document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1006317>Question 1006317</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The clean-up crew of a medium-sized firm has an average clean-up time of 72.0 total hours with a variance of 64 hours squared. The clean-up times are bell-shaped. (a) What proportion of the time will it take the clean-up crew 56 or more hours for cleaning the plant? (b) Ninety-five percent of the time the total clean-up time will fall within what interval about the mean? State the lower and upper limits of the interval.<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 #622474 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=622474\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Mean=72 hours<BR>\n" );
document.write( "SD=8 hours, square root of variance<BR>\n" );
document.write( "56 hours is 2 SDs below the mean.<BR>\n" );
document.write( "0.0228 or 2.28% of the time, it will be fewer than 56 hours<BR>\n" );
document.write( "97.72% of the time, it will be greater than or equal to 56 hours.<BR>\n" );
document.write( "95% of the time is +/- 1.96 SDs or 15.68 hours<BR>\n" );
document.write( "CI is (56.32,87.68) units are hours. </SPAN>\n" );
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