document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1092901>Question 1092901</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The mean amount purchased by a typical customer at Churchill's Grocery Store is $21.50 with a standard deviation of $7.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 44 customers, answer the following questions.<BR>\n" );
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document.write( "a.	<BR>\n" );
document.write( "What is the likelihood the sample mean is at least $24.50? (Round z value to 2 decimal places and final answer to 4 decimal places.)<BR>\n" );
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document.write( "  Probability	____<BR>\n" );
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document.write( "b.	<BR>\n" );
document.write( "What is the likelihood the sample mean is greater than $19.50 but less than $24.50? (Round z value to 2 decimal places and final answer to 4 decimal places.)<BR>\n" );
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document.write( "  Probability	____<BR>\n" );
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document.write( "c.	<BR>\n" );
document.write( "Within what limits will 90 percent of the sample means occur? (Round your answers to 2 decimal places.)<BR>\n" );
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document.write( "  Sample mean ____and _____<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 #707506 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=707506\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The mean of the sample distribution is the same as the mean of the population.  The standard error is the standard deviation of the population/sqrt(sample size)<BR>\n" );
document.write( "Therefore, SE is 7/sqrt (44)=1.055<BR>\n" );
document.write( "now use z-test<BR>\n" );
document.write( "z>=(x-pop mean)/1.055=(24.50-21.50)/1.055, so z > 2.84 or 0.0023.<BR>\n" );
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document.write( "z for 19.50 is -2/1.055 or -1.90<BR>\n" );
document.write( "z for 24.50 is 2.84.  <BR>\n" );
document.write( "Want the probability of z being between those two values, and that is 0.9690<BR>\n" );
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document.write( "90% sample means are between z -1.65 and +1.65<BR>\n" );
document.write( "z=(x-mean)/sd<BR>\n" );
document.write( "-1.65*1.055=x-mean<BR>\n" );
document.write( "-1.74+mean=x and that value would be $19.76<BR>\n" );
document.write( "+1.74 would be on the other side and that is $23.24<BR>\n" );
document.write( "($19.76, $23.24)<BR>\n" );
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