document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1133150>Question 1133150</A>: <I><SPAN STYLE=\"word-break: break-all;\"> As manager of a company you know that the distribution of completion times for an assembly operation is a normal distribution with a mean of 120 seconds and a standard deviation of 20 seconds. If you have to award bonuses to the top 10% of your workers what time would you use as a cut-off time? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #750349 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=750349\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the z value for 0.90 is 1.282.  Here, however, we want the people with the LEAST time needed, so we will use z=-1.282, since the top 10 workers will be at the left end of the curve for time.<BR>\n" );
document.write( "z=(x-mean)/sd<BR>\n" );
document.write( "-1.282=(x-120)/20<BR>\n" );
document.write( "-25.64=x-120<BR>\n" );
document.write( "x=94.36 seconds or fewer </SPAN>\n" );
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