Question 1

Assume that the weight loss for the first month of a diet program varies between 6 pounds and 12 pounds, and is spread evenly over the range of possibilities, so that there is a uniform distribution. Find the probability of the given range of pounds lost. More than 11 pounds

2/3

1/6

2/3

1/6

5/6

1/7

5/6

1/7

Question 2

If z is a standard normal variable, find the probability. The probability that z lies between 0 and

3.01

0.1217

0.1217

0.9987

0.5013

0.4987

0.9987

0.5013

0.4987

Question 3

Solve the problem. Round to the nearest tenth unless indicated otherwise. In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. Find , which is the consumption level separating the bottom 45% from the top 55%.

1078.3

1021.7

1148.1

1087.8

1078.3

1021.7

1148.1

1087.8

Question 4

Assume that X has a normal distribution, and find the indicated probability. The mean is 60.0 and the standard deviation is 4.0. Find the probability that X is less than 53.0.

0.0802

0.5589

0.0802

0.5589

0.9599

0.0401

0.9599

0.0401

Question 5

Solve the problem. The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 94 inches, and a standard deviation of 14 inches. What is the probability that the mean annual snowfall during 49 randomly picked years will exceed 96.8 inches?

0.0808

0.0026

0.0808

0.0026

0.4192

0.5808

0.4192

0.5808

Question 6

Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution. With n = 18 and p = 0.30, estimate P(6).

0.1015

0.8513

0.1958

0.1239

0.1015

0.8513

0.1958

0.1239

Question 7

Use the normal distribution to approximate the desired probability. A coin is tossed 20 times. A person, who claims to have extrasensory perception, is asked to predict the outcome of each flip in advance. She predicts correctly on 14 tosses. What is the probability of being correct 14 or more times by guessing? Does this probability seem to verify her claim?

0.4418, no

0.0582, no

0.4418, yes

0.0582, yes

Question 8

Solve the problem. The following confidence interval is obtained for a population proportion, p: 0.689 < p < 0.723. Use these confidence interval limits to find the margin of error, E.

0.017

0.706

0.017

0.706

0.018

0.034

.018

0.034

Question 9

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places. 95% confidence; n = 320, x = 60

0.0449

0.0514

0.0449

0.0514

0.0428

0.0385

0.0428

0.0385

Question 10

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n = 51, x = 27; 95% confidence

0.414 < p < 0.644

0.392 < p < 0.666

0.391 < p < 0.667

0.413 < p < 0.645

Question 11

Use the given data to find the minimum sample size required to estimate the population proportion. Margin of error: 0.004; confidence level: 95%; unknown

60,148

60,018

60,025

50,024

60,148

60,018

60,025

50,024

Question 12

Solve the problem. Round the point estimate to the nearest thousandth. Find the point estimate of the proportion of people who wear hearing aids if, in a random sample of 304 people, 20 people had hearing aids.

0.063

0.063

0.066

0.934

0.062

Question 13

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval.

0.471 < p < 0.472

0.435 < p < 0.508

0.438 < p < 0.505

0.444 < p < 0.500

Question 14

Solve the problem. A newspaper article about the results of a poll states: “In theory, the results of such a poll, in 99 cases out of 100 should differ by no more than 5 percentage points in either direction from what would have been obtained by interviewing all voters in the United States.” Find the sample size suggested by this statement.

544

544

664

27

385

664

27

385

Question 15

Use the confidence level and sample data to find the margin of error E. Round your answer to the same number of decimal places as the sample mean unless otherwise noted. Weights of eggs: 95% confidence n = 53,

0.36 oz

0.13 oz

0.02 oz

0.16 oz

Question 16

Use the confidence level and sample data to find a confidence interval for estimating the population mu. Round your answer to the same number of decimal places as the sample mean. Test scores: n = 92, mean = 90.6, sigma = 8.9; 99% confidence

88.4 < mu < 92.8

88.8 < mu < 92.4

88.2 < mu < 93.0

89.1 < mu < 92.1

Question 17

Use the given information to find the minimum sample size required to estimate an unknown population mean mu. Margin of error: $120, confidence level: 95%, sigma = $593

133

133

94

83

66

94

83

66

Question 18

Assume that a sample is used to estimate a population mean mu. Use the given confidence level and sample data to find the margin of error. Assume that the sample is a simple random sample and the population has a normal distribution. Round your answer to one more decimal place than the sample standard deviation. 95% confidence; n = 91; x-bar = 16, s = 9.1

1.71

4.10

1.71

4.10

1.63

1.90

1.63

1.90

Question 19

Use the given degree of confidence and sample data to construct a confidence interval for the population mean mu. Assume that the population has a normal distribution. A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 225 milligrams with s = 15.7 milligrams. Construct a 95% confidence interval for the true mean cholesterol content of all such eggs.

215.0 mg < mu < 235.0 mg

216.9 mg < mu < 233.1 mg

214.9 mg < mu < 235.1 mg

215.1 mg < mu < 234.9 mg

Question 20

Solve the problem. Find the critical value corresponding to a sample size of 3 and a confidence level of 95 percent.

7.378

5.991

0.103

0.0506

7.378

5.991

0.103

0.0506