document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1071705>Question 1071705</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The mean amount purchased by a typical customer at Churchill's Grocery Store is $26.00 with a standard deviation of $6.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 62 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 $27.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)<BR>\n" );
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document.write( "b.	<BR>\n" );
document.write( "What is the likelihood the sample mean is greater than $25.00 but less than $27.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)<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( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #686654 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=686654\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The mean amount purchased by a typical customer at Churchill's Grocery Store is $26.00 with a standard deviation of $6.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 62 customers, answer the following questions.<BR>\n" );
document.write( "a.	<BR>\n" );
document.write( "What is the likelihood the sample mean is at least $27.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)<BR>\n" );
document.write( "z(27) = (27-26)/(6/sqrt(62)) = 1.31<BR>\n" );
document.write( "P(x-bar >=27) = P(z > 1.31) = normalcdf(1.31,100) = 0.0951<BR>\n" );
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document.write( "b.	<BR>\n" );
document.write( "What is the likelihood the sample mean is greater than $25.00 but less than $27.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)<BR>\n" );
document.write( "z(25) = (25-26)/(6/sqrt(62)) = -1.31<BR>\n" );
document.write( "P(25< x-bar < 27) = P(-1.31< z < 1.31) = normalcdf(-1.31,1.31) = 0.8098<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( "Left tail of that interval has 0.05 of the area under the curb<BR>\n" );
document.write( "Left boundary of the interval:: z = invNorm(0.05) = -1.6449<BR>\n" );
document.write( "Right boundary of the interval:: z = invNorm(0.95) = 1.6449<BR>\n" );
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document.write( "Corresponding sample mean interval ::<BR>\n" );
document.write( "x-bar = -1.6449*(6/sqrt(62)) + 26 = $24.75<BR>\n" );
document.write( "x-bar = +1.6449*(6/sqrt(62))+26 = $27.25<BR>\n" );
document.write( "Ans: (24.75,27,25)<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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