document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1123352>Question 1123352</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Scores on a test have a mean of 71 and Q3 is 82. The scores have a distribution that is approximately normal. Find P90. (You will need to first find the standard deviation.) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #739653 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=739653\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I assume that Q3 means the third quartile<BR>\n" );
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document.write( "the percentile associated with Q3 is 75%<BR>\n" );
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document.write( "the z-score associated with 75% is approximately 0.67<BR>\n" );
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document.write( "0.67 = (82 - 71)/standard deviation<BR>\n" );
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document.write( "standard deviation = 11/0.67 is approximately 16.42<BR>\n" );
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document.write( "z-score = (90 - 71)/16.42 is approximately 1.16<BR>\n" );
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document.write( "P ( X < 90 ) is approximately 0.8770 of 87.70 percentile<BR>\n" );
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