document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1198234>Question 1198234</A>: <I><SPAN STYLE=\"word-break: break-all;\"> John calculates the average height of the trees in his yard to be 6 feet, with a standard deviation of 1 foot. The tree closest to his house is 7 feet tall. According to the empirical rule, what percentage of the trees should he expect to fall between 6 feet and 7 feet?\r<BR>\n" );
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document.write( "One of the following is correct.  Which one?\r<BR>\n" );
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document.write( "A) 0.68 (68%)<BR>\n" );
document.write( "B) 0.95 (95%)<BR>\n" );
document.write( "C) 0.00 (0%)<BR>\n" );
document.write( "D) 0.34 (34%) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #831799 by </B><A HREF=/tutors/aboutme.mpl?userid=ewatrrr><B>ewatrrr(24785)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ewatrrr><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=ewatrrr><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=831799\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> For the normal distribution: <BR>\n" );
document.write( "one  standard deviation from the mean accounts for about 68.2% of the set ****<BR>\n" );
document.write( "two standard deviations from the mean account for about 95.4%<BR>\n" );
document.write( "and three standard deviations from the mean account for about 99.7%.<BR>\n" );
document.write( "µ = 6 and σ = 1\r<BR>\n" );
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document.write( "what percentage of the trees should he expect to fall between 6 feet and 7 feet?<BR>\n" );
document.write( " <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=68.2%2F2 ALIGN=MIDDLE  ALT=\"68.2%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > 0r 34% (Nearest whole number) </SPAN>\n" );
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