80
H
0
:
ρ
≤
≠
of
correlation
X
and
X
≤
=
the
20:
0
null
and
alter native
hypotheses.

n − 2
t
between
Y
8 − 2
⇒
r
0.989426
=
2
16.71
Step
2:
Find
t.
Step
3:
Find
the
2
1 − r
1 − 0.989426
p-value,
and
conr m
*Unsaved
1.1
using
the
stats
test
menu
of
the
GDC.
tCdf(–9.E999, –16.71, 6)+tCdf(16.71, 9.E999, 6)
0.000003
|
1/99
p
=
P (T
≤
0.000003
−16.71) + P (T
<
0.,
hence
≥ 16.71)
we
reject
=
0.000003
H
,
since
there
is
Step
4:
State
conclusion.
0
signicant
random
evidence
variables
X
of
correlation
and
between
the
Y.
Chapter
4
147
To
calculate
the
squares
of
the
In
other
the
of
the
squares
of
words,
conditional
probability
regression
line
horizontal
the
the
linear
distribution
X
on
distances
distances
distribution
of
is
of
X
X
we
from
nd
the
the
least
points,
sum
i.e.
the
of
area
minimized.
regression
of
Y,
given
given
of
Y,
X
on
rather
Y
focuses
than
on
on
the
the
joint
Y.
y
x
(x,
on
y
y)
x
The
formula
for
the
X
on
Y
regression
n
∑
(x
−
x )( y
i
−
⎞
⎛
⎟
⎜
y )
i
i =1
x
− x
=
⎞
∑
⎟
(y
−
y )
=
∑
⎝
It
is
the
on
Y
discussed
148
the
Y
)
important
X
y
−
gradient
and
to
Var(Y
Statistical
the
⎜
⎟
⎜
⎠
⎝
of
⋅
y
⎟
⎟
2
(y
−
y )
line
beginning
modeling
the
X
on
∑
y
⎟
2
− ny
⎟
i
⎠
i =1
Y
regression
line
is
the
quotient
).
remember
regression
in
⎟
y
i
i =1
Cov(X,
− nx
i
n
2
⎜
of
y
i
⎜
n
that
x
i =1
⎜
⎜
Notice
is
n
⎛
⎜
line
that
pass
of
both
the Y
through
this
the
chapter
on
X
mean
in
regression
point ( x ,
drawing
line
y ),
scatter
and
as
diagrams.
Example
A
teacher
given
in

gave
two
dierent
Statistics
tests
to
his
class
with
the
following
paired
percentages.
X-Test
65
88
83
92
50
67
00
00
73
90
83
94
Y-Test2
52
57
78
76
30
67
96
74
65
87
78
89
The
teacher
was
absent
a
Justify
test,
b
would
for
that
and
the
a
nd
Determine
two
results,
if
like
rst
linear
this
to
use
test,
these
but
results
scored
regression
can
for
52%
be
repor t
on
used
the
to
card
second
estimate
grades.
One
student,
however,
test.
this
student’s
result
on
the
rst
value.
there
is
signicant
correlation
at
the
5%
signicance
level
between
the
variables.
a
Since
=
r
=
0.837,
there
positive
cor relation,
nd
line
is
a
hence
strong
we
can
=TwoVar(‘xvalues, ‘yvalues, 1):
2
4
50
30
Σx
5
67
67
sx
6
96
σx
7
74
n
83465.
:=
sn–...
:=
the
of
best
t.
15.4123
σn...
14.7561
12.
8
73
65
y
9
90
87
Σy
10
83
78
Σy
11
94
89
sy
12
83
78
σy
70.75
849.
2
63693.
:=
sn–...
:=
13
Σxy
14
r
15
MinX
18.1565
σn...
17.3835
72266.
0.837269
50.
C12
Find
E
B
the
X
on
Y
line
by
letting
Y
xvalues
=
be
=LinRegMx(‘yvalue
1
65
52
2
88
57
3
92
76
m
4
50
30
b
5
67
67
r
6
100
96
r
7
100
74
Res...
8
73
65
9
90
87
10
83
78
11
94
89
12
83
78
T itle...
Linear
Regression...
the
the
independent
variable,
and
X
dependent.
m*x+b
0.71072
31.7999
2
E1
x
=
=“Linear
Regression
0.707(52)
test’s
result
0.701019
0.837269
{–3.75732506032...
(mx+b)”
+
3.8
would
=
have
68.8,
been
hence
about
the
rst
Substitute
f or
y
to
nd
x.
69%.
Chapter
4
149
b
H
ρ
:
=
0;
H
0
:
ρ
≠
t
=
0
Step
1:
State
alter native
n − 2
t
=
the
null
and

hypotheses.
12 − 2
⇒
r
0 .837
≈
2
4.84
2
1 − r
Step
2:
Find
t.
Step
3:
Find
the
1 − (0.837 )
p-value
using
a
*Unsaved
1.1
GDC.
tCdf(–9.E999, –4.84, 10)+tCdf(4.84, 9.E999, 10)
0.000681
|
1/99
p
=
P (T
The
≤
−4.84 ) + P (T
p-value
0.00068
≥
is
4.84 )
less
=
0.000681
than
5%,
so
we
reject
H
,
Step
4:
State
the
conclusion
0
hence
there
random
is
a
signicant
variables
X
and
correlation
between
the
Y
Alter natively
B
xs
E
D
use
=
the
stats
88
57
3
83
78
4
92
76
Alternate...
β
&
ρ
above
methods,
‘LinReg
t-test’
under
the
RegEqn
a+b*x
tests
can
see
equation
t
50
30
6
67
67
7
100
96
8
100
74
9
73
65
10
90
87
11
83
78
12
94
89
=“Linear
Since
Reg
p-value
PVal
0.000679
df
10.
a
31.7999
b
0.71072
s
8.83862
SESlope...
0.146776
2
r
0.701019
r
0.837269
t Test”
=
0.00679
<
0.05,
we
reject
H
0
Statistical
modeling
of
4.8422
one
5
menu
of
your
GDC,
and
≠...
you
150
the
=LinRegtT
2
E1
to
F
ys
.
screen.
r,
t,
p-value,
regression
and
line
all
in
Exercise
1
The
4D
scores
shown
of
below
,
tness.
As
includes
a
height
the
in
in
school
higher
the
cm,
a
scores
table,
the
weight
in
physical
indicating
data
kg,
tness
gathered
and
age
in
on
62
0
45
44.9
2
94

50
42.
3
8
8
39
30.2
4
62
9
48
4.4
5
58
8
30
4.8
6
86
0
50
38.4
7
59

54
54.
8
72
40
38.6
9
54
53
52.4
0
60
20
38.6
regression
of
line
9
is
students
to
by
be
used
tting
to
predict
S
with
S
and
one
W-Weight
the
of
students
years.

