6.11: Solutions
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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. about 68%
33. about 4%
35. between –5 and –1
37. about 50%
39. about 27%
41. The lifetime of a Sunshine CD player 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.
57.0.154 0.874
59. 0.693
60. 0.346
61. 0.110
62. 0.946
63. 0.071
64. 0.347
66. c
68.
- 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.
- 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.
- 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.
- iv
- 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.
- X = 720 720 – 52015720 – 52015 = 1.74 The exam score of 720 is 1.74 standard deviations above the mean of 520.
- 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. - X – μσ𝑋 – 𝜇σ = 700 – 514117700 – 514117 ≈ 1.59, the z-score for the SAT. Y – μσ𝑌 – 𝜇𝜎 = 30 – 215.330 – 215.3 ≈ 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
79.
- X ~ N(66, 2.5)
- 0.5404
- No, the probability that an Asian male is over 72 inches tall is 0.0082
81.
- X = number of hours that a Chinese four-year-old in a rural area is unsupervised during the day.
- X ~ N(3, 1.5)
- The probability that the child spends less than one hour a day unsupervised is 0.0918.
- The probability that a child spends over ten hours a day unsupervised is less than 0.0001.
- 2.21 hours
83.
- X = the distribution of the number of days a particular type of criminal trial will take
- X ~ N(21, 7)
- The probability that a randomly selected trial will last more than 24 days is 0.3336.
- 22.77
85.
- mean = 5.51, s = 2.15
- Check student's solution.
- Check student's solution.
- Check student's solution.
- X ~ N(5.51, 2.15)
- 0.6029
- The cumulative frequency for less than 6.1 minutes is 0.64.
- 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.
- 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.
- 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=√(100)(0.1)(0.9)=3
i. z=±1:x1=μ+zσ=10+1(3)=13 and x2=μ−zσ=10−1(3)=7.68% of the defective cars will fall between seven and 13 .
ii. z=±2:x1=μ+zσ=10+2(3)=16 and x2=μ−zσ=10−2(3)=4.95% of the defective cars will fall between four and 16
iii. z=±3:x1=μ+zσ=10+3(3)=19 and x2=μ−zσ=10−3(3)=1.99.7% of the defective cars will fall between one and 19.
90.
- n=190;p=15=0.2;q=0.8
- μ=np=(190)(0.2)=38
- σ=√npq=√(190)(0.2)(0.8)=5.5136
a. For this problem: P(34<x<54)=0.7641
b. For this problem: P(54<x<64)=0.0018
c. For this problem: P(x>64)=0.0000012 (approximately 0 )
92.
- 24.5
- 3.5
- Yes
- 0.67
93.
- 63
- 2.5
- Yes
- 0.88
94. 0.02
95. 0.37
96. 0.50