Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $135.000, This distribution follows the normal distribution with a standard deviation of $42.000. a. If we select a random sample of 76 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? O Answer is complete and correct. The distribution will be normal c. What is the likelihood of selecting a sample with a mean of at least $139,000? (Round your 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 $129,000? (Round your 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 $129,000 but less than $139,000. (Round your z value to 2 decimal places and final answer to 4 decimal places.) Probability

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I need help answering A, C, D, and E please

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is
$135.000, This distribution follows the normal distribution with a standard deviation of $42.000.
a. If we select a random sample of 76 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?
O Answer is complete and correct.
The distribution will be
normal
c. What is the likelihood of selecting a sample with a mean of at least $139,000? (Round your 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 $129,000? (Round your 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 $129,000 but less than $139,000. (Round your z value to 2
decimal places and final answer to 4 decimal places.)
Probability
Transcribed Image Text:Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $135.000, This distribution follows the normal distribution with a standard deviation of $42.000. a. If we select a random sample of 76 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? O Answer is complete and correct. The distribution will be normal c. What is the likelihood of selecting a sample with a mean of at least $139,000? (Round your 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 $129,000? (Round your 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 $129,000 but less than $139,000. (Round your z value to 2 decimal places and final answer to 4 decimal places.) Probability
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