document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=240859>Question 240859</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 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.<BR>\n" );
document.write( "a.	If we select a random sample of 50 households, what is the standard error of the mean?<BR>\n" );
document.write( "b.	What is the expected shape of the distribution of the sample mean?<BR>\n" );
document.write( "c.	What is the likelihood of selecting a sample with a mean of at least $112,000?<BR>\n" );
document.write( "d.	What is the likelihood of selecting a sample with a mean of more than $100,000<BR>\n" );
document.write( "e.	What is the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #176449 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=176449\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 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.<BR>\n" );
document.write( "a. If we select a random sample of 50 households, what is the standard error of the mean?<BR>\n" );
document.write( "The standard error depends on the level of significance.\r<BR>\n" );
document.write( "\n" );
document.write( "If alpha is 5%, SE = 2.0096*40,000/sqrt(50) = 11369.0143<BR>\n" );
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document.write( "b. What is the expected shape of the distribution of the sample mean?<BR>\n" );
document.write( "Normal with mean = 110,000 and std = 40,000/sqrt(25) = 5656.85<BR>\n" );
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document.write( "c. What is the likelihood of selecting a sample with a mean of at least $112,000?<BR>\n" );
document.write( "t(112,000)= (112,000-110,000)/[40,000/sqrt(50)] = 0.3536..<BR>\n" );
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document.write( "P(x >= 112000) = P(t >= 0.3536 with df = 49) = 0.3626<BR>\n" );
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document.write( "d. What is the likelihood of selecting a sample with a mean of more than $100,000<BR>\n" );
document.write( "Same process as \"c\"<BR>\n" );
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document.write( "e. What is the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000?<BR>\n" );
document.write( "Combine the t-values from \"c\" and \"d\".<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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