document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1112696>Question 1112696</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hello , Cant Solve This , Someone help me ?\r<BR>\n" );
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document.write( "The mean amount purchased by a typical customer at Churchill's Grocery Store is $23.50 with a standard deviation of $6.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 52 customers, answer the following questions.\r<BR>\n" );
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document.write( "Q1)What is the likelihood the sample mean is at least $25.50? (Round your z value to 2 decimal places and final answer to 4 decimal places.)<BR>\n" );
document.write( "Probability ?\r<BR>\n" );
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document.write( "Q2)What is the likelihood the sample mean is greater than $22.50 but less than $25.50? (Round your z value to 2 decimal places and final answer to 4 decimal places.)<BR>\n" );
document.write( "Probability ?\r<BR>\n" );
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document.write( "Q3) Within what limits will 95 percent of the sample means occur? (Round your answers to 2 decimal places.)<BR>\n" );
document.write( "Sample mean ....... and ...... </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #727729 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=727729\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> z=(x bar-mean)/sigma/sqrt(n)<BR>\n" );
document.write( "=(25.50-23.50)/6/sqrt(52)<BR>\n" );
document.write( "=2*sqrt(52)/6<BR>\n" );
document.write( "=2.40<BR>\n" );
document.write( "probability z is greater than that is 0.0082<BR>\n" );
document.write( "---------for 22.50<BR>\n" );
document.write( "z=-1*sqrt(52)/6<BR>\n" );
document.write( "= -1.20<BR>\n" );
document.write( "probability z is between these two values, -1.20 and 2.40, is 0.8767<BR>\n" );
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document.write( "95% will occur between +/- 1.96*6/sqrt(52)<BR>\n" );
document.write( "This is 1.63 on each side of the given mean of 23.50<BR>\n" );
document.write( "(21.87, 25.13)\r<BR>\n" );
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