document.write( "Question 862717: (Please help me with this question?)\r
\n" ); document.write( "\n" ); document.write( "Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $120,000. This distribution follows the normal distribution with a standard deviation of $38,000.\r
\n" ); document.write( "\n" ); document.write( "(a)
\n" ); document.write( "If we select a random sample of 58 households, what is the standard error of the mean? (Round your answer to the nearest whole number.)\r
\n" ); document.write( "\n" ); document.write( " 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?
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\n" ); document.write( "
\n" ); document.write( " Not normal, the standard deviation is unknown.
\n" ); document.write( " Unknown.
\n" ); document.write( " Uniform
\n" ); document.write( " Normal.
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\n" ); document.write( "(c)
\n" ); document.write( "What is the likelihood of selecting a sample with a mean of at least $124,000? (Round z value to 2 decimal places and final answer to 4 decimal places.)\r
\n" ); document.write( "\n" ); document.write( " Probability=? \r
\n" ); document.write( "\n" ); document.write( "(d)
\n" ); document.write( "What is the likelihood of selecting a sample with a mean of more than $112,000? (Round z value to 2 decimal places and final answer to 4 decimal places.)\r
\n" ); document.write( "\n" ); document.write( " Probability=? \r
\n" ); document.write( "\n" ); document.write( "(e)
\n" ); document.write( "Find the likelihood of selecting a sample with a mean of more than $112,000 but less than $124,000. (Round z value to 2 decimal places and final answer to 4 decimal places.)\r
\n" ); document.write( "\n" ); document.write( " Probability=?
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Algebra.Com's Answer #519978 by rothauserc(4718)\"\" \"About 
You can put this solution on YOUR website!
a) standard error of the mean = 38000 / square root(56) = 5078
\n" ); document.write( "b) normal
\n" ); document.write( "c) P(x > or = 124000) = 1 - P(x<124000)
\n" ); document.write( "we calculate the z score for P(x<124000)
\n" ); document.write( "124000 - 120000 / 5078 = 0.78771169751870815282 = 0.79
\n" ); document.write( "consult z table for probability assocated with z score = .79
\n" ); document.write( "P(x > or = 124000) = 1 - .79 = 0.21
\n" ); document.write( "d) P(x > or = 112000) = 1 - P(x<112000)
\n" ); document.write( "we calculate the z score for P(x<112000)
\n" ); document.write( "112000 - 120000 / 5078 = -1.58
\n" ); document.write( "consult -z table for probability assocated with z score = -1.58
\n" ); document.write( "P(x > 112000) = 1 - P(x<112000) = 1 - 0.06 = 0.94
\n" ); document.write( "e) P(x > 112000 and x < 124000) = P(x<124000) - P(x<112000) = 0.79 - 0.06 = 0.73\r
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