document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1053453>Question 1053453</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Last year the number of skate boards produced per day at a certain factory were normally distributed with a mean of 20,500 and a standard deviation of 55 skateboards. Answer all of the following <BR>\n" );
document.write( "A. On what percent of the days last year did the factory produce 20,555 skateboards or fewer. (I think it's 84.1 percent) <BR>\n" );
document.write( "B. On what percent of days last year did the factory produce 20,610 skateboards or more ( I think it's 97.5 percent) <BR>\n" );
document.write( "C. On what percent of days last year did the factory produce 20,445 skateboards or fewer (I think it's 2.5 percent)  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #668682 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=668682\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> z=(x-mean)/sd<BR>\n" );
document.write( "=(20555-20500)/55=1 and that would translate to 84.1%<BR>\n" );
document.write( "20610-20500=110, and 110/55=2, and that translates to 97.7%<BR>\n" );
document.write( "20445-20500=-55 and -55/55=-1, and that translates to 15.9% (the complement of 84.1%) </SPAN>\n" );
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