0
are
physical
the
H-Height
9
test
better
A-Age
physical
the
tness
variables
A,
H,
W.
By
nding
which
the
r
regression
b
Find
c
Determine
The
the
body
S
Justify
at
and
A,
be
used
H,
and
with
S
S
to
and
W,
nd
the
determine
equation
of
line.
the
and
mass
Body
S
should
equation
one-year-old
Hear t
for
variable
between
2
with
in
S-Score
scores
or
students
shown
Student
A
10
of
5%
the
in
the
level
of
variable
grams
mice
regression
are
and
given
signicance
you
the
in
line.
chose
hear t
the
is
the
correlation
signicant.
mass
table
if
in
mg
of
ten
below
.
30
37
38
32
36
32
33
38
44
38
36
56
50
40
55
57
43
60
70
44
the
use
one-year-old
of
linear
mouse
regression
whose
body
to
estimate
mass
is
35g,
the
and
hear t
nd
mass
this
of
a
value.
Chapter
4
151
3
The
table
below
millimetres
of
the
form
correlation
a
State
b
Determine
and
a
c
Inter pret
d
Justify
An
needed,
Y,
90
92
chemical
is
at
pressure
the
rest
from
a
of
10
adults
consists
hear t
muscle
between
bivariate
of
in
two
contracts)
beats).
and
Assume
distribution
that
with
35
40
45
50
55
60
65
98
00
03
05
08
0
0
hypothesis.
product
moment
correlation
coecient
for
this
data
p-value.
p-value
the
value
to
when
values
ρ
80
nding
experiment
Blood
sample
30
the
its
pressure
hear t
25
the
diastolic
4
random
suitable
state
the
20
Y-Diastolic
a
Hg.
(pressure
when
coecient
X-Systolic
blood
mm
systolic
(pressure
values
the
mercur y
,
measurements:
diastolic
shows
is
of
in
line
an
the
of
best
adult
performed
dissolve
compound.
a
context
who
ve
cer tain
The
t
of
of
the
Y
has
on
a
times
X,
and
systolic
by
substance
summar y
problem.
use
value
measuring
that
statistics
it
to
of
138 mm
the
weighs X
estimate
the
Hg.
temperatures
grams
in
a
are:
2
∑
x
= 182;
Find
the
∑
y
=
200;
regression
substance,
x
grams,
∑
line
y
= 9850;
of
that
X
will
on
∑
Y,
xy
and
dissolve
= 8390
use
in
a
it
to
litre
nd
of
the
the
weight
chemical
of
at
the
90
degrees.
5
The
summar y
statistics
for
a
given
data
set
2
n
= 10;
Find
∑
the
x
= 30;
1
It
with
the
220;
the
y
∑
= 86;
regression
∑
y
= 1588;
∑
lines a
Y
on
determined
speed
of
the
model.
that
car,
The
Y
the
(in
temperature
km/h),
summar y
in
a
= 8;
Find
of
b
∑
x
=
the
440;
car
speeds
driving
Statistical
x
xy
and
= 580
b
X
on
Y
28, 400;
the
data
regression
at
an
modeling
∑
y
coecient,
correlation
in
car
are
tires, X
to
given
a
(in
°C),
varies
linear
as:
2
=
correlation
appropriate
152
∑
signicant
The
X
according
statistics
2
n
follows:
QUESTIONS
been
regression
=
of
as
exercise
STYLE
has
x
equations
Review
EXAM
∑
is
2
at
var y
line,
average
the
=
606;
and
5%
from
of
y
=
determine
49, 278;
if
∑
there
is
xy
= 37,000
evidence
level.
20
km/h
estimate
speed
∑
55
the
to
tire
km/h.
90
km/h.
Using
temperature
of
a
an
2
If
Var(X )
=
determine
3
Find
a
4
the
and
Var(Y )
largest
smallest
positive
The
15
the
r
value
correlation
independent
=
7
for
possible
at
that
the
random
two
random
covariance
will
10%
give
of
X
signicant
signicance
variables
variables
and
X,
Y,
Z
and
evidence
level
and
X
Y,
Y.
when n
each
of
=
have
20.
a
mean
2
of
0
and
a
variance
determine
if
the
the
value
5
The
data
st
in
c
d
ten
time
time
and
cer tain
at
two
b
+
Y,
V
=
following
the
V
dr ug
+
Z,
and
variables
variables
and
X
in
each
W
and
=
X
−
Z,
determine
case
are
W
absorbed
dierent
by
times,
the
and
adult
the
body
was
following
20.
55.0
39.6
24.
3.2
9.5
22.3
35.
4.2
.
8.7
52.3
4.2
26.5
29.3
.2
25.
33.2
the
its
your
of
An
11th
product
p-value
the
patient
was
rate
of
absor ption
rate
the
next
for
rates
to
time,
this
the
Show
mutually
Show
a
correlation
coecient
for
this
data,
two
if
=
and
if
of
it
0,
1st
be
time.
25
the
problem
is
two
have
found
times
a
the
2nd
Predict
children
and
was
what
been
the
0.532.
times
nd
the
for
this
dr ug
Y
a
the
same
1%
in
was
on
had
a
dr ug
patient.
correlation
Use
lines,
but
the
correlation
the
regression
time,
underwent
that
positive
dierent
the
of
tested
would
was
there
two
to
time
dierent
the
r
2nd
context
line.
the
group
determine
that
the
unable
19.8
dr ug,
between
in
regression
absor ption
The
moment
p-value.
equation
level
b
a
the
that
X
0.3
Inter pret
between
a
of
V
=
44.5
state
tests
6
and
U
of
implies
patients
Determine
and
b
U
If
recorded.
2nd
a
a
percentage
tested
.
correlation
found
independent:
σ
of
absor ption
coecient
signicance
the
absor ption
administered.
X
and
X
on
Y
are
per pendicular.
that
if
r
=
±1,
the
regression
lines
of
Y
on
X
and
X
on
Y
are
identical.
c
Given
that
the
equation
of
the
Cov ( X , Y
y
=
a
+
bx,
where
b
regression
of
Y
on
X
is
)
,
=
line
and
the
regression
line
of
Var ( X )
Cov ( X , Y
X
on
Y
is
x
=
c
+
dy,
where
d
,
Var (Y
if
d
If
b
and
the
and
nd
are
least
the
r,
d
positive,
squares
least
the
both
product
and r
regression
squares
)
=
line
regression
correlation
=

of
line
if
bd
Y
of
show
that
r
=
+
bd
)
on
X
b
X
on
and
is
Y
d
are
given
is
given
both
by
by
y
x
=
=
negative.
12
+
0.19x,
−4.4
+
0.77y,
coecient.
Chapter
4
153
Chapter
Denition:

