# 4.12: Chapter 4 Solution (Practice + Homework)

1.

$$x$$$$P(x)$$
00.12
10.18
20.30
30.15
40.10
50.10
60.05

Table4.6

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.

$$x$$$$P(x)$$
00.03
10.04
20.08
30.85

Table4.7

13.

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

15.

$$x$$$$P(x)$$
00.05
10.05
20.10
30.20
40.25
50.35

Table4.8

17.

1 – 0.05 = 0.95

18.

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

19.

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

20.

$$X =$$ the number that reply “yes”

22.

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

24.

5.7

26.

0.4151

28.

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

30.

1,2,…

32.

1.4

35.

0, 1, 2, 3, 4, …

37.

0.0485

39.

0.0214

41.

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

43.

0, 1, 2, 3, 4, ...

45.

No

48.

1. $$X =$$ the number of pages that advertise footwear
2. 0, 1, 2, 3, ..., 20
3. 3.03
4. 1.5197

50.

1. $$X =$$ the number of Patriots picked
2. 0, 1, 2, 3, 4
3. Without replacement

53.

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

$$X = 0, 1, 2, ...25$$

55.

0.0165

57.

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

59.

4. 4.43

61.

4

63.

• $$X =$$ number of questions answered correctly
• $$X \sim B(32, 13)(32, 13)$$
• 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."
• $$P(x > 24) = 0$$
• The probability of getting more than 75% of the 32 questions correct when randomly guessing is very small and practically zero.

65.

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

67.

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

69.

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

71.

1. $$X =$$ the number of matches
2. $$0, 1, 2, 3$$
3. In dollars: $$−1, 1, 2, 3$$
4. $$\frac{1}{2}$$
5. The answer is $$−0.0787$$. You lose about eight cents, on average, per game.
6. The house has the advantage.

73.

1. $$X \sim B(15, 0.281)$$

Figure 4.4

1. $$\text{Mean }= \mu=n p=15(0.281)=4.215$$
2. $$\text{Standard Deviation } =\sigma=\sqrt{n p q}=\sqrt{15(0.281)(0.719)}=1.7409$$
2. $$P(x > 5)=1 – 0.7754 = 0.2246$$
$$P(x = 3) = 0.1927$$
$$P(x = 4) = 0.2259$$
It is more likely that four people are literate that three people are.

75.

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

77.

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

79.

0, 1, 2, and 3

81.

1. $$X \sim G(0.25)$$
• $$\text{Mean }=\mu=\frac{1}{p}=\frac{1}{0.25}=4$$
• $$\text{Standard Deviation }=\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.25}{0.25^{2}}} \approx 3.4641$$
2. $$P(x = 10) = 0.0188$$
3. $$P(x = 20) = 0.0011$$
4. $$P(x \leq 5) = 0.7627$$

82.

1. $$X \sim P(5.5); \mu = 5.5; \sigma=\sqrt{5.5} \approx 2.3452$$
2. $$P(x \leq 6) \approx 0.6860$$
3. There is a 15.7% probability that the law staff will receive more calls than they can handle.
4. $$P(x > 8) = 1 – P(x \leq 8) \approx 1 – 0.8944 = 0.1056$$

84.

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

Using the Poisson distribution:

• $$\mu = np = 100(0.03) = 3$$
• $$X \sim P(3)$$
• $$P(x \leq 4) \approx 0.8153$$

Using the binomial distribution:

• $$X \sim B(100, 0.03)$$
• $$P(x \leq 4) = 0.8179$$

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

86.

1. $$X =$$ the number of children for a Spanish woman
2. $$0, 1, 2, 3,...$$
3. $$0.2299$$
4. $$0.5679$$
5. $$0.4321$$

88.

1. $$X =$$ the number of fortune cookies that have an extra fortune
2. $$0, 1, 2, 3,... 144$$
3. $$4.32$$
4. $$0.0124$$ or $$0.0133$$
5. $$0.6300$$ or $$0.6264$$
6. As $$n$$ gets larger, the probabilities get closer together.

90.

1. $$X =$$ the number of people audited in one year
2. $$0, 1, 2, ..., 100$$
3. $$2$$
4. $$0.1353$$
5. $$0.3233$$

92.

1. $$X =$$ the number of shell pieces in one cake
2. $$0, 1, 2, 3,...$$
3. $$1.5$$
4. $$0.2231$$
5. $$0.0001$$
6. Yes

94.

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