document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=47667>Question 47667</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The average weight of a box at a store is 10lbs. with a standard deviation of 2 lbs. If 200 boxes are at the store approximately how many boxes will be between 10 and 13 lbs. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #31485 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=31485\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The average weight of a box at a store is 10lbs. with a standard deviation of 2 lbs. If 200 boxes are at the store approximately how many boxes will be between 10 and 13 lbs.<BR>\n" );
document.write( "10 AND 13 CORRESPOND TO A DEVIATION OF +3 FROM MEAN OF 10 SINCE 13=10+3<BR>\n" );
document.write( "DEVIATION OF 3 IS 1.5 TIMES THE STANDARD DEVIATION OF 2 SINCE 2*1.5=3<BR>\n" );
document.write( "FROM NORMAL DISTRIBUTION IT IS EXPECTED THAT ABOUT 86% WILL BE BETWEEN -1.5 S.D TO +1.5 S.D AROUND THE MEAN <BR>\n" );
document.write( "THAT IS 86% ARE BETWEEN 10-3=7 AND 10+3=13...OR..43% WILL BE BETWEEN 10 AND 13<BR>\n" );
document.write( "HENCE APPROXIMATELY 200*43/100=86 BOXES MAY BE BETWEEN 10 AND 13 LBS. </SPAN>\n" );
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