Answers to Selected Exercises

Probability Distribution Function (PDF) for a Discrete Random Variable

x	P(x)
0	0.12
1	0.18
2	0.30
3	0.15
4	0.10
5	0.10
6	0.05

3. 0.10 + 0.05 = 0.15

5. 1

7. 0.35 + 0.40 + 0.10 = 0.85

9. 1(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45

11.

x	P(x)
0	0.03
1	0.04
2	0.08
3	0.85

13. Let X = the number of events Javier volunteers for each month.

15.

x	P(x)
0	0.05
1	0.05
2	0.10
3	0.20
4	0.25
5	0.35

17. 1 – 0.05 = 0.95

Mean or Expected Value and Standard Deviation

20. 0.2 + 1.2 + 2.4 + 1.6 = 5.4

22.The values of P(x) do not sum to one.

24.Let X = the number of years a physics major will spend doing post-graduate research.

27. 1(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) = 0.35 + 0.40 + 0.45 + 0.60 + 0.50 + 0.30 = 2.6 years

29. X is the number of years a student studies ballet with the teacher.

31. 0.10 + 0.05 + 0.10 = 0.25

33. The sum of the probabilities sum to one because it is a probability distribution.

35. [latex]-2\left(\frac{40}{52}\right)+30\left(\frac{12}{52}\right)=-1.54+6.92=5.38[/latex]

38. The variable of interest is X, or the gain or loss, in dollars.

The face cards jack, queen, and king. There are (3)(4) = 12 face cards and 52 – 12 = 40 cards that are not face cards.

We first need to construct the probability distribution for X. We use the card and coin events to determine the probability for each outcome, but we use the monetary value of X to determine the expected value.

Card Event	X net gain/loss	P(X)
Face Card and Heads	6	[latex]\left(\frac{{12}}{{52}}\right)\left(\frac{{1}}{{2}}\right)=\left(\frac{{6}}{{52}}\right)[/latex]
Face Card and Tails	2	[latex]\left(\frac{{12}}{{52}}\right)\left(\frac{{1}}{{2}}\right)=\left(\frac{{6}}{{52}}\right)[/latex]
(Not Face Card) and (H or T)	–2	[latex]\left(\frac{{40}}{{52}}\right)\left(1\right)=\left(\frac{{40}}{{52}}\right)[/latex]

Expected Value = (6)[latex]\frac{{6}}{{52}}+\left(2\right)\left(\frac{{6}}{{52}}\right)+\left(-2\right)\frac{{40}}{{52}}=-\frac{{32}}{{52}}[/latex]
Expected value = –$0.62, rounded to the nearest cent
If you play this game repeatedly, over a long string of games, you would expect to lose 62 cents per game, on average.
You should not play this game to win money because the expected value indicates an expected average loss.

40.

42.

Software Company

x P(x)

5,000,000 0.10

1,000,000 0.30

–1,000,000 0.60

Hardware Company

x P(x)

3,000,000 0.20

1,000,000 0.40

–1,000,00 0.40

Biotech Firm

x P(x)

6,00,000 0.10

0 0.70

–1,000,000 0.20
$200,000; $600,000; $400,000
third investment because it has the lowest probability of loss
first investment because it has the highest probability of loss
second investment

Software Company
5,000,000	0.10
1,000,000	0.30
–1,000,000	0.60

Hardware Company
3,000,000	0.20
1,000,000	0.40
–1,000,00	0.40

Biotech Firm
6,00,000	0.10
0	0.70
–1,000,000	0.20

44. 4.85 years

46. b

48.

Let X = the amount of money to be won on a ticket. The following table shows the PDF for X.

x	P(x)
0	0.969
5	[latex]\frac{{250}}{{10000}}[/latex]= 0.025
25	[latex]\frac{{50}}{{10000}}[/latex] = 0.005
100	[latex]\frac{{10}}{{10000}}[/latex] = 0.001

Calculate the expected value of X.

0(0.969) + 5(0.025) + 25(0.005) + 100(0.001) = 0.35

A fair price for a ticket is $0.35. Any price over $0.35 will enable the lottery to raise money.

Binomial Distribution

49. X = the number that reply “yes”

50. 0, 1, 2, 3, 4, 5, 6, 7, 8

52. 5.7

54. 0.4151

57.

X = the number of patients calling in claiming to have the flu, who actually have the flu.

X = 0, 1, 2, …25

61.

X = the number of DVDs a Video to Go customer rents
0.12
0.11
0.77

63. 4.43

65. 0.4734

67.

X = number of questions answered correctly
X ~ B(32, 13)
We are interested in MORE THAN 75% of 32 questions correct. 75% of 32 is 24. We want to find P(x > 24). The event “more than 24” is the complement of “less than or equal to 24.”
Using your calculator’s distribution menu: 1 – binomcdf(32, 13, 24)
P(x > 24) = 0
The probability of getting more than 75% of the 32 questions correct when randomly guessing is very small and practically zero.

69.

X = the number of college and universities that offer online offerings.
0, 1, 2, …, 13
X ~ B(13, 0.96)
12.48
0.0135
P(x = 12) = 0.3186 P(x = 13) = 0.5882 More likely to get 13.

