iPad 9:52 PM Chapter 7 Assignment Question 2 (of 4) value 40.00 points The mean
ID: 3335224 • Letter: I
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iPad 9:52 PM Chapter 7 Assignment Question 2 (of 4) value 40.00 points The mean amount of life insurance per household is $110,000. This distribution is positively skewed. The standard deviation of the population is $40,000. Use Appendix B.1 for the z-values A random sample of 50 households revealed a mean of $112,000. What is the standard error of the mean? (Round the final answer to 2 decimal places.) a. Standard error of the mean b. Suppose that you selected 50 samples of households. What is the expected shape of the distribution of the sample mean? Sample mean (Click to select) c. What is the likelihood of selecting a sample with a mean of at least $112,000? (Round the final answer to 4 decimal places.) Sample mean d. What is the likelihood of selecting a sample with a mean of more than $100,000? (Round the final answer to 4 decimal places. Sample mean e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000. Round the final answer to 4 decimal places.) Sample meanExplanation / Answer
a) SE of mean = 40,000 / sqrt(50) = 5656.854
b) The expected shape of the distribution of the sample mean is
Populatio mean = 110,000
c) P(sample mean > 112000) = P(Z > (112000 - 110000)/5656.854) = P(Z > 0.35355) = 0.36184
d) P(sample mean > 100000) = P(Z > (100000 - 110000)/5656.854) = P(Z > -1.7678) = 0.96145
e) P(100000 < Sample Mean < 112000) = P(-1.7678 < Z < 0.35355) = P(Z<0.35355) - P(Z<-1.7678) =0.63816-0.03855 = 0.59961
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