document.write( "Question 472858: 32. 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.\r
\n" ); document.write( "\n" ); document.write( "a. If we select a random sample of 50 households, what is the standard error of the mean?\r
\n" ); document.write( "\n" ); document.write( "b. What is the expected shape of the distribution of the sample mean?\r
\n" ); document.write( "\n" ); document.write( "c. What is the likelihood of selecting a sample with a mean of at least $112,000?\r
\n" ); document.write( "\n" ); document.write( "d. What is the likelihood of selecting a sample with a mean of more than $100,000?\r
\n" ); document.write( "\n" ); document.write( "e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000.\r
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Algebra.Com's Answer #324280 by edjones(8007)\"\" \"About 
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a)\"sigma%2Fsqrt%28n%29=40000%2Fsqrt%2850%29\"=5,657
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\n" ); document.write( "b) Normal curve
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\n" ); document.write( "c) z=(112000-110000)/5657=2000/5657=.354
\n" ); document.write( "Area above .354 on the normal curve is .362 the likelihood of selecting a sample with a mean of at least $112,000.
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