Asked by Student5
Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $120,000. This distribution follows the normal distribution with a standard deviation of $38,000.
(A) If we select a random sample of 58 households, what is the standard error of the mean? (Round your answer to the nearest whole number.)
Standard error of the mean=?
(B) What is the expected shape of the distribution of the sample mean?
Not normal, the standard deviation is unknown.
Unknown.
Uniform
Normal.
(C) What is the likelihood of selecting a sample with a mean of at least $124,000? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability=?
(D) What is the likelihood of selecting a sample with a mean of more than $112,000? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability=?
(E) Find the likelihood of selecting a sample with a mean of more than $112,000 but less than $124,000. (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability=?
(A) If we select a random sample of 58 households, what is the standard error of the mean? (Round your answer to the nearest whole number.)
Standard error of the mean=?
(B) What is the expected shape of the distribution of the sample mean?
Not normal, the standard deviation is unknown.
Unknown.
Uniform
Normal.
(C) What is the likelihood of selecting a sample with a mean of at least $124,000? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability=?
(D) What is the likelihood of selecting a sample with a mean of more than $112,000? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability=?
(E) Find the likelihood of selecting a sample with a mean of more than $112,000 but less than $124,000. (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability=?
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