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Chapter 8 Solutions
Page 1 of 15
CHAPTER 8
EXERCISE SOLUTIONS
8.1
a.
b.
c.
d.
e.
Continuous.
Discrete.
Continuous.
Discrete.
Discrete.
8.2
a.
b.
c.
d.
Discrete.
Continuous.
Discrete.
Discrete.
8.3
a. 3/6 = 1/2 makes the probabilities sum to 1.
b. .1 makes the probabilities sum to 1. Calculation is 1 – (.1 + .2 + .3 + .3) = .1.
8.4
1 – (.05 + .20 + .50 + .15) = .10. (The sum of all probabilities must equal 1.)
8.5
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 8.5c
c.
k
0
1
2
.1
.2
.5
P(X≤k)
As an example, P(X≤1) = P(X=0) + P(X=1) = .1+.1=.2.
3
1
8.6
8.6
k
P(X=k )
8.7
0
1
2
3
1/8
3/8
3/8
1/8
a. .80. Find this by adding the probabilities for X = 0, 1, and 2.
P(X ≤2) = P(X =0) + P(X=1) + P(X=2) = .14 + .27 + .39 = .80.
b. P(X = 1 or X = 2) = .37 + .29 = .66.
c. P(X > 0) = 1 – P(X = 0) = 1 − .14 = .86.
Chapter 8 Solutions
8.8
Page 2 of 15
The probability that X=0 are left-handed is (7/10)(6/9) = 42/90 (see Section 7.4, Rule 3).
The probability that X=2 are left-handed is (3/10)(2/9) = 6/90.
P(X=1) must be such that probabilities add to 1, so it is 1−(42/90) −(6/90) = 42/90.
A summary of the distribution (pdf) is:
k
P(X=k )
0
42/90
1
42/90
2
6/90
8.9
The answer is 18/36=1/2. Add the probabilities given in Example 8.5 for X = 2, 4, 6, 8, 10, and 12.
8.10
The sample space of all possible sequences is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
a. For each sequence in the sample space, count the number of tails. Then count how many
outcomes there are for each possible number of tails. Divide by 8 because all outcomes in the
sample space are equally likely. The distribution (pdf) is:
k
P(X =k)
0
1/8
1
3/8
2
3/8
3
1/8
b. For each sequence in the sample space, take the difference (Y) between the number of heads
and the number of tails. For instance, for HHT, Y = 2 (heads) − 1 (tails) = 1. List all possible
differences, count how many sequences in the sample space have each possible value, then divide
by 8 to get the probability of that value. The distribution (pdf) is:
−3
1/8
k
P(Y =k )
−1
3/8
1
3/8
3
1/8
c. The sum of heads and tails is 3 for every sequence which means P(Z=3) = 1 so the distribution
(pdf) is:
k
P(Z =k)
8.11
The probability that X=1 is the probability that the first child is a girl, so P(X=1)=.5. For X=2, the
sequence must be Boy, Girl so P(X=2)= (.5)(.5)=.25. For X=3, the sequence is Boy, Boy, Girl and
this probability is (.5)(.5)(.5)=.125. The probability that X=4 can be found be subtracting the sum
of the other probabilities from 1, and this gives P(X=4)=.125. A summary of the distribution is:
k
P(X=k)
8.12
3
1
1
.5
2
.25
3
.125
4
.125
This will differ for each student. An example of another continuous random variable is the
amount of rainfall (in inches) during the event time. Rainfall can be any number between 0 and
some maximum value. One example of another discrete random variable is the number of police,
ambulance, and fire vehicles driving by with sirens. The number of vehicles can be 0,1,2, …, and
so on up to the logical maximum for the situation.