The
summary
obser ved
value
r
of
the
sample
linear
correlation
coecient
is
dened
as
n
∑
(x
−
x )( y
i
y )
−
i
r
d
∑
i =1
=
d
x
y
=
n
2
n
2
(x
∑
x )
−
∑
i
i =1
The
d
∑
2
( y
2
∑
x
d
y
y )
−
i
i =1
sample
product
moment
cor relation
coecient R,
for
n
paired
obser vations
(x,
y)
n
(X
∑
−
X )(Y
i
−
Y
)
i
i =1
on
X
and
Y,
is
R
=
n
n
2
∑
(X
−
X )
2
∑
i
i =1
(Y
−
Y
)
i
i =1
n
∑
X Y
i
nX Y
i
i =1
An
alter native
formula
for
R
is
R
=
n
2
∑
2
(X
− n X
2
2
)(Y
i
−
nY
)
i
i =1
We
can
classify
by
using

±

0.5

–

0.

–0.5
<
R
≤
–0.

–0.
<
R
<
0.
The
the
the
following
indicates
≤
≤
R
R
≤
<
≤
R

a
<
0.5
population
of
general
perfect
indicates
–0.5
E[( X
ρ
strength
a
indicates
a
a
indicates
positive
strong
weak
a
between
the
random
variables X
and
Y
classication:
strong
indicates
− E [ X ])(Y
correlation
positive/negative
indicates
product
the
a
correlation.
negative
positive
weak
highly
moment
correlation.
correlation.
correlation.
negative
weak
correlation.
correlation,
coecient
or
no
correlation.
is
− E[
Y ])]
=
2
E [( X
− E [ X ])
2
] E[
Y
− E [
Y ])
]
or
Cov ( X , Y
ρ
Var [( X )
Denition:
X
)
=
and
of
the
Y
of
two
Cov(X,
Var (Y
For
a
)
each
joint
=
E[(X
of
n
paired
distribution,
variables’
Y )
set
–
Cov(X,
deviations
E(X ))(Y
–
obser vations,
from
Y
),
their
E(Y ))]
=
is
a
=
E(X ),
x
or,
=
μ
E(Y )
E[(X
y
Cov(X,
Y )
=
E(XY )
–
μ
x
154
Statistical
modeling
y
covariance
measure
respective
of
the
means,
)(Y
–
x
where
the
)],
y
of
the
mean
i.e.
random
value
of
variables
the
product
Properties
of

Symmetr y:
2
For
two
covariance
Cov(X,
Y )
independent
=
Cov(Y,
events
Theorem
:
If
X
and
Y
are
Theorem
2:
If
X
and
Y
have
Theorem
3:
For
if
If
and
X
only
and
Y
ρ
if
X
=
have
and
and
and
Y,
Cov(X,
independent
a
with
linear
a
Var(X )
Y )
variables,
relationship,
bivariate
normal
=
Cov(X,
=
0.
ρ
then
i.e.
Y
=
=
X )
0.
mX
+
c,
distribution, X
ρ
then
and
Y
=
are
±.
independent
0.
a
bivariate
n
distribution
Y
X
X ),
normal
distribution
that ρ
such
=
0,
then
the
sampling
2
has
R
the
student’s
t-distribution
with
(n
–
2)
degrees
of
freedom.
2
1
Using
the
R
principle
of
least
squares,
the Y
on
X
regression
line
is
given
by
the
following
formulae:
n
n
⎛
⎜
⎞
∑
[( x
−
x )( y
i
−
⎛
y )]
⎟
i
⎜
i =1
y
−
y
=
⎞
⎟
(x
−
x )
or
y
−
y
=
2
∑
⎝
(x
−
⎟
⎜
⎟
⎟
⎜
⎠
⎝
∑
of
the
Y
on
X
regression
line
for
the
X
on
Y
regression
n
(x
−
x )( y
i
−
⎞
⎛
⎟
⎜
y )
i
i =1
− x
=
⎟
(y
−
y )
=
line
− nx
⎟
i
⎠
(X
)
)
is:
⎞
∑
x
y
i
− nx
⋅
y
⎟
i
⎟
∑
( y
−
⎟
⎜
⎟
⎜
⎠
⎝
y )
i
i =1
gradient
(y
−
y )
n
2
2
∑
( y
of
the
X
on
Y
regression
⎟
2
− ny
)
⎟
i
⎠
i =1
Cov
The
x
( X ,Y
⎜
n
⎜
⎝
x )
i =1
⎜
⎜
−
n
⎛
x
(x
⎟
is
Var
formula
2
i =1
Cov
∑
⎟
2
x )
i
i =1
gradient
⎜
y )]
n
⎜
The
− nx
i
⎜
n
The
y
i
i =1
⎜
⎜
x
∑
line
( X ,Y
is
)
.
Var
(Y
)
Chapter
4
155
+∞
Answers
Chapter
6
a
The
function
b
The
modal
1
Mode (X )
Exercise
=
−1,
1
b
Mode (X )
=
m
=
1
Median,
µ
0.95
=
m
=
σ
a
=
σ
.72
Mode (X )
Median,
(x)dx
=
1
=
m
0
=
b
=
doesn’t
exist.
B
a
P(X
=
2)
=
0.24
b
P(X
=
3)
=
c
P(X
=
4)
=
0.0625
d
P(X
=
5)
=
0.104
0.000182
a
P(X
≤
4)
=
0.684
b
P(X
>
6)
=
0.000729
c
P(5
d
P(1<
2

