800000008800000080000
sampling
if I
sampling distribution
,
is the
of a random sample of size
mean
sampling distribution
the
large
the
distribution
sampling
Notation
:
REMARKS
:
X
-
N
Cm
,
CLT
*
it the distribution of
does not
he 30
,
the
since it is
:
it is slightly
,
M
we can use
and Varus
=
%
When n is
,
then
sufficiently
the normal
distribution
approximate
to
normally distributed population
a
then
,
the
sampling distribution will also be exactly normal
,
the normal approximation in the theorem
Of the
no matter
only if
=
limiting
different
formula
I
=
-
.
.
%
,
the
population
Central
when
n
,
is
is not too
Limit Theorem
if
nz 30
it can also be
.
M
original transformation formula
different from the
that X
sufficiently large
:
in the
will be good
regardless of the
population
through
established
variance var (X )
z
:
/
,
approximately normally distributed
we now use the
note
=
and variance 02
of the sample
approximation is good
M and
size is sufficiently large
population is normal
the
Now , we have already
ECI )
the sample
get the sample from
to
shape (symmetric of skewed
mean
mean E- (F)
with mean M
of I.
is not normal
if the population
if
population
%)
require us
the size
how small
*
taken from a ( large or infinite )
approximately normally distributed with
simply states that if
*
*
is
n
.
theorem
The
& CLT ( central limit theorems
is
normal
approximately normally distributed with
.
transformed
into standard normal distribution
.
with that,
CLT
example 1
:
sample of size 100 is taken from
A random
Approximate
a. I
b.
>
I
-
selecting
of
the probability
large population
a
sample that satisfies
a
with
mean
:
gg8
M
E 1
According
to CLT
I will be approximately normally distributed with
,
E (F)
notation
I
:
answer
(M
N
-
M
=
1000 and var (F)
=
1000
=
=
,
=
%
100
=
.
.
.
6.25
6.25 )
:
given
M
:
1000
:
Ñ= 625
n
→
0=25
100
=
approximate the probability of selecting
a) I
625
=
sample that satisfies
a
:
gg8
>
solution
find Pfx
:
I
=
2
=
pc _×< 990 )
-
M
¥
1
=
-
X
> 998 )
-
p
990-1000
z<
25
Too
1-
=
=
b) IX
solution
PC 2<-0.8 )
1- 0.2119
=
0.7081
-
or
78.81
El
M
Find Pci-x
:
=
%
P C- Is -×
=p C- I
=P
(
-
<
I
-
-
§=n<
-
M 1<-11
P
=
.
.
.
ME 1)
me
4-
< z<
=p C- 0.4<2
1)
<
¥)
=P
=
=
Czso 4)
.
-
<
0.4 I
Plz < -0.41
0.6554 -0.3446
0.3108
Or
¥ )
✓ 100
31.08
%
M
=
1000 and
variance 02=625
CLT
2
Example
:
solution
since
:
✗
( which
is
length of life of
the
this population is also
Hence , I is normally
Notation
:
I
-
N
(m
=
M
:
=
or
h
solve for
=
.
.
.
500
50
=
15
PC -×< 475)
(
=P
=
=
:
about Pt×< 4751
:
given
C. LT
.
)
,
For this problem , we are asked
answer
electric light bulb ) is
distributed
=
soo
normal
an
Example
Notation
:
I
-
.
.
.
475-500
z s
E-
)
plz c- 1.94 )
0.0262 Or 2.62%
3
N
(
M
=
3.2
j% /
'
,
=
For this problem , we are asked about PCX
> 3.
1)
.
.
.
normally distributed
,
then any sample from
answer
:
given
:
µ
:
3.2
0=1.8
he 36
solve for PIX
1)
.
.
.
PCIe 3. 1)
1-
=
z
> 3.
FI
=
÷
=\
=
=
=
I
-
-
•
(zj÷ )
plz c-
3. I -3.2
0.33 )
I -0.3707
0.6293
or
62.93%
CLT
example 4
:
P ( IX M /
< 1. 5)
-
For this
answer
I
-
problem , we are
asked what sample size must they choose
.
.
.
:
N(m=
given
0.95
=
:
?
1
=
,
µ= ?
0=6
n
P ( IX
Pl
"
"
P
-
?
=
-
M
1<1.51--0.95
1.5 < I
f- ¥
1- 5
-
M <
I
<
-
1.51=0.95
M
¥n
1.5
sign 1--0.95
f- j÷<z< jÉn 1=0.95
'
'
Simplifying ,
J¥<z< ¥ )
Pf
P
(2
,
<2 <
Zz )
=
=
0.95
0.95
we need to find the two
middle area
equal
to
2-
Cz , and zz )
scores
95%
~
( ¥n<z<¥n 1=0.95
Zz
21
From here , we can say that
2,
→
is
the
2.
5th percentile
.
.
.
.
area to the left equal to 0.025
Zz is the 97.5
→ area
th
percentile
to the left equal to
.
0.975
who
bound the
know that since
we also
Furthermore ,
the normal
curve is
symmetric ,
z
and zz
,
OPPOSITE SIGNS
From the 2- table
t
An
area
* An area to
Recall that
.
the
.
Therefore
Answer
we
,
h
left equal
=
P
to 0.975
to a
corresponds to
f
P
(2 ,
< 2 <
→
<
now
look
Jn
7.84
=
Zz )
g)
=
=
0.95
0.95
2<1.961--0.95
the
for
→
n
=
SAMPLE SIZE ,
61.4656=62
62 ( always round up for sample size )
n
2- score
2-
a
Jn
< z
PC-1.96
,
1.96
:
to 0.025 corresponds
equal
.
substituting
=
,
to the left
.
Score
of
2
,
=
-
l
-
96
of 22=1.96
have
the
SAME MAGNITUDE but
The normal
distribution
as an
approximation
to the
poison
distribution
REMEMBER !
*
if
*
A
*
✗ npo
(X )
and if A > 15 then ✗ may reasonably be
continuity correction
the
larger
Example 1.6.1
d
the
must
be
better the
applied
approximation
approximated by the
normal
distribution y
-
N
C d , d)