The mean amount of life insurance per household is $122 000. This distribution is positively skewed. The standard deviation of the population is $42 000. Use Appendix 8.1 for the zvalues. a. A random sample of 60 households revealed a mean of $127 000. What is the standard error of the mean? (Round the final answer to the nearest whole number.) Standard error of the mean b. Suppose that you selected 60 samples of households. What is the expected shape of the distribution of the sample mean? Shape c. What is the likelihood of selecting a sample with a mean of at least $127 000? (Round the 1 values to 2 decimal places and the final answers to 4 decimal places.) Probability d. What is the likelihood of selecting a sample with a mean of more than $115 000? (Round the 2 values to 2 decimal places and the final answers to 4 decimal places.) Probability e. Find the likelihood of selecting a sample with a mean of more than $115 000 but less than $127 000. (Round the 2 values to 2 decimal places and the final answers to 4 decimal places.) Probability