The mean amount of life insurance per household is $127 000. This distribution is positively skewed. The standard de population is $41 000. Use Appendix B1 for the z values.

a. A random sample of 50 households revealed a mean of $133 000. What is the standard error of the mean? (Round to the nearest whole number.)

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

Shape

(Click to select)

c. What is the likelihood of selecting a sample with a mean of at least $133 000? (Round the z values to 2 decimal ple final answers to 4 decimal pleces.)

Probability

d. What is the likelihood of selecting a sample with a-mean of more than $121 000? Probability

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e. Find the likelihood of selecting a sample with a mean of more than $121 000 but less than $133 000. (Round the z values to 2 decimal places end the final answers to 4 decimal pleces.)

Probability ?