4.11: Chapter 6 Solution (Practice + Homework)
- Page ID
- 51793
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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. \(z \approx –0.33\)
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 – μσ – σ = 700 – 514117700 – 514117 ≈ 1.59, the z-score for the SAT. Y – 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.
- \(n=100 ; p=0.1 ; q=0.9\)
- \(\mu=n p=(100)(0.10)=10\)
- \(\sigma=\sqrt{n p q}=\sqrt{(100)(0.1)(0.9)}=3\)
i. \(z= \pm 1: x_1=\mu+\mathrm{z} \sigma=10+1(3)=13\) and \(x 2=\mu-\mathrm{z} \sigma=10-1(3)=7.68 \%\) of the defective cars will fall between seven and 13 .
ii. \(z= \pm 2: x_1=\mu+\mathrm{z} \sigma=10+2(3)=16\) and \(x 2=\mu-\mathrm{z} \sigma=10-2(3)=4.95 \%\) of the defective cars will fall between four and 16
iii. \(z= \pm 3: x_1=\mu+\mathrm{z} \sigma=10+3(3)=19\) and \(x 2=\mu-\mathrm{z} \sigma=10-3(3)=1.99 .7 \%\) of the defective cars will fall between one and 19.
90.
- \(n=190 ; p=\frac{1}{5}=0.2 ; q=0.8\)
- \(\mu=n p=(190)(0.2)=38\)
- \(\sigma=\sqrt{n p q}=\sqrt{(190)(0.2)(0.8)}=5.5136\)
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