document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=889447>Question 889447</A>: <I><SPAN STYLE=\"word-break: break-all;\"> According to Chebyshev's theorem, what proportion of a distribution will be within k = 4 standard deviations of the mean?  Show all work as to how to find this. please work the problem out. Thanks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #538165 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=538165\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> note that 4^2 = 16<BR>\n" );
document.write( "For K = 4 we have 1 – 1/K^2 = 1 - 1/16 = 15/16 = 93.75%. So Chebyshev’s theorem (inequality) says that at least 93.75% of the data values of any distribution must be within 4 standard deviations of the mean.<BR>\n" );
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