document.write( "Question 1165213: Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $136,000. This distribution follows the normal distribution with a standard deviation of $36,000.\r
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\n" ); document.write( "\n" ); document.write( "If we select a random sample of 78 households, what is the standard error of the mean? (Round your answer to the nearest whole number.)\r
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\n" ); document.write( "\n" ); document.write( "What is the expected shape of the distribution of the sample mean?\r
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\n" ); document.write( "\n" ); document.write( "What is the likelihood of selecting a sample with a mean of at least $141,000? (Round your z value to 2 decimal places and final answer to 4 decimal places.)\r
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\n" ); document.write( "\n" ); document.write( "What is the likelihood of selecting a sample with a mean of more than $131,000? (Round your z value to 2 decimal places and final answer to 4 decimal places.)\r
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\n" ); document.write( "\n" ); document.write( "Find the likelihood of selecting a sample with a mean of more than $131,000 but less than $141,000. (Round your z value to 2 decimal places and final answer to 4 decimal places.)\r
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Algebra.Com's Answer #789774 by Boreal(15235)\"\" \"About 
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a. SEM is 36000/sqrt(78)=$4076.20
\n" ); document.write( "b. A sample from a normal distribution may be considered normal as well. The curve is narrower and steeper, as the variance is much less than for an individual.
\n" ); document.write( "c.the z would be 5000/4076.20 or 1.23. That would be 0.1093
\n" ); document.write( "d. c would also be the answer for < 131000, so the answer to d is 0.8997, the complement
\n" ); document.write( "e. Here, the z value is between -1.23 and +1.23, and that probability is 0.7813
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