document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1152710>Question 1152710</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The mean amount purchased by each customer at Churchill’s Grocery Store is $35 with a standard deviation of $9. The population is positively skewed. For a sample of 42 customers, answer the following questions:\r<BR>\n" );
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document.write( "a. What is the likelihood the sample mean is at least $39? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.)\r<BR>\n" );
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document.write( "Sample mean            \r<BR>\n" );
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document.write( "b. What is the likelihood the sample mean is greater than $33 but less than $39? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.)\r<BR>\n" );
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document.write( "c. Within what limits will 98% of the sample means occur? (Round the final answers to 2 decimal places.)\r<BR>\n" );
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document.write( "Sample mean      </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #775565 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=775565\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Assuming I can use normality, since otherwise the problem can't be done without knowing the skew. <BR>\n" );
document.write( "z=(xbar-35)/9/sqrt(42)<BR>\n" );
document.write( "at least 39 for sample mean  <BR>\n" );
document.write( "z>(4*sqrt(42)/9)= 2.88<BR>\n" );
document.write( "that probability is 0.0020\r<BR>\n" );
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document.write( "between 33 and 39 <BR>\n" );
document.write( "z is -2*sqrt42/9 for $33, or -1.44<BR>\n" );
document.write( "probability z is between -1.44 and 2.88 is 0.9231\r<BR>\n" );
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document.write( "half-interval for 98%CI is z=2.33 and the interval is z sigma/sqrt(n)=3.24<BR>\n" );
document.write( "98% sample means will be ($31.76, $38.24)<BR>\n" );
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