Mode (X )
0.695
value
2
2
2
f
1
1
Median,
=
if
∞
check
µ
well-dened

Ski lls
a
is
Median,
=
m
≤
X
≤
7)
=
0.158
X
≤
7)
=
0.009
0
=
3
P(X
4
a
P(X
≤
3)
=
=
5)
0.980
=
0.0000377
b
P(X
≥
4)
=
0.000512
5
a
P(X
=
4)
=
0.0921
b
P(X
≥
7)
=
0.377
0
3
µ
=
σ
=
µ
=
0
σ
=
0.342
4
0.487
Investigation
c
Mode (X )
=
3
a
Median,
m
µ
=
4.6
σ
=
0.839
=
0.36
b
0.0081
c
0.409
4
Exercise
1
For
C
1B,
Question
1:
2
3
a
b
2
+
2
2
a
E(X )
=
1.67,
Var(X )
=
1.11
b
E(X )
=
7.14,
Var(X )
=
43.9
c
E(X )
=
2,
d
E(X )
=
1.14,
3
1
4
a
f
=
′(x)
,
x
≠
2;
2
(2
f
(x)dx
x )
=
−ln(2
−
x)
+
c,
x
1
f
′(x)
=
3e
;
f (x)dx
=
2
For
3 x +1
3x + 1
b
<
e
+
B,
Var(X )
=
2
Var(X )
Question
=
0.155
2:
c
3
⎛ π
c
f
=
′(x)
− cos
2x
⎜
⎟
3
⎝
f
(x)dx
;
=
cos
4
E(X )
=
4,
b
E(X )
=
1.43,
Var(X )
c
E(X )
=
3.33,
d
E(X )
=
1.01,
a
E(X )
=
1.37
b
Mario
c
6
=
12
Var(X )
=
0.612
⎠
⎛ π
9
a
⎞
2x
=
7.78
⎞
⎜
⎟
3
⎝
Var(X )
+
c
Var(X )
=
0.00916
⎠
2
5
x
2
d
f
′(x)
=
4x(x
−
2);
f
⎪
k
=
3
=
F (x )
b
x
+ 7x
=
four
shots
to
destroy
the
balloon.
x
= 0, 1
,
D
1
a
P(X
=
2)
=
0.16
b
P(X
1
,
x
=
4)
=
c
P(X
=
9)
=
0.235
d
P(X
=
32)
2
a
P(X
≤
4)
=
0.508
b
P(X
>
6)
c
P(5
≤
3
a
p
0.4
4
a
X
0.188
2
=
0.089
> 2
=
0.109
10
0,
⎧
⎪
⎪
b
< 0
24
⎩
a
make
students
Exercise
x
+ 6
,
⎨
⎪
a
must
2
⎪
2
c
A
⎪
a
4x
3
0,
⎧
1
3
x
5
Exercise
4
(x)dx
F (x )
=
<
x
X
≤
7)
=
0.525
d
P(8
<
X
≤
11)
=
0.326
1
=
b
P(3
≤
X
≤
5)
=
0.503
2
9x
x
,
⎨
= 1
,
x
2,
3,
4
~
⎜ 2,
NB
b
⎝
⎪
1
,
⎩
>
x
47
3
1
⎛
20
c
⎪
128
32
4
4
5
a
0.313
6
a
0.264
b
b
0.999999
0.0473
7
c
P(X
≤
2)
=
10
⎧ x
3
a
P( X
=
x)
=
,
x
= 0, 1
,
need
2
7
0,
⎩
Mode
=
P( X
=
x)
=
a
0.0829
Exercise
⎨
so
,
x
= 1
,
3,
5,
7,
⎩
0,
1
9
b
25
⎪
b
it
is
almost
more
cer tain
than
a
that
dozen
he
will
students.
0.589
E
1
x
⎪
a
1,
inter view
otherwise
2
⎧
4
to
6
⎨
⎪
b
≈
+ 1
⎪
m
=
G (t )
1
=
3
t
+
16
7
1
2
t
+
4
1
3
t
+
8
4
t
+
4
16
otherwise
x
0

2
3
4
i
5
a
b
=
1
0,
⎧
x
1
4
6
4
1
16
16
16
16
16
p
< 0
i
⎪
π
b
F (x )
=
⎪
2
2
cos x ,
0
≤
x
⎨
≤
3
2
⎪
x
π
⎪
1
,
x

2
3
k
…
k
3
⎩
⎛
1
c
P
(
X
≥
)
6
156
Answers
5
25
p
π
=
0.732
…
i
>
…
i
6
36
216
−
1
k −1
5 ⎞
⎜
⎟
⎝
6 ⎠
1
5
=
×
k
6

not
one
one
6
…
not
2
3
a
P(X
=
0)
8
=
b
P(X
≤
1)
9
1
c
P(X
≥
3)
5
a
E(2X
6
Var(5X
7
E(3X
−
3)
=
45
b
Var(2X
−
11)
=
192
=
3
+
3)
=
90
1
=
d
+
2)
=
6
2
+
2,
Var(3X
+
2)
=
3
k
27
3
Exercise
4
a
E(X )
=
b
E(X )
=
p,
Var(X )
=
p(1−
r
Var(X )
=
Question
E(X )
=
Question
Question
Var(X )
=
1
E(X )
=
2
6,
Var(X )
=
30
3
a
E(X
b
E(2Y
=
,
Var(X )
−Z )
c
E(2Z
−7X )
d
E(X
−Y
+
Z )
=
e
E(X
+Y
−
Z )
=
f
E(3Z
−
Var(3Z
2
P(X
=
2,
=
Var(X
22,
+
Y )
=
3,
Var(2Y
Var(2Z
20,
−
=
1.9
−
Var(X
14,
Z )
=
8.4
7X )
=
35.7
−Y
Var(X
+
+Y
Z )
−
=
Z )
4.7
=
2X
+
4Y )
=
4.7
10
4
4t
4
a
Y )
=
2
6
+
3
1
E(X )
1
2
p
1
2,
B
pq]
rq
,
p
5
p)[=
=
1)
=
b
G (t)
+
2
7
7
c
The
expected
d
The
maximum
number
of
shots
is
−
2X
+
4Y )
=
49.6
2t
=
2
a
3
Var(2X
E(3X
4
Var(X
+
5Y )
=
31
b
Var(11Y
−
7X )
=
703
t
−
3Y )
=
72
72p
2.
20
number
of
shots
is
5.
−
Y )
=
20
p
Exercise
F
Exercise
C
1
=
2
2
⎛ 2 + 7t
1
a
G
(t)
X
+
=
⎞
+ 3t
⎜
Y
⎟
⎝
12
2
b
P(X
+
Y
≤
1)
a
X
x
0

