Probability Distribution Function (PDF) for a Discrete Random Variable — Practice

1.

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.

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

15.

17. 1 – 0.05 = 0.95

Mean or Expected Value and Standard Deviation — Practice

19. 0.2 + 1.2 + 2.4 + 1.6 = 5.4

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

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

25. 1 – 0.35 – 0.20 – 0.15 – 0.10 – 0.05 = 0.15

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]

Binomial Distribution — Practice

37. X = the number that reply “yes”

39. 0, 1, 2, 3, 4, 5, 6, 7, 8

41. 5.7

43. 0.4151

Geometric Distribution — Practice

45. 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.

47. 1,2,…

49. 1.4

Hypergeometric Distribution — Practice

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

53. 2, 3, 4, 5, 6, 7, 8, 9

55. 6.26

57. 0, 1, 2, 3, 4, …

59. 0.0485

61. 0.0214

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

65. 0, 1, 2, 3, 4, …

67. No

Mean or Expected Value and Standard Deviation — Homework

71. The variable of interest is X, or the gain or loss, in dollars. The face cards are 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.

• 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.

73.

1. 0.1
2. 1.6

75.

1. 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

2. $200,000; $600,000; $400,000
3. third investment because it has the lowest probability of loss
4. first investment because it has the highest probability of loss
5. second investment

77. 4.85 years

79. b

81. 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 — Homework

83. X = the number of patients calling in claiming to have the flu, who actually have the flu. X = 0, 1, 2, …2585. 0.016587.

1. X = the number of DVDs a Video to Go customer rents
2. 0.12
3. 0.11
4. 0.77

89. d. 4.43

91. c

93.

• 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.

95.

1. X = the number of college and universities that offer online offerings.
2. 0, 1, 2, …, 13
3. X ~ B(13, 0.96)
4. 12.48
5. 0.0135
6. P(x = 12) = 0.3186 P(x = 13) = 0.5882 More likely to get 13.

97.

1. X = the number of fencers who do not use the foil as their main weapon
2. 0, 1, 2, 3,… 25
3. X ~ B(25,0.40)
4. 10
5. 0.0442
6. The probability that all 25 not use the foil is almost zero. Therefore, it would be very surprising.

99.

1. X = the number of audits in a 20-year period
2. 0, 1, 2, …, 20
3. X ~ B(20, 0.02)
4. 0.4
5. 0.6676
6. 0.0071

101.

1. X = the number of matches
2. 0, 1, 2, 3
3. X ~ B(3,16)
4. In dollars: −1, 1, 2, 3
5. 12
6. 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.
7. The house has the advantage.

103.

a. X ~ B(15, 0.281)

b. Mean = μ = np = 15(0.281) = 4.215

Standard Deviation = σ = [latex]\sqrt{npq}=\sqrt{15\left(0.281\right)\left(0.719\right)}=1.7409[/latex]

c. 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 than three people are.

Geometric Distribution — Homework

105.

1. X = the number of adults in America who are surveyed until one says he or she will watch the Super Bowl.
2. X ~ G(0.40)
3. 2.5
4. 0.0187
5. 0.2304

107.

1. X = the number of pages that advertise footwear
2. X takes on the values 0, 1, 2, …, 20
3. X ~ B(20, [latex]\frac{{29}}{{192}}[/latex])
4. 3.02
5. No
6. 0.9997
7. X = the number of pages we must survey until we find one that advertises footwear. X ~ G([latex]\frac{{29}}{{192}}[/latex])
8. 0.3881
9. 6.6207 pages

109. 0, 1, 2, and 3

111.

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

Hypergeometric Distribution — Homework

113.

1. X = the number of pages that advertise footwear
2. 0, 1, 2, 3, …, 20
3. X ~ H(29, 163, 20); r = 29, b = 163, n = 20
4. 3.03
5. 1.5197

115.

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

Poisson Distribution – Homework

117.

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

119.

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.

121.

1. X = the number of children for a Spanish woman
2. 0, 1, 2, 3,…
3. X ~ P(1.47)
4. 0.2299
5. 0.5679
6. 0.4321

123.

1. X = the number of fortune cookies that have an extra fortune
2. 0, 1, 2, 3,… 144
3. X ~ B(144, 0.03) or P(4.32)
4. 4.32
5. 0.0124 or 0.0133
6. 0.6300 or 0.6264
7. As n gets larger, the probabilities get closer together.

125.

1. X = the number of people audited in one year
2. 0, 1, 2, …, 100
3. X ~ P(2)
4. 2
5. 0.1353
6. 0.3233

127.

1. X = the number of shell pieces in one cake
2. 0, 1, 2, 3,…
3. X ~ P(1.5)
4. 1.5
5. 0.2231
6. 0.0001
7. Yes

129. d