document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1180572>Question 1180572</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 2.	The weight distribution of a sample of adults with physical inabilities is approximately normal, with mean 72 and standard deviation 8. Find an interval of values around the mean such that it<BR>\n" );
document.write( "i.	Includes 95 percent of the observed values                                      <BR>\n" );
document.write( "ii.	Includes almost all observed values (and thus coincides with the range, min-max).         <BR>\n" );
document.write( "iii.	Includes 50 percent of the observed values.                                     <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 #810687 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=810687\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The first interval would be z=+/-1.96*sigma, to 1.96*8=15.68<BR>\n" );
document.write( "This is +/- 15.68 or (56.32, 87.68) some weight units, prob kg.<BR>\n" );
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document.write( "The second one would be +/-3 sd or +/- 24 , so the interval would be (48, 96)<BR>\n" );
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document.write( "The third would be +/- 0.674*8 (the 25th percentile is z=-0.674 and the 75th is z=+0.674)<BR>\n" );
document.write( "+/-5.39<BR>\n" );
document.write( "(66.61, 77.39)\r<BR>\n" );
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