1. mean = 4 hours; standard deviation = 1.2 hours; sample size = 16
3.
1. Check student’s solution.
2. 3.5, 4.25, 0.2441
5. The fact that the two distributions are different accounts for the different probabilities.
7.
- Χ = amount of change students carry
- Χ ~ E(0.88, 0.88)
- X⎯⎯⎯ = average amount of change carried by a sample of 25 sstudents.
- X⎯⎯⎯ ~ N(0.88, 0.176)
- 0.0819
- 0.1882
- The distributions are different. Part a is exponential and part b is normal.
9.
- 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, 13)
- Yes. I would be surprised, because the probability is almost 0.
- No. I would not be totally surprised because the probability is 0.2312
11.
- 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, 0.0220)
- 2.74 minutes
- 0.03 minutes
13.
- 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.
15.
- 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
- [latex]\overline{X}[/latex] ∼ N[latex]\left(2000,\frac{{8000}}{{\sqrt{1000}}}\right)[/latex]
- 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 < [latex]\overline{X}[/latex]< 2100) = 0.1537P(2100 < [latex]\overline{X}[/latex]< 2200) = 0.1317
17. [latex]\displaystyle\overline{{X}}[/latex] ~ N(4.59, [latex]\frac{{0.10}}{{\sqrt{16}}}[/latex])
The Central Limit Theorem for Sums
18. 0.3345
20. 7,833.92
22. 0.0089
24. 7,326.49
26. 77.45%
28. 0.4207
30. 3,888.5
32. 0.8186
34. 5
36. 0.9772
38. The sample size, n, gets larger.
40. 49
42. 26.00
44. 0.1587
46. 1,000
49.
- the total length of time for nine criminal trials
- N(189, 21)
- 0.0432
- 162.09; ninety percent of the total nine trials of this type will last 162 days or more.
51.
- X = the salary of one elementary school teacher in the district
- X ~ N(44,000, 6,500)
- ΣX ~ sum of the salaries of ten elementary school teachers in the sample
- ΣX ~ N(44000, 20554.80)
- 0.9742
- $52,330.09
- 466,342.04
- Sampling 70 teachers instead of ten would cause the distribution to be more spread out. It would be a more symmetrical normal curve.
- If every teacher received a $3,000 raise, the distribution of X would shift to the right by $3,000. In other words, it would have a mean of $47,000.
Using the Central Limit Theorem
53. N(25, 0.0577)
55. 0.0003
57. 25.07
59.
- N(2,500, 5.7735)
- 0
61. 10
63. N[latex]\left(10,\frac{{10}}{{8}}\right)[/latex]
65. 0.7799
67. 1.69
69. 0.0072
74.
-
- X = the closing stock prices for U.S. semiconductor manufacturers
- i. $20.71; ii. $17.31; iii. 35
- Exponential distribution, Χ ~ Exp[latex]\left(\frac{{1}}{{20.71}}\right)[/latex]
- Answers will vary.
- i. $20.71; ii. $11.14
- Answers will vary.
- Answers will vary.
- Answers will vary.
- N[latex]\overline{X}~\left(60,\frac{{9}}{{\sqrt{25}}}\right)[/latex]
- 0.5000
- 59.06
- 0.8536
- 0.1333
- N(1500, 45)
- 1530.35
- 0.6877
87.
- $52,330
- $46,634
89. We have
- μ = 17, σ = 0.8, [latex]\overline{x}[/latex]= 16.7, and n = 30. To calculate the probability, we use normalcdf (lower, upper, [latex]\mu,\frac{{\sigma}}{{\sqrt{n}}}[/latex]) = normalcdf [latex]\left(E-99,16.7,17,\frac{{0.8}}{{\sqrt{30}}}\right)[/latex] = 20
- If the process is working properly, then the probability that a sample of 30 batteries would have at most 16.7 lifetime hours is only 2%. Therefore, the class was justified to question the claim.
91.
- For the sample, we have n = 100, [latex]\overline{x}[/latex]= 0.862, s = 0.05
- [latex]\overline{x}[/latex]= 85.65, Σs = 5.18
normalcdf
(396.9,E99,(465)(0.8565),(0.05)([latex]\sqrt{465}[/latex])) ≈ 1- Since the probability of a sample of size 465 having at least a mean sum of 396.9 is appproximately 1, we can conclude that Mars is correctly labeling their M&M packages.
93. Use normalcdf
[latex]\left(E-99,1.1,1\frac{{1}}{{\sqrt{70}}}\right)[/latex]
= 0.7986. This means that there is an 80% chance that the service time will be less than 1.1 hours. It could be wise to schedule more time since there is an associated 20% chance that the maintenance time will be greater than 1.1 hours.