document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1113615>Question 1113615</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In a recent&#8203; year, the mean number of strokes per hole for a famous golfer was approximately 3.3<BR>\n" );
document.write( "(a) Find the variance and standard deviation using the fact that the variance of a Poisson distribution is  sigma squared equals mu&#963;2=&#956;. b) How likely is this golfer to play an 18-hole round and have more than 72 strokes?\r<BR>\n" );
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document.write( "I have the answer to the Variance =3.3<BR>\n" );
document.write( "I also have the answer to the standard deviation = 1.8\r<BR>\n" );
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document.write( "I just can not come up with eh answer to how likely is the golfer to play an 18 hole round and have more than 72 strokes.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #728696 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=728696\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> mean = variance = 3.3<BR>\n" );
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document.write( "standard deviation = square root(3.3) = 1.8<BR>\n" );
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document.write( "P(X > 72) = 1 - P(X< or = 72) = 1 - summation i = 0 to 72 of P(i, 3.3)<BR>\n" );
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document.write( "P(X > 72) = 1 - 1 = 0<BR>\n" );
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