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3.The average number of moves a person makes in his or her lifetime is 12 and the standard deviation is 3.3. Assume that the sample

3.The average number of moves a person makes in his or her lifetime is 12 and the standard deviation is 3.3. Assume that the sample is taken from a large population and the correction factor can be ignored. Round the final answers to four decimal places and intermediate z value calculations to two decimal places.

Part 1 of 3

Find the probability that the mean of a sample of 25 people is less than 10.

P(X<10)=?

Part 2 of 3

Find the probability that the mean of a sample of 25 people is greater than 10.

P(X>10)=?

Part 3 of 3

Find the probability that the mean of a sample of 25 people is between 11 and 12.

P(11

4.Medicare Hospital Insurance The average yearly Medicare Hospital Insurance benefit per person was $4064 in a recent year. Suppose the benefits are normally distributed with a standard deviation of$460. Round the final answers to at least four decimal places and intermediate z-value calculations to two decimal places.

Part 1 of 2

Find the probability that the mean benefit for a random sample of 24 patients is less than $3870.

P(X<3870)=?

Part 2 of 2

Find the probability that the mean benefit for a random sample of 24 patients is more than $4200.

P(X>4200)=?

5.Ages of Proofreaders At a large publishing company, the mean age of proofreaders is 36.2 years and the standard deviation is 3.7 years. Assume the variable is normally distributed. Round intermediate z-value calculations to two decimal places and the final answers to at least four decimal places.

Part 1 of 2

f a proofreader from the company is randomly selected, find the probability that his or her age will be between 35.5 and 37 years.

P(35.5

6.Mail Order A mail order company has a 11% success rate. If it mails advertisements to 582 people, find the probability of getting less than 57 sales. Round z-value calculations to 2 decimal places and final answer to at least 4 decimal places.

P(X<57)=?

7. Small Business Owners Seventy-six percent of small business owners do not have a college degree. If a random sample of 60 small business owners is selected, find the probability that exactly 49 will not have a college degree. Round the final answer to at least 4 decimal places and intermediate z-value calculations to 2 decimal places.

P(X=49)=?

8.Population of College Cities College students often make up a substantial portion of the population of college cities and towns. State College, Pennsylvania, ranks first with 71.1% of its population made up of college students. What is the probability that in a random sample of 175 people from State College, more than 50 are not college students? Round the final answer to at least 4 decimal places and intermediate z-value calculations to 2 decimal places.

P(X>50)=?

9.Residences of U.S. Citizens According to the U.S. Census, 67.5% of the U.S. population were born in their state of residence. In a random sample of 194 Americans, what is the probability that fewer than 118 were born in their state of residence? Round the final answer to at least four decimal places and intermediate z-value calculations to two decimal places.

P(X<118)=?

10.Overweight Men For a random sample of 60 overweight men, the mean of the number of pounds that they were overweight was 28. The standard deviation of the population is 4.4 pounds.

Part 2 of 4

(b) Find the 95% confidence interval of the mean of these pounds. Round intermediate answers to at least three decimal places. Round your final answers to one decimal place.

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11.Freshmen GPAs First-semester GPAs for a random selection of freshmen at a large university are shown. Estimate the true mean GPA of the freshman class with 90% confidence. Assume =0.62. Round intermediate and final answers to two decimal places. Assume the population is normally distributed. 3.2 1.9 2.8 2.2 4.0 1.9 2.8 2.0 2.7

3.9 3.8 3.5 2.7 3.1 3.8 3.0 3.2 2.8 2.7 2.5 1.9

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12.Day Care Tuition A random sample of 53 four-year-olds attending day care centers provided a yearly tuition average of $3922 and the population standard deviation of $630.

Part 1 of 2

Find the 99% confidence interval of the true mean. Round your answers to the nearest whole number.

$ < <$

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