Normal Probability & Confidence Interval
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Chapter 8
34. Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000.
a) If we select a random sample of 50 households what is the standard error of the mean?
b) What is the expected shape of the distribution of the sample mean?
c) What is the likelihood of selecting a sample with a mean of a least $112,000
d) What is the likelihood of selecting a sample with a mean of more than $100k
e) Find the likelihood of selecting a sample with a mean of more than look but less than $112,000
Chapter 9
35. Marty Rowatti recently assumed the position of the director of the YMCA of South Jersey. He would like some current data on how long current members of the YMCA have been members. To investigate, suppose he selects a random sample of 40 current members. The mean length of membership of those included in the sample is 8.32 years and the standard deviation is 3.07 years.
a. What is the mean of the population?
b. Develop a 90% confidence interval for the population mean.
c. The previous director, in the summary report she prepared as she retired indicated the mean length of membership was now "almost 10 years." Does the sample information substantiate this claim? Cite evidence.
33. A recent study of 50 self-service gasoline stations in the Greater Cincinnati- Northern Kentucky metropolitan area revealed that the mean price of unleaded gas was $2.029 per gallon. The sample standard deviation was $0.03 per gallon.
a. Determine a 99 percent confidence interval for the population mean price.
b. Would it be reasonable to conclude that the population mean was $1.50? Why or why not?
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Solution Summary
The solution provides step by step method for the calculation of normal probabilities and confidence interval for population mean. Formula for the calculation and Interpretations of the results are also included.
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