document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=747863>Question 747863</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Heights of women have a bell-shaped distribution with a mean of 159cm and a standard deviation of 7cm. Using Chebyshev's theorem, what do we know about the percentage of women with heights that are within 3 standard deviations of the mean? What are the minimum and maximum heights that are within 3 standard deviations of the mean? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #455204 by </B><A HREF=/tutors/aboutme.mpl?userid=FrankM><B>FrankM(1040)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=FrankM><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=FrankM><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=455204\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Chebyshev stated that no more than 1/k2 of the distribution's values can be more than k standard deviations away from the mean. \r<BR>\n" );
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document.write( "So 3 standard deviations away means 1/9 of the values fall outside the 3 STD range. \r<BR>\n" );
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document.write( "159cm+/-21cm = 148cm min, 180cm max. \r<BR>\n" );
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document.write( "The maximum height is 5 ft 11 inches. I can't confirm it's true, but the suggestion that the high end, 1/2 of the outliers, or just 1/18 of the women, are this tall passes the common sense test.   </SPAN>\n" );
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