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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

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

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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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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