8.13
k
P(X≤k)
2
1
36
3
3
36
4
6
36
5
10
36
6
15
36
7
21
36
8
26
36
9
30
36
10
33
36
11
35
36
12
1
Chapter 8 Solutions
Page 3 of 15
8.14
This may differ for each student and the explanation should support the response. Following are
some suggestions.
a. The cdf would be of more interest, since she would want to know if the number (and percent)
of constituents who oppose the new law is less than or equal to some standard (like 50%).
b. The pdf may be of more interest because it reveals what proportion of people become infected
on the first exposure, what proportion of people become infected on the second exposure, etc. The
cdf may be of interest to someone who has had k exposures and wants to know the probability of
infection so far.
c. The cdf would be of most interest, because the company would want to know things such as the
probability that the drug will lower the blood pressure of less (or more) than half of the sample.
d. The cdf would have more interest if the caterer wants to have a very high probability of having
enough food for all guests. The caterer would probably want to choose k to be large enough that
P(X>k) would be very small. A caterer who wishes to make very exact plans might be interested
in the pdf, which would give a probability for each possible number, but it's more likely that the
cdf would be of interest.
8.15
a. $30, the cost of the insurance.
b. X = $0 with probability .90 (the probability he doesn’t need to be towed during a year), and X
= $100 with probability .10 (the probability he needs to be towed during a year).
c. If he buys insurance, E(X) = $30.
If he does not buy insurance, E(X) = $0 × .90 + $100 × .10 = $10.
In the long run, he’s better off to not buy the insurance. The average cost per year, without
insurance will be $10, which is less than the $30 per year cost of the insurance.
8.16
a. E(X) = (0 × .05) + (1 × .60) + (2 × .30) + (3× .05) = 1.35.
b. The average number of children per family over many families similar to the Brann family.
c. No. The expected value of 1.35 is an impossible number of children for one family.
8.17
E(X) = ($100)( .01) + (−$5) (.99) = −$3.95.
8.18
a. E(X) = −1(.25) + 0(.50) + 1(.25) = 0.
b. σ 2 = (−1−0)2 (.25) + (0−0)2 (.50) + (1−0)2 (.25) = .50.
c.
8.19
0.50 = 0.707
a. E(X) = −2(.25) + 0(.50) + 2(.25) = 0.
b. σ 2 = (−2 − 0)2 (.25) + (0 − 0)2 (.50) + (2 − 0)2 (.25) = 2.
c. σ = 2 = 1 .414
8.20
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 8.4)
is:
k
P(X=k)
0
1/8=.125
1
3/8 =.375
2
3/8=.375
3
1/8=.125
The expected value is calculated as the sum of "value×probability" and the calculations here are,
E(X) =(0×.125) +(1×.375)+(2×.375) + (3×.125) = 1.5 girls.
8.21
The distribution of the sum is given in Example 8.5. The expected value is
µ = E( X ) = 2 ×
1
2
3
3
2
1
252
+3×
+ 4×
+ ... + 10 ×
+ 11 ×
+ 12 ×
=
=7
36
36
36
36
36
36
36
Chapter 8 Solutions
8.22
Page 4 of 15
σ = 2 .415 . The formula is σ =
∑(x −µ)2 p(x) where the sum is over all values of x. Here, the
mean is µ = 7 (see Exercise 8.10). Details of the calculations are:
x
x−µ
(x−µ)2
p(x)
σ = 25
2
−5
25
1/36
3
−4
16
2/36
4
−3
9
3/36
5
−2
4
4/36
6
−1
1
5/36
7
0
0
6/36
8
1
1
5/36
9
2
4
4/36
10
3
9
3/36
11
4
16
2/36
12
5
25
1/36
1
2
3
4
5
6
5
4
3
2
1
+ 16 + 9
+4
+1 + 0
+1 + 4
+9
+ 16 + 25
36
36
36
36
36
36
36
36
36
36
36
= 210 / 36 = 2 .415.
