document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1119062>Question 1119062</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A physical fitness association is including the mile run in its&#8203; secondary-school fitness test. The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 450450 seconds and a standard deviation of 6060 seconds. Between what times do we expect most&#8203; (approximately 95%) of the boys to run the&#8203; mile? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #734514 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=734514\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 95% would be a z between -1.96 and +1.96<BR>\n" );
document.write( "z=(x-mean)/sd<BR>\n" );
document.write( "1.96=(x-450)/60<BR>\n" );
document.write( "117.6=x-450<BR>\n" );
document.write( "x=567.6 seconds one end and 332.4 seconds other end.<BR>\n" );
document.write( "(332.4, 567.6) units are seconds. </SPAN>\n" );
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