71.

X = the number of fencers who do not use the foil as their main weapon
0, 1, 2, 3,… 25
X ~ B(25,0.40)
10
0.0442
The probability that all 25 not use the foil is almost zero. Therefore, it would be very surprising.

73.

X = the number of audits in a 20-year period
0, 1, 2, …, 20
X ~ B(20, 0.02)
0.4
0.6676
0.0071

75.

X = the number of matches
0, 1, 2, 3
X ~ B(3,16)
In dollars: −1, 1, 2, 3
12
Multiply each Y value by the corresponding X probability from the PDF table. The answer is −0.0787. You lose about eight cents, on average, per game.
The house has the advantage.

77.

X ~ B(15, 0.281)
1. Mean = μ = np = 15(0.281) = 4.215
2. Standard Deviation = σ = [latex]\sqrt{npq}=\sqrt{15\left(0.281\right)\left(0.719\right)}=1.7409[/latex]
P(x > 5) = 1 – P(x ≤ 5) = 1 – binomcdf(15, 0.281, 5) = 1 – 0.7754 = 0.2246

P(x = 3) = binompdf(15, 0.281, 3) = 0.1927

P(x = 4) = binompdf(15, 0.281, 4) = 0.2259

It is more likely that four people are literate that three people are.

Geometric Distribution

78. X = the number of freshmen selected from the study until one replied “yes” that same-sex couples should have the right to legal marital status.

80. 1,2,…

82. 1.4

85.

X = the number of adults in America who are surveyed until one says he or she will watch the Super Bowl.
X ~ G(0.40)
2.5
0.0187
0.2304

87.

X = the number of pages that advertise footwear

X takes on the values 0, 1, 2, …, 20

X ~ B(20, [latex]\frac{{29}}{{192}}[/latex])

3.02

0.9997

X = the number of pages we must survey until we find one that advertises footwear. X ~ G([latex]\frac{{29}}{{192}}[/latex])

0.3881

6.6207 pages

89.0, 1, 2, and 3

91.

X ~ G(0.25)
Mean = [latex]\mu=\frac{{1}}{{p}}={{1}}{{0.25}}=4[/latex], Standard Deviation = [latex]\sigma=\sqrt{\frac{{1-p}}{{{p^{2}}}}}[/latex]≈ 3.4641
P(x = 10) = geometpdf(0.25, 10) = 0.0188
P(x = 20) = geometpdf(0.25, 20) = 0.0011
P(x ≤ 5) = geometcdf(0.25, 5) = 0.7627

92. X = the number of business majors in the sample.

94. 2, 3, 4, 5, 6, 7, 8, 9

96. 6.26

98.

X = the number of pages that advertise footwear
0, 1, 2, 3, …, 20
X ~ H(29, 163, 20); r = 29, b = 163, n = 20
3.03
1.5197

100.

X = the number of Patriots picked
0, 1, 2, 3, 4
X ~ H(4, 8, 9)
Without replacement

103. 0, 1, 2, 3, 4, …

105. 0.0485

107. 0.0214

109. X = the number of U.S. teens who die from motor vehicle injuries per day.

111.0, 1, 2, 3, 4, …

113. No

115.

X ~ P(5.5); μ = 5.5; [latex]=\sqrt{5.5}{\approx}2.3452[/latex]
P(x ≤ 6) = poissoncdf(5.5, 6) ≈ 0.6860
There is a 15.7% probability that the law staff will receive more calls than they can handle.
P(x > 8) = 1 – P(x ≤ 8) = 1 – poissoncdf(5.5, 8) ≈ 1 – 0.8944 = 0.1056

117.

Let X = the number of defective bulbs in a string.

Using the Poisson distribution:

μ = np = 100(0.03) = 3

X ~ P(3)

P(x ≤ 4) = poissoncdf(3, 4) ≈ 0.8153

Using the binomial distribution:

X ~ B(100, 0.03)

P(x ≤ 4) = binomcdf(100, 0.03, 4) ≈ 0.8179

The Poisson approximation is very good—the difference between the probabilities is only 0.0026.

119.

X = the number of children for a Spanish woman
0, 1, 2, 3,…
X ~ P(1.47)
0.2299
0.5679
0.4321

121.

X = the number of fortune cookies that have an extra fortune
0, 1, 2, 3,… 144
X ~ B(144, 0.03) or P(4.32)
4.32
0.0124 or 0.0133
0.6300 or 0.6264
As n gets larger, the probabilities get closer together.

123.

X = the number of people audited in one year
0, 1, 2, …, 100
X ~ P(2)
2
0.1353
0.3233

125.

X = the number of shell pieces in one cake
0, 1, 2, 3,…
X ~ P(1.5)
1.5
0.2231
0.0001
Yes

127. 0.0671

Module 4: Discrete Random Variables

Probability Distribution Function (PDF) for a Discrete Random Variable

Mean or Expected Value and Standard Deviation

Binomial Distribution

Geometric Distribution

Candela Citations