1
1
2
2
⎠
13
=
c
E(X
+
Y
)
=
P{X
9
=
x}
6
3
2
⎛
2
a
G
(t)
X
=
⎞
t
⎜
1
⎟
2
1
+Y
⎝ 6 − 7t
+ 2t
E(X )
⎠
=
,
Var(X )
=
2
b
E(X
+
Y
)
=
15
c
Var(X
+
Y
)
=
24
b
0.25(t−1)
3
a
G
(t)
b
0.989
=
e
0.15(t−1)
,
G
X
(t)
=
e
4
Since
we
have
6
independent
G
Y
(t)
=
e
coin,
we
add
six
instances
⎜
qt
Review
2
⎠
a
a
exercise
E(Y
)
=
P(1
≤
X
≤
X
=
3,
Var(Y
x
P{X
4)
b
P(X
≥
2)
)
=
d
0
=
∈ [
0.67,
6.67]

2
1
3
3
2
1
25
E(X
)
=
,
Var(X
)
=
9
3
c
E(X )
5
=
d
Var(X )
=
4
a
G
(t)
=
b
16
0.6(t – 1
e
0.12(t − 1),
),
G
X
(t)
=
e
0.28(t – 1)
G
Y
(t)
=
Since
die,
e
we
we
have
add
4
independent
four
instances
b
1
a
i
c
ii
0.264
0.105
c
iii
0.0625
iv
Eight
a
=
0
2
f
b
(x )
=
⎨
=
≤
z
c
0
⎩
E(X )
x
≤
2
⎪
b
E(Y
)
=
a
E(X
+
,
b
E(3X
c
Var(X
X
)
Var(Y
)
=
X
)
=
9,
Var(X
9,
Var(3X
)
=
+
X
+
X
)
=
12
36
2
+
X
+
X
+
3X
)
=
48
a
a
E(X
+
)
X
=
144
+
X
+
X
+
X
)
=
10

Var(X
+
X
+
b
E(Y
Y
+
Y
c
E(X
X
+
X
+
X
)
=
5
check
P(X
=
2)
=
0.311
=
−
1
b
P(1
≤
X
≤
3)
=
0.786
,
a
=
−
,
2
2
+
+
X
Var(X
1
1
a
X
otherwise
4
Chapter
a
same
9
+
=
Var(6X
Ski lls
the
⎧ x
⎪
a
of
variable
0.922
3
b
rolls
the
8
3
0.0117
of
Z
4
2
y
x}
=
5
1
X
1
=
625
5
same
2
624
4
the
⎟
1
⎝
3
of
variable
3
c
2
the
Z
r
r
1
of
pt
⎛
4
ips
0.05(t−1)
,
a
=
+
+
X
)
X
+
=
+
X
15
X
+
+
X
Var(Y
X
+
+
X
Y
+
+
Y
+
Y
+
Y
+
Y
+
Y
+
Y
)
Y
=
)
)
=
9
25
=
14
2
3
3
5
a
X
=
x

2
3
4
i
Exercise
1
a
A
E(3X )
=
15.9,
Var(3X )
=
P{X
10.8
=
1
1
1
1
4
4
4
4
x }
i
b
E(X
c
E(4X
+
+
3)
1)
=
d
E(2X
−
5)
e
E(kX
+
2
a
E(3X
+
3
E(2Y
4
a
8.3,
Var(X
+
3)
=
1.2
5
=
22.2,
Var(4X
+
1)
=
19.6
E(X )
=
2
=
5.6,
Var(2X
−
5)
=
4.8
2
p)
=
2)
=
5.3k
+
p,
14
Var(kX
b
+
p)
Var(3X
−
=
1.2k
2)
=
Y
=
E(3
1)
−
=
2,
2Y )
Var(2Y
=
1
−
1)
b
=

3
1
1
1
2
3
6
21.6
Y
=
y }
i
3
Var(3
2
i
P{
−
y
−
2Y )
=
8
5
E(Y
)
=
3
Answers
157
b
Z
=
z

