Homework 5 Solutions
8.1
a.
Continuous
b.
Discrete
c.
Continuous
d.
Discrete
e.
Discrete
8.7
1 – (.05 + .20 + .50 + .15) = .10.
(The sum of all probabilities must equal 1.)
8.8
a.
Condition 1:
Sum =
.1 + .1 + .3 + .5 = 1.
Condition 2:
Each of the four probabilities is between 0 and 1.
b.
Figure for Exercise 8.8c
c.
k
0
1
2
3
P(X
≤
k)
.1
.2
.5
1
As an example,
P(X
≤
1
)
=
P
(
X
=0) +
P
(
X
=1) = .1+.1=.2.
8.17
a.
.06 + .13 = .19
b.
P
(
X
> 0) = 1 –
P
(
X
= 0) = 1− .73 = .27.
Another method is to add the probabilities for X = 1, 2, 3, 4.
c.
k
0
1
2
3
4
P
(
X
≤
k
)
.73
.89
.95
.98
1
8.22
a.
E(
X
) =

2(.25) + 0(.50) + 2(.25) = 0.
b.
2
σ
=
(

2

0)
2
(.25) + (0

0)
2
(.50) + (2

0)
2
(.25) = 2.
c.
414
.
1
2
=
=
σ
8.23
An intuitive solution (and a correct one) is that the expected value for the number of girls among
three children is 3
×
.5 = 1.5.
A more elaborate solution is to determine the probability distribution
for X =number of girls, and then use the formula for E(
X
). The distribution given in Example 8.6
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.
 Spring '08
 BARROSO,JOAOR
 Statistics, Probability, Probability theory

Click to edit the document details