8.23
σ = $29 .67 . The standard deviation is calculated as σ =
∑(x − µ)2 p(x) where the sum is over all
values of x. The expected value (mean) was found in Exercise 8.6 as µ = −$0.35. The details for
finding σ are:
x
x−µ
p(x)
$4,999
$4,999.35
.000035
$49
$49.35
.00168
$4
$4.35
.0303
$0
$0.35
.242
−$1
−$0.65
.726
4999 . 35 2 (.000035 ) + $ 49 . 35 2 (. 00168 ) + $ 4. 35 2 (. 0303 `) + $ 0. 35 2 (. 242 ) + $ 0. 65 2 (.726 )
8.24
= $ 29 .67
µ = E(X) = ∑ xp(x) = (500×.1) + (0×.9) = $50 per person (in the long run).
8.25
Answer = $60. In Exercise 8.13, the result is that the company will pay out $50 per person in the
long run so they should charge each person $50 + $10 = $60 to make a $10 per person profit.
8.26
µ = E(X) =
∑ xp(x) = (0×.03) + (1×.25) + (2×.72) = 1.69.
(The probability for living with one
parent is the sum of probabilities for mother only and father only.) Here, the expected value is the
average number of parents in the household over the population of all children. This value is not
an "expected" value for an individual child because it is not possible to live with 1.69 parents. It is
meaningful only as a long run or population average.
8.27
µ = E(X) =
∑ xp(x) =
(3×.7) + (4×.3) = 3.3. You would not expect to have this grade point
average every quarter or semester. Instead, it is your expected grade point average in the long run.
8.28
a. µ = E(X) =
∑ xp(x) = (15×.8) + (20×.2) = 16 minutes.
b. No, the expected value will never be your actual commute time (which is always either 15 min.
or 20 min.)
8.29
a. Yes. n = 200 and p = .5.
b. E(X) = np = (200)(.5) = 100.
8.30
a. Yes. n = 10 and p = .5.
b. No. p is not the same from trial to trial.
c. No. The “trials” (cities) are not independent of each other as they will tend to have the same
weather.
d. No. The “trials” (children) are not independent of each other because they are in the same
class and flu is contagious.
Chapter 8 Solutions
Page 5 of 15
8.31
a. n = 30 and p = 1/6.
b. n = 10 and p = 1/100.
c. n = 20 and p = 3/10.
8.32
a. µ = E(X) = np = (30)(1/6) = 5.
b. µ = E(X) = np = (10)(1/100) = 0.10.
c. µ = E(X) = np = (20)(3/10) = 6.
8.33
The answers can be found using any of the methods discussed in Section 8.4, including the use of
Minitab or Excel.
a. P(X = 5) = .0264.
b. P(X = 2) = .3020.
c. P(X = 1) = .2684.
d. P(X = 9) = .000004.
8.34
a. µ = 10(.50) = 5; σ = 10 (.5)(1 − .5) = 1.581 .
b. µ = 1(.40) =0.40; σ = 1(.40 )(1 − .40) = 0.490 .
c. µ= 100(.90) = 90; σ = 100 (.9)(1 − .9 ) = 3.
d. µ = 30(.01) = 0.3; σ = 30 (.01)(1 − .01) = 0.545.
8.35
a. The probability of success does not remain the same from one trial (game) to the next. The
probability of winning a game against a good team is not the same as the probability of winning a
game against a poor team.
b. The number of trials is not specified in advance.
c. The probability of success does not remain the same from one trial to the next because whether
or not the first card is an ace affects the probability that the next card is an ace, and so on. This
also means that trials are not independent.
8.36
Use a binomial distribution with n = 10, p = .5. Let X = number of games won by human. The
answers can be found using any of the methods discussed in Section 8.4, including the use of
Minitab or Excel.
a. P(X = 5) = .2461.
b. P(X = 3) = .1172. If computer wins 7 games, then human wins 3 games.
c. P(X ≥ 7) = .1719, calculated as P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) =
.1172 + .0439 + .0098 + .0010 = .1719. Equivalently, P(X ≥ 7) = 1 − P(X ≤ 6) = 1 −.8281.