2
3
4
6
1
5
8
24
8
9
2
1
5
1
1
1
1
6
24
8
12
24
24
Chapter

i
P{Z
Ski lls
check
z }
=
i
25
E(Z
)
µ
=
x
2
a
P(X
3
a
P(Y
=
σ
3.62,
=
=
5)
=
2.82
0.03125
b
P(3
≤
X
<
8)
=
0.322
b
P(3
≤
X
<
8)
=
0.569
=
6
6
2
1
=
0)
=
0.670
2
Exercise
Exercise
A
D
2
1
1
a
P(Y
−
Z
−W
<
0)
=
a
x
P(X
c
P(3X
+
Y
+
+
Y
Z
>
+
Z
W
+
d
P(X
−
3Z
≤
2Y
+
e
P(X
−
4Z
≤
3X
−
2Y
+
=
W
)
2Z
3W
=
=
)
=
33
x
b
=
=
29.75,
s
=
x
67,
0.892
s
=
=
25.3
1518
2
2
x
3
a
=
0.65,
s
=
0.864
2
x
=
33.7
b
s
=
23.8
b
[
b
[0.104,
b
n
0.671
=
)
2
s
0.5
Exercise
0.303
f
P(W
a
0.551
b
0.162
3
a
0.159
b
0.201
c
0.564
4
a
0.0186
b
0.0000155
c
0.149
d
0.118
1
≤
0)
)
2
Exercise
−Y
>
W
11,
2
c
b
=
0.5
0.994
1
2
B
a
[4.19,
c
[2819,
a
[3.90,
c
[320,
3
n
4
[72.3 g,
=
5
a
5.81]
12.8,
9.20]
2889]
6.10]
0.167]
325]
30
77.2 g]
E
a
P(1
b
P(2
≤
X
≤
3)
=
x
=
15.8
=
10
0.777
Investigation
2
0.053
3
a
≤
≤
X
8)
=
0.992
c
P( X
≥
0.8)
=
0.639
a
P( X
≥
37)
=
iii
0.228
b
b
P(X
+
X
+
X
+
X
+
i
X
>
180)
=
At
[95.0,
105.0]
ii
[96.9,
[97.8,
102.2]
iv
[98.7,
the
a
≤
P( X
70)
=
0.920
b
P(T
≤
800)
=
we
P(1.5
b
P(1.25
c
P( X
d
P( X
e
P
≤
≤
X
≤
≤
X
≥
2.5)
<
397)
=
0.834
3
a
take,
a
=
=
the
the
condence
sample
inter val
1
a
[14.45,
c
[3430,
2
a
[1.94,
c
[319.6,
a
[67.6
b
[
25.81,
b
[0.112,
20.19]
3526]
0.421
3
)
15.55]
0.923
0.0353
<
X
=
8.06]
0.159]
324.9]
g,
82.5
g]
0.0983
2
X
<
<
2.2)
0.193
b
0.579
=
4
a
5
a
[483.4,
c
For
x
=
13.35
0.00469
b
b
The
condence
level
is
90%.
0.280
c
the
588.6]
same
b
set
of
[463.0,
data,
the
609.0]
higher
the
signicant
0.917
the
wider
the
condence
inter val
we
get.
62
Exercise
Review
narrower
C
level
4
the
0.639
1
<
2
P(1.8
=
1.35)
0.48)
1
(
2
101.3]
larger
F
a
f
the
get.
Exercise
1
level,
0.782
we
Exercise
signicance
0.355
size
4
same
103.1]
D
exercise
1
1
0.796
2
b
P(X
d
E(Z )
≥
6)
=
=
0.384
c
P(X
+
Y
<
5)
=
a
[
2.18,
1.98]
c
[
0.0609,
b
[
13.73,
7.44]
0.0859]
0.173
2
7
a
d
=
Bob−Rick
3
−8
−3
9
6
−9
3
2
−4
−3
3
−7
i
Var(Z )
=
77
b
e
The
random
variable
Z
has
no
[−4.27,
Exercise
distribution
3
a
0.00621
c
N(720,
d
0.299
since
7
=
E(Z )
b
≠
4.60]
Poisson
Var(Z )
=
1
0.0787
E
77
a
Since
the
p-value
is
0.000008
<
0.1
we
reject
the
2
2
4
a
20
or
5
a
0.377
20
2
+
12
)
=
N(720,
(4
34 )
null
)
b
47
b
16
b
0.196
Answers
Since
the
sucient
c
the
45
c
158
hypothesis
5%
Since
at
p-value
the
is
evidence
p-value
is
>
the
0.05
null
level.
we
have
no
hypothesis
level.
evidence
at
signicance
1%
reject
0.010566
sucient
the
signicance
0.095581
to
signicance
the
10%
to
reject
level.
>
the
0.01
null
we
have
no
hypothesis
at
2
a
Since
no
the
p-value
sucient
hypothesis
b
Since
have
the
no
Since
the
3
a
Since
the
at
is
the
is
p-value
reject
0.029673
1%
p-value
>
we
the
>
to
0.005283
the
we
b
the
c
we
level.
reject
4
H
p-value
the
null
0.743755
evidence
10%
Since
reject
we
the
Since
null
level.
0.05
the
sucient
the
0.01
<
Since
at
signicance
0.036819
a
null
0.01
<
1%
3
have
level.
reject
signicance
at
is
0.1
signicance
evidence
hypothesis
the
to
10%
p-value
the
null
0.131776
evidence
at
sucient
hypothesis
c
is
to
the
p-value
is
p-value
evidence
signicance
:
“The
mean
the
to
volume
we
have
null
no
hypothesis
level.
the
<
5%
0.115078
at
1%
0.1
0.004763
at
sucient
the
reject
signicance
hypothesis
>
>
reject
0.05
we
reject
signicance
0.01
the
we
null
level.
have
no
hypothesis
level.
is
( µ
120 ml.”
=
120)
0
the
null
hypothesis
at
the
5%
signicance
level.
H
:
“The
mean
volume
is
not
( µ
120 ml.”
≠
120)
1
b
Since
the
p-value
sucient
at
c
the
Since
the
4
a
H
10%
the
null
:
is
evidence
reject
signicance
p-value
is
hypothesis
“The
0.109391
to
mean
>
we
null
have
no
Since
hypothesis
1%
the
weight
is
<
1%
0.01
we
reject
signicance
( µ
26 g.”
=
:
0.283654
signicance
a
to
level
correct
>
reject
and
0.01
the
we
null
conclude
volume
of
a
have
no
hypothesis
that
the
at
the
factor y
par ticular
ice-cream
product.
5
H
0
H
p-value
evidence
adver tised
level.
26)
the
sucient
level.
0.000153
at
0.1
the
:
“The
mean
life
expectancy
is
30,000
hours.”
0
“The
mean
weight
is
not
( µ
26 g.”
≠
( µ
26)
=
30, 000)
1
b
Use
the
z-test.
Since
the
p-value
is
0.00001
H
:
“The
mean
life
expectancy
is
less
than
30,000
1
<
0.01
1%
we
reject
signicance
har vested
5
a
H
:
the
level
snails
“The
mean
null
are
hypothesis
and
not
level
conclude
from
of
fat
in
at
the
that
( µ
hours.”
the
Since
the
population.
hypothesis
the
drink
that
the
30,000)
p-value
the
is
<
at
0.094543
the
10%
manufacturer
<
0.1
we
signicance
claims
a
reject
level
longer
the
and
life
null
conclude
expectancy
0
( µ
1.4 g.”
H
:
“The
=
mean
of
1.4)
level
of
fat
in
the
drink
the
LED
lamps.
is
1
Exercise
more
than
1.4
g.”
( µ
>
1
b
Use
the
z-test.
Since
the
p-value
is
0.335687
H
>
we
have
no
sucient
evidence
to
reject
:
“There
hypothesis
at
the
5%
signicance
level
the
“There
conclude
that
H
mean
:
“The
the
company
volume
of
claim
juice
is
in
a
dierence
dierence
in
in
nishing
nishing
( µ
times.”
( µ
times.”
=
≠
0)
0)
d
the
t-test.
