# Solutions

1.

ounces of water in a bottle

3.

2

5.

–4

7.

–2

9.

The mean becomes zero.

11.

z = 2

13.

z = 2.78

15.

x = 20

17.

x = 6.5

19.

x = 1

21.

x = 1.97

23.

z = –1.67

25.

z ≈ –0.33

27.

0.67, right

29.

3.14, left

31.

33.

35.

between –5 and –1

37.

39.

41.

The lifetime of a wearable fitness device measured in years.

43.

P(x < 1)

45.

Yes, because they are the same in a continuous distribution: P(x = 1) = 0

47.

1 – P(x < 3) or P(x > 3)

49.

1 – 0.543 = 0.457

51.

0.0013

53.

0.1186

55.
2. 3, 0.1979
57.

0.154

58.

0.874

59.

0.693

60.

0.346

61.

0.110

62.

0.946

63.

0.071

64.

0.347

66.

c

68.
1. Use the z-score formula. z = –0.5141. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average.
2. Use the z-score formula. z = 1.5424. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average.
3. Height = 79 + 3.5(3.89) = 92.615 inches, which is taller than 7 feet, 8 inches. There are very few NBA players this tall so the answer is no, not likely.
70.
1. iv
2. Kyle’s blood pressure is equal to 125 + (1.75)(14) = 149.5.
72.

Let X = an SAT math score and Y = an ACT math score.

1. X = 720 : $z=720-520115=1.74z=720-520115=1.74$. The exam score of 720 is 1.74 standard deviations above the mean of 520.
2. z = 1.5
The math SAT score is 520 + 1.5(115) ≈ 692.5. The exam score of 692.5 is 1.5 standard deviations above the mean of 520.
3. = ≈ 1.59, the z-score for the SAT. = ≈ 1.70, the z-scores for the ACT. With respect to the test they took, the person who took the ACT did better (has the higher z-score).
75.

d

77.
1. X ~ N(66, 2.5)
2. 0.5404
3. No, the probability that an Asian male is over 72 inches tall is 0.0082
79.
1. X ~ N(36, 10)
2. The probability that a person consumes more than 40% of their calories as fat is 0.3446.
3. Approximately 25% of people consume less than 29.26% of their calories as fat.
81.
1. X = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day.
2. X ~ N(3, 1.5)
3. The probability that the child spends less than one hour a day unsupervised is 0.0918.
4. The probability that a child spends over ten hours a day unsupervised is less than 0.0001.
5. 2.21 hours
83.
1. X = the distribution of the number of days a particular type of criminal trial will take
2. X ~ N(21, 7)
3. The probability that a randomly selected trial will last more than 24 days is 0.3336.
4. 22.77
85.
1. mean = 5.51, s = 2.15
5. X ~ N(5.51, 2.15)
6. 0.6029
7. The cumulative frequency for less than 6.1 minutes is 0.64.
8. The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one.
9. The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30.
10. The approximation would have been less accurate, because the smaller sample size means that the data does not fit normal curve as well.
88.
• n = 100; p = 0.1; q = 0.9
• μ = np = (100)(0.10) = 10
• σ = $npq npq$ = $(100)(0.1)(0.9) (100)(0.1)(0.9)$ = 3
1. $z=±1: x1=µ+zσ=10+1(3)=13z=±1:x1=µ+zσ=10+1(3)=13$ and $x2=µ–zσ=10–1(3)=7.68%x2=µ–zσ=10–1(3)=7.68%$ of the defective cars will fall between seven and 13.
2. $z=±2: x1=µ+zσ=10+2(3)=16z=±2:x1=µ+zσ=10+2(3)=16$ and $x2=µ–zσ=10–2(3)=4. 95 %x2=µ–zσ=10–2(3)=4. 95 %$ of the defective cars will fall between four and 16
3. $z=±3: x1=µ+ zσ=10+3(3)=19z=±3:x1=µ+zσ=10+3(3)=19$ and $x2=µ–zσ=10–3(3)=1. 99.7%x2=µ–zσ=10–3(3)=1. 99.7%$ of the defective cars will fall between one and 19.
90.
• n = 190; p = $1 5 1 5$ = 0.2; q = 0.8
• μ = np = (190)(0.2) = 38
• σ = $npq npq$ = $(190)(0.2)(0.8) (190)(0.2)(0.8)$ = 5.5136
1. For this problem: P(34 < x < 54) = 0.7641
2. For this problem: P(54 < x < 64) = 0.0018
3. For this problem: P(x > 64) = 0.0000012 (approximately 0)
92.
1. 24.5
2. 3.5
3. Yes
4. 0.67
93.
1. 63
2. 2.5
3. Yes
4. 0.88
94.

0.02

95.

0.37

96.

0.50