Chapter 5
Discrete Probability
Distributions
Solutions:
1.
a.
Head, Head (H,H)
Head, Tail (H,T)
Tail, Head (T,H)
Tail, Tail (T,T)
b.
x
=
number of heads on two coin tosses
c.
Outcome
Values of
x
(H,H)
2
(H,T)
1
(T,H)
1
(T,T)
0
d.
Discrete.
It may assume 3 values: 0, 1, and 2.
6.
a.
values:
0,1,2,...,20
discrete
b.
values:
0,1,2,...
discrete
c.
values:
0,1,2,...,50
discrete
d.
values:
0
≤
x
≤
8
continuous
e.
values:
x
>
0
continuous
7.
a.
f
(
x
)
≥
0
for all values of
x
.
Σ
f
(
x
)
=
1
Therefore, it is a proper probability distribution.
b.
Probability
x
=
30
is
f
(30)
=
.25
c.
Probability
x
≤
25
is
f
(20) +
f
(25)
=
.20 + .15
=
.35
d.
Probability
x
>
30
is
f
(35)
=
.40
12.
a.
Yes;
f
(
x
)
≥
0 for all
x
and
Σ
f
(
x
)
=
.15 + .20 + .30 + .25 + .10
=
1
b.
P(1200 or less)
=
f
(1000) +
f
(1100) +
f
(1200)
=
.15 + .20 + .30 = .65
13.
a.
Yes, since
f
(
x
)
≥
0 for
x
=
1,2,3 and
Σ
f
(
x
)
=
f
(1) +
f
(2) +
f
(3)
=
1/6 + 2/6 + 3/6
=
1
b.
f
(2)
=
2/6
=
.333
c.
f
(2) +
f
(3)
=
2/6 + 3/6
=
.833
17.
a/b.
x
f
(
x
)
x f
(
x
)
x

μ
(
x

μ
)
2
(
x

μ
)
2
f
(
x
)
0
.10
.00
2.45
6.0025
.600250
1
.15
.15
1.45
2.1025
.315375
2
.30
.60
 .45
.2025
.060750
3
.20
.60
.55
.3025
.060500
4
.15
.60
1.55
2.4025
.360375
5
.10
.50
2.55
6.5025
.650250
2.45
2.047500
E
(
x
) =
μ
=
2.45
σ
2
=
2.05
σ
=
1.43
19.
a.
E
(
x
)
=
Σ
x f
(
x
)
=
0 (.56) + 2 (.44)
=
.88
b.
E
(
x
)
=
Σ
x f
(
x
)
=
0 (.66) + 3 (.34)
=
1.02
c.
The expected value of a 3  point shot is higher.
So, if these
probabilities hold up, the team will make more points in the long run
with the 3  point shot.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '10
 oba
 Statistics, Probability, Probability theory, probability density function, Insurance coverage, Discrete probability distribution, Discrete Probability Distributions

Click to edit the document details