8.37
a. The number of trials is specified in advance. There are two possible outcomes—either the
subject guesses correctly or not. If the subject merely guesses, the probability of success remains
the same from trial to trial. Whether a subject guesses correctly or not on a trial is independent
from the results of previous trials.
b. Yes, X is a binomial random variable with n = 10 and p = .25.
c. The number correct is either 6 or more or 5 or less, so P(X≥6) =1−P(X≤5) =1−.9803 = .0197.
d. With p = .5, P(X≥6) = 1−P(X≤5) = 1−.6230 = .3770.
e. This answer will differ for each student. A factor to consider is that among all people who
merely guess, .0197(about 2%) will be able to get 6 or more correct. If many people are tested, a
few who just guess will be able to get 6 or more right. Another factor to consider is the possible
proportion in the population tested that actually has psychic ability. If few people have psychic
ability, a result of 6 or more correct might reasonably be considered to have been the result of
lucky guessing. If many people actually have psychic ability, it might be reasonable to think the
result was obtained from one of those with psychic ability. The discussion of the confusion of the
inverse in Section 7.7 is relevant to this discussion.
Chapter 8 Solutions
Page 6 of 15
8.38
The answers for this exercise can be found using any of the methods discussed in Section 8.4,
including the use of Minitab or Excel.
a. P(X = 4) = .2051
b. P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − .6496 = .3504
c. P(X ≤ 3) = .6496
d. P(X = 0) = .5905
e. P(X ≥ 1) = 1 − P(X = 0) = 1 − .5905 = .4095
8.39
a. Note that 1/4 of 1,000 is 250 so the desired probability is P(X≥250). n = 1000 and p = the
proportion of adults in the United States living with a partner, but not married at the time of the
sampling. The value of p is not known.
b. The desired probability is P(X≥110), n = 500, and p = .20.
c. Note that 70% of 20 is 14 so the desired probability is P(X≥14). n = 20, and p = .50.
8.40
The formulas are µ =np and σ = np (1 − p )
a. µ = 10(1/2) = 5 and σ = 10(. 5)(1 − .5 ) = 1.5811
b. µ = 100(1/4) = 25 and σ = 100 (.25 )(1 − .25 ) = 4 .33
c. µ= 2500(1/5) = 500 and σ = 2500 (.2)(1 − . 2) = 20
d. µ = 1(1/10) = .1 and σ = 1(.1)(1 − .1) = .3
e. µ = 30(.4) = 12 and σ = 30 (. 4 )(1 − .4 ) = 2 .683
8.41
For n = 2 and p = .5, P(X = 0) = .25, P(X = 1) = .5, and P(X = 2) = .25. These can be found in
several ways. One way is to list possible outcomes, which are {SS, SF, FS, and FF}, recognize
that all outcomes are equally likely, and then tabulate the distribution of X = number of successes.
µ = E(X) = ∑ xp(x) = (0×.25) + (1×.5)+ (0×.25) = 1.
σ=
∑( x − µ)2 p( x) =
( 0 − 1) 2 (. 25 ) + (1 − 1) 2 (. 5) + (2 − 1) 2 (.25 ) = 0 .5 = .7071 .
8.42
a. P(0≤ X≤ 30) = .5 (because the interval from 0 to 30 is one-half of the interval of possible
outcomes (0 to 60) and the distribution is uniform.
b. P(30≤X≤ 60) = .5 by the same reasoning as in part (a).
8.43
a.
8.44
Table A.1 can be used to find the answers.
a. .5000.
b. .3632.
c. .6368.
d. .9750.
e. .0099.
f. .9951.
g. .9505.
1 .5 − 0
= 1 .5 .
1
4 − 10
b.
= −1 .
6
0 − 10
c.
= −2 .
5
−25 − ( −10 )
d.
= −1 .
15