Since
the
p-value
0.874063
>
0.05
we
and
no
sucient
evidence
to
reject
the
null
hypothesis
correct.
at
a
is
1
have
6
no
d
:
Use
null
is
0
H
0.05
G
1.4)
the
the
5%
signicance
level
and
conclude
that
there
bottle
0
is
is
300
( µ
ml.”
=
no
dierence
in
nishing
times
on
the
two
Rubik’s
300)
Cubes.
H
:
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mean
volume
of
juice
in
the
bottle
1
2
is
less
than
300
ml.”
( µ
<
H
300)
b
Use
the
z-test.
Since
the
p-value
is
:
“There
0.1
we
reject
the
null
hypothesis
0.004612
“There
at
the
level
and
conclude
that
the
less
volume
than
dierence
dierence
in
in
the
the
( µ
scores.”
( µ
scores.”
t-test.
Since
the
p-value
0.085622
<
0.1
reject
the
and
null
hypothesis
conclude
that
at
players
the
10%
score
a
signicance
better
result
stated.
using
the
new
type
of
dar t.
F
H
:
“There
is
no
dierence
in
the
weights.”
( µ
0
a
0)
0)
bottles
3
1
=
≠
d
the
when
Exercise
a
1
level
contain
is
10%
we
signicance
no
d
:
Use
<
is
0
H
Since
the
p-value
0.6382
>
0.05
we
have
no
H
:
=
0)
d
“Students
who
join
the
programme
drop
some
1
sucient
evidence
to
reject
the
null
weight.”( µ
hypothesis
>
0)
d
at
b
the
Since
5%
the
sucient
at
c
the
Since
signicance
p-value
0.10858
evidence
10%
the
level.
to
reject
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p-value
>
Use
0.1
the
we
have
null
no
have
hypothesis
at
level.
0.013122
>
we
have
no
the
no
0.01
the
evidence
to
at
signicance
reject
the
null
2
a
Since
null
b
1%
the
p-value
hypothesis
Since
the
the
hypothesis
5%
<
0.05
we
signicance
0.007494
at
the
<
0.01
we
reject
Since
the
sucient
p-value
1%
signicance
0.225675
evidence
to
the
10%
and
before
the
>
conclude
and
after
0.05
null
that
the
we
hypothesis
there
is
programme.
H
a
0.0455
2
a
E(X )
3
a
0.133
b
reject
>
0.1
the
we
null
=
0.773
signicance
25;
α
=
0.00130
b
b
0.978
0.972
level.
reject
the
exercise
level.
have
a
[602,
c
We
702]
b
notice
that
[586,
718]
no
a
higher
signicance
level
hypothesis
means
at
weight
0.121576
reject
the
1
c
in
level
to
hypothesis
Review
null
signicance
dierence
p-value
level.
0.027947
at
p-value
the
evidence
no
1
the
Since
sucient
5%
Exercise
sucient
t-test.
a
wider
condence
inter val.
level.
Answers
159
3
2
a
Dierence
2
9
6
8
5
4
5
8
7
r
=
–0.970;
Exercise
b
H
:
“There
is
no
dierence
in
there
is
a
strong
negative
correlation
5
C
potassium
0
(µ
levels.”
=
1
0)
p
=
0.00229.
Since
p
<
0.01,
we
reject
the
null
d
H
:
“There
is
a
≠
0)
dierence
in
hypothesis.
potassium
There
is
evidence
of
signicant
1
(µ
levels.”
correlation
between
the
two
variables
at
the
d
Since
the
p-value
0.46297
sucient
evidence
at
signicance
the
that
the
1%
there
two
is
no
types
to
reject
level
dierence
of
>
0.01
the
null
and
in
we
we
have
hypothesis
2
p
=
level.
0.394.
evidence
conclude
measurement
biochemical
1%
no
is
of
not
the
analyzers.
Since
to
reject
evidence
two
p
0.05,
the
of
variables
>
there
null
the
not
enough
hypothesis,
signicant
at
is
5%
i.e.
correlation
there
between
level.
2
3
a
x
=
5.33
b
[3.72,
c
In
s
=
3
4.38
=
0.0343.
6.94]
of
the
sample
of
15
cases,
the
mean
obser vations
will
fall
value
taken
within
the
of
from
correlation
a
10%
the
[3.72,
5
a
x
=
a
30 s
c
H
:
51.2
b
is
0.1,
we
reject
evidence
between
the
two
of
the
null
signicant
variables
at
the
condence
D
a
Use
W,
because
it
has
the
strongest
correlation,
95%
average
r
2
x
b
“The
There
<
6.94].
1
4
p
level.
Exercise
inter val
Since
hypothesis.
99%
population
p
=
30.97
time
is
s
=
–0.546
0.298
( µ
30 s.”
=
=
b
S
c
p
=
–1.07W
+
114
30)
0
H
:
“The
average
time
is
more
than
=
0.102.
Since
p
>
0.05,
there
is
no
evidence
30 s.”
1
to
( µ
>
reject
the
evidence
d
We
null
hypothesis,
i.e.
there
is
not
30)
use
t-test
since
the
standard
deviation
of
signicant
correlation
between
is
the
two
variables
at
the
5%
level.
unknown.
Since
the
the
null
p-value
0.000163
hypothesis
average
time
is
and
more
<
0.05
conclude
than
30 s,
we
2
r
=
3
a
H
b
r
sets
up
the
the
therefore
speedometers
=
to
:
ρ
=
0;
a
x
=
7
a
i
b
i
c
8
n
=
ρ
≠
0
:
1
p
=
0.000008
There
is
ver y
strong
evidence
to
indicate
a
positive
association
between
the
random
speed.
15.37,
x
=
thus
the
sample
23.5
[21.8,
25.2]
24.9]
⊂
[21.8,
condence
inter val
condence
inter val.
a
i
]– ∞,
b
i
0.569
size
=
should
ii
s
ii
[22.0,
25.2],
is
a
be
y
=
X
4
x
=
0.6y
and
Y
the
so
a
subset
5
a
Y
b
X
]– ∞,
ii
0.705
of
a
+
12.4;
66.4
on
X:
on
Y
y
:
=
x
2.48x
=
from
10%
to
5%,
a
Chapter
II
error
increases
is
approximately
97.
29.6
exercise
8.96[
Type
the
from
1.17;
–
95%
a
I
r
=
0.975,
p
=
sucient
0.000039;
evidence
of
p
a
<
0.05,
strong
hence
there
positive
error
probability
0.569
to
between
the
two
random
variables.
of
b
Type
+
0.380y
relationship
decreases
y
grams
90%
of
ii
probability
10.8;
24.9]
is
When
+
10.5
Review
9.19[
0.623x
16.
1
a
150 mg
210
[22.0,
c
82.8;
the
show
d
b
+
H
0.962;
variables
6
1.91x
0
=
a
higher
y
reject
that
c
company
0.756;
75.8 °C
0.705.

