# Chapter 7 Solution (Practice + Homework)

- Page ID
- 6058

**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