5.12: Chapter 5 Solution (Practice + Homework)
- Page ID
- 50580
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)
1.
- \(U(24, 26), 25, 0.5774\)
- \(N(25, 0.0577)\)
- \(0.0416\)
3.
0.0003
5.
25.07
7.
- \(N(2,500, 5.7735)\)
- \(0\)
9.
2,507.40
11.
- \(10\)
- \(\frac{1}{10}\)
13.
\(N(10, \frac{10}{8}))\)
15.
0.7799
17.
1.69
19.
0.0072
21.
391.54
23.
405.51
25.
Mean = 25, standard deviation = 2/7
26.
Mean = 48, standard deviation = 5/6
27.
Mean = 90, standard deviation = 3/4
28.
Mean = 120, standard deviation = 0.38
29.
Mean = 17, standard deviation = 0.17
30.
Expected value = 17, standard deviation = 0.05
31.
Expected value = 38, standard deviation = 0.43
32.
Expected value = 14, standard deviation = 0.65
33.
0.23
34.
0.060
35.
1/5
36.
0.063
37.
1/3
38.
0.056
39.
1/10
40.
0.042
41.
0.999
42.
0.901
43.
0.301
44.
0.832
45.
0.483
46.
0.500
47.
0.502
48.
0.519
49.
- \(Χ\) = amount of change students carry
- \(Χ \sim E(0.88, 0.88)\)
- \(\overline X\) = average amount of change carried by a sample of 25 students.
- \(\overline X \sim N(0.88, 0.176)\)
- \(0.0819\)
- \(0.1882\)
- The distributions are different. Part 1 is exponential and part 2 is normal.
51.
- length of time for an individual to complete \(IRS\) form 1040, in hours.
- mean length of time for a sample of 36 taxpayers to complete \(IRS\) form 1040, in hours.
- \(N(10.53, \frac{1}{3})\)
- Yes. I would be surprised, because the probability is almost 0.
- No. I would not be totally surprised because the probability is 0.2312
53.
- the length of a song, in minutes, in the collection
- \(U(2, 3.5)\)
- the average length, in minutes, of the songs from a sample of five albums from the collection
- \(N(2.75, 0.066)\)
- 2.74 minutes
- 0.03 minutes
55.
- True. The mean of a sampling distribution of the means is approximately the mean of the data distribution.
- True. According to the Central Limit Theorem, the larger the sample, the closer the sampling distribution of the means becomes normal.
- The standard deviation of the sampling distribution of the means will decrease making it approximately the same as the standard deviation of X as the sample size increases.
57.
- \(X\) = the yearly income of someone in a third world country
- the average salary from samples of 1,000 residents of a third world country
- \(\overline X \sim N\left(2000, \frac{8000}{\sqrt{1000}}\right)\)
- Very wide differences in data values can have averages smaller than standard deviations.
- The distribution of the sample mean will have higher probabilities closer to the population mean.
\(P(2000 < \overline X < 2100) = 0.1537 \)
\(P(2100 < \overline X < 2200) = 0.1317\)
59.
b
60.
64
61.
- Yes
- Yes
- Yes
- 0.6
62.
400
63.
2.5
64.
25
65.
0.0087
66.
0.0064, 0.0064
67.
- It has no effect.
- It is divided by \(\sqrt(2)\).
- It is divided by 2.
68.
- 4/5
- 0.04
- 0.0016
69.
- Yes
- No
70.
0.955
71.
0.927
72.
0.648
73.
0.101
74.
0.273