BUS210:
Business
Statistics
Homework
Answers
Chapter
5
5.2
5.5
5.10
5.13
5.14
X
P(x)
0
0.10
1
0.20
2
0.45
3
0.15
4
0.05
5
0.05
(a)
Mean
=
X*P(X)
0.00
0.20
0.90
0.45
0.20
0.25
2.00
(X‐µX)
4
1
0
1
4
9
Variance
=
(b)
Stdev
=
(a)
E(X)
=
2.9
(b)
σ
=
1.772
(a)
Let
X
=
number
of
failure
P(X
=
0)
=
0.7351.
(b)
If
the
probabilities
of
the
six
events
are
not
all
equal,
then
you
would
calculate
the
probability
in
(a)
by
multiplying
together
the
probability
of
no
failure
in
each
of
the
six
events.
(a)
mean
=
1
standard
deviation
=
0.9747
(b)
P(X
=
0)
=
0.3585
(c)
P(X
=
1)
=
0.3774
(d)
P(X
≥
2)
=
0.2642
(a)
P(X
≤
5)
=
9.9185
x
10‐5
(b)
P(X
≤
10)
=
0.0719
(c)
P(X
≤
15)
=
0.8173
5.16
(a)
Using the equation, if λ = 2.5, P(X = 2) =
€
€
5.17
€
€
5.23
(X‐µX)*P(X)
0.40
0.20
0.00
0.15
0.20
0.45
1.40
1.183
(b)
If λ = 8.0, [using table] P(X = 8) = 0.1396 (c)
If λ = 0.5, [using table] P(X = 1) = 0.3033 (d)
If λ = 3.7, [using table] P(X = 0) = 0.0247 e-2.5 (2.5) 2
= 0.2565 2!
(a)
If
λ=
2.0,
P(X
≥
2)
=
1
–
[P(X
=
0)
+
P(X
=
1)]
=
1
–
[0.1353
+
0.2707]
=
0.5940
(b)
If
λ=
8.0,
P(X
≥
3)
=1
–
[P(X=0)
+
P(X=1)
+
P(X=2)]
=
1–[0.0003+0.0027+0.0107]=
1–0.0137=
0.9863
(c)
If
λ=
0.5,
P(X
≤
1)
=
P(X
=
0)
+
P(X
=
1)
=
0.6065
+
0.3033
=
0.9098
(d)
If
λ=
4.0,
P(X
≥
1)
=
1
–
P(X
=
0)
=
1
–
0.0183
=
0.9817
(a)
λ=
1.42
P(X
=
0)
=
0.2417
(b)
P(X
≥
1)
=
0.7583
(c)
P(X
≥
2)
=
0.4151
BUS210:
Business
Statistics
Homework
Answers
Chapter
5
5.34
5.35
5.37
5.44
(a)
0.74
(b)
0.74
p
=
0.74,
n
=
5
(c)
P(X
=
4)
=
0.3898
(d)
P(X
=
0)
=
0.0012
(e)
Stock
prices
tend
to
rise
in
the
years
when
the
economy
is
expanding
and
fall
in
the
years
of
recession
or
contraction.
Hence,
the
probability
that
the
price
will
rise
in
one
year
is
not
independent
from
year
to
year.
(a)
0.0060
(b)
0.0403
(c)
0.1673
(d)
0.000105
(e)
The
probability
is
essentially
zero
that
all
ten
automatically
opt
to
talk
to
a
live
operator.
Hence,
the
40%
figure
given
in
the
article
does
not
appear
to
apply
to
this
particular
system.
(a)
0.006047
(b)
0.040311
(c)
0.953643
(d)
mean
=
4
;
stdev
=
1.5492
(a)
The
assumptions
needed
are
(i)
the
probability
that
a
golfer
loses
a
golf
ball
in
a
given
interval
in
a
game
is
constant,
(ii)
the
probability
that
a
golfer
loses
more
than
one
golf
ball
in
this
interval
approaches
zero
as
the
interval
gets
smaller,
(iii)
the
probability
that
a
golfer
loses
a
golf
ball
is
independent
from
interval
to
interval.
λ= 4.5
(b)
P(X
=
0)
=
0.0111
(c)
P(X
≤
5)
=
0.70293
(d)
P(X
≥
6)
=
0.29707