Information from the American Institute of Insurance indicates the mean amount o
ID: 2669754 • Letter: I
Question
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 at least $112,000?
d. What is the likelihood of selecting a sample with a mean of more than $100,000?
e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000?
Explanation / Answer
Hi there, a. If we select a random sample of 50 households, what is the standard error of the mean? SE = sd/sqrt(N) = 40000/sqrt(50) = 5656.85 b.What is the expected shape of the distribution of the sample mean? It will be normally distributed. c. What is the likelihood of selecting a sample with a mean of at least $112,000? z = (112000-110000)/5656.85 z = 0.353553 from a z table... prob(z > 0.353553) = 0.3618 d. What is the likelihood of selecting a sample with a mean of more than $100,000? z = (100000-110000)/5656.85 z = -1.767768 from a table: prob(z > -1.767768) = 0.9615 e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000 = Part D - Part C: = 0.9615 - 0.3618 = 0.5997