2
10.2
3
r
4
a
=
0.299
1
Ski lls
check
,
no
b
0,
no
2
–8
1
2
a
2E(Z )
b
4Var(Z
c
E(X
a
x
=
a
3E(Y
)
–
)E(Y
)
0.382
=
)
+
9Var(Y
2E(X )
)
+
5
4Var(X )
a
r
b
The
)E(Z )
y
43.5,
Var(X
3
–
=
0.991;
–31.75,
Var(Y
0.951
c
y
)
=
98.69
0.938
d
0.732
=
c
20
d
p
=
p
0.905x
6
1
r
=
–0.382;
2
r
=
0.794;
160
Answers
there
there
is
is
a
a
weak
strong
negative
positive
correlation
correlation
d
r
=
+
0.00620;
A
0.383
2.78x10
suggests
the
correlation
Exercise
=
p-value
between
46.25,
b
=
two
a
strong
random
relationship
variables;
2.59
p
<
0.01,
between
hence
the
there
random
is
a
strong
variables.
Index
Page
numbers
in
italics
indicate
the
Answers
section.
negative
A
correlation
125–6
G
no
actuarial
science
correlation
geometric
5
positive
alter native
126
hypotheses
correlation
distribution
sample
linear
correlation
expected
coecient
128,
weak
26,
and
strong
value
and
variance
154
of
distribution
27,
38
B
Ber noulli
12–13,
124–5
100–1
correlation
geometric
random
128,
38
variables
18–20,
38
131
Ber noulli
trials
12,
13,
21,
23
Investigation
covariance
Ber noulli,
Jacob
135–6
13
negative
denition
Bessel,
Friedrich
136,
binomial
154
68
distribution
proper ties
binomial
18
distribution
26,
29,
of
covariance
20–3
136–7,
38
geometric
random
variables
18
155
bivariate
distributions
122,
123–4,
Gosset,
critical
value
William
Sealy
91
101
154–5
cumulative
bivariate
distributive
function
5,
H
normal
38
distribution
Halley
,
138–9
discrete
correlation
and
continuous
quantities
correlation
and
covariance
causation
Edmund
hypothesis
124–8
5–6
bivariate
128–30
100–1,
120
normal
distribution
135–6
5
testing
138–9
D
hypothesis
testing
four
138–9
D’Alember t,
linear
regression
of
testing
covariance
exercise
152–3,
Moivre,
for
and
Y
of
dependence
freedom
is
known
μ
when
120
90
variables
2,
is
testing
unknown
for
μ
when
105–8,
120
5–6
signicance
James
for
101–4,
68
σ
random
120
testing
hypothesis
Gustav
139–40
Auguste
101,
68
of
Dodson,
Bravais,
37,
130–4
discrete
X
5,
σ
Dirichlet,
t-statistic
Abraham
162
distributions
hypothesis
85
hypothesis
degree
sampling
Charles
136–7
de
review
in
25
141–50
Darwin,
proper ties
Jean
steps
testing
for
matched
5
pairs
133
109–11,
121
E
I
C
empirical
Cauchy
,
Augustin
r ule
independent
68
Equitable
Central
Limit
67
Theorem
41,
Life
variables
estimates
80,
independent
estimators
condence
inter vals
for
79,
84,
39,
80,
and
Chebyshev
,
estimators
Pafnuty
variance
for
80–1
the
mean
L
of
a
normal
Laplace,
Pierre-Simon
Law
of
Large
68
Numbers
level
of
signicance
41
68
random
variable
81
85,
101
100
well-dened
condence
inter vals
for
mean
denitions
algebra
condence
inter val
for
combination
97–9,
Lévy
,
Maurice
Lindeberg,
Jarl
68
Waldemar
condence
inter val
for
μ
85–9,
inter val
for
μ
90–5,
68
independent
random
normal
random
variables
linear
regression
141–4,
58
linear
transformation
148
of
a
single
42–6
variable
42–4,
of
a
single
76
transformation
of
two
or
linear
transformation
of
two
or
120
variables
47–57
more
variables
47–50,
76
90
review
value
exercises
74–5,
159
Ludo
12–16
85
sampling
random
variables
3,
distribution
of
the
Lyapunov
,
Aleksandr
68
6
mean
correlation
normal
58–63
transformation
more
continuous
of
when
linear
unknown
investigation
combination
120
variable
condence
linear
when
linear
known
condence
of
120
variables
is
76
matched
independent
is
42,
84–5
linear
pairs
82–3
the
expectation
σ
81
estimates
71
and
σ
32,
76
71
unbiased
website
76
variables
91
estimators
Theorem
random
120
the
denitions
mean
random
58–63,
120
77
claims
normal
5
68–72,
64–7
124–8
M
correlation
and
causation
correlation
coecient
128–30
F
Maclaurin’s
formula
27
126–7,
Fisher,
Sir
Ronald
Aylmer
85,
87
Markov
,
Andrei
Andreyevitch
68
133–4
Index
161
matched
pairs
signicance
pairs
moment
probability
97
testing
109–11,
for
review
121
and
hypothesis
25–35
exercise
probability
50
estimators
generating
functions
matched
36–7,
generating
making
157
functions
25,
38
estimates
testing
sense
of
review
exercise
Type
and
I
data
79
117–19,
Type
80–1
100–11
II
161
errors
N
proper ties
negative
binomial
20–2,
Neyman,
27,
distribution
sum
38
Jerzy
of
112–16
39
unbiased
independent
variables
and
32–4
87
estimators
variance
random
of
for
a
variable
the
mean
normal
81
Q
normal
null
random
variables
hypotheses
failure
to
Type
and
81
well-dened
100–1
reject
qualitative
and
quantitative
data
5
sum
101
of
denitions
independent
82–3
variables
32–4
R
I
Type
II
errors
112,
121
random
samples
random
variables
T
64–5
40–1
t-distribution
91–3
O
estimators
one-tail
tests
lower)
normal
(upper
100,
Poisson
and
120,
regression
121
distribution
graph
for
80–1
t-statistics
distribution
analysis
53–4
41,
101
t-statistic
141–50
and
a
trac
Y
for
dependence
of
X
139–40
analysis
35
S
one-tail
test
115
two-tail
sampling
distribution
of
tests
100,
120,
121
the
normal
distribution
graph
for
a
P
mean
64–5,
76
two-tail
p-value
sampling
101
distributions
Blaise
sample
20
product
I
and
distribution
correlation
20
Karl
87,
133
R
130–1,
cur ves
signicance
72
distribution
27,
29,
38
skills
levels
85,
random
Pólya,
variables
Denis
George
Pólya’s
probability
20,
68
distribution
population
64–5,
population
sample
probability
77,
4,
40,
100
distribution
analysis
two-tail
2–3,
methods
modeling
error
statistical
65,
analysis
122,
test
78,
161
76
p-value
von
methods
value
value
function
geometric
162
Index
condence
5–12
distribution
12–24
for
mean
inter vals
84–99
for
101
Mises,
Richard
78,
the
z-distribution
z-statistics
85
101
120–1
distribution
graph
116
80
condence
Z
cumulative
for
V
critical
statistical
standard
38–9
graph
115
157
159
79
73
5,
algebra
distributions
values
statistical
20
121
101
156
parameters
size
expectation
53–4
68
test
checks
a
Poisson,
112,
112
distribution
one-tail
normal
Poisson
errors
154
a
percentile
II
research
coecient
normal
Pearson,
Type
moment
medical
Pascal’s
116
130–4
Type
Pascal,
test
101
92
Edler
68
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