document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=973667>Question 973667</A>: <I><SPAN STYLE=\"word-break: break-all;\"> For the first 7 months of the year, the average precipitation in Toledo, Ohio is 19.32 inches. If the average precipitation is normally distributed with a standard deviation of 2.44 inches, find the following probabilities: \r<BR>\n" );
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document.write( "A) a randomly selected year will have precipitation greater than 18 inches for the first 7 months <BR>\n" );
document.write( "B) five randomly selected years will have an average precipitation greater than 18 inches for the first 7 months  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #595775 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=595775\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> mean=19.32 inches<BR>\n" );
document.write( "sd=2.44 inches\r<BR>\n" );
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document.write( "z>(18-19.32)/2.44  ;;; This should have a probability of 50-84%<BR>\n" );
document.write( "z>(-1.32)/2.44 or z> -0.540  <BR>\n" );
document.write( "P=0.706\r<BR>\n" );
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document.write( "z>(-1.32)*sqrt(5)/2.44     The SE is 2.44/sqrt(5).  I inverted in division and put sqrt (5) in numerator.\r<BR>\n" );
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document.write( "z>-1.209 \r<BR>\n" );
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document.write( "P=0.887.  This is expected, because now we are looking at 5 random years ALL of which have an expected greater than 18 inch amount.<BR>\n" );
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