document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1081726>Question 1081726</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A recent study of the lifetimes of cell phones found the average is 24.3 months. The standard deviation is 2.6 months. If a company provides its 35 employees with a cell phone, find the probability that the mean lifetime of these phones will be less than 23.9 months. Assume cell phone life is a normally distributed variable, the sample is taken from a large population and the correction factor can be ignored. . Round final answer to four decimal places and intermediate z-value calculations to two decimal places. \r<BR>\n" );
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document.write( "P(X <23.9)=<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 #695805 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=695805\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> t df=34<(xbar-mean)/s/sqrt(n)<BR>\n" );
document.write( "<(23.9-24.3)/2.6/sqrt(35)<BR>\n" );
document.write( "<-0.4*sqrt(35)/2.6=-0.91<BR>\n" );
document.write( "that probability is 0.1845. </SPAN>\n" );
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