document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1187527>Question 1187527</A>: <I><SPAN STYLE=\"word-break: break-all;\">  Poisson Distribution. A healthy female rabbit typically produces an average of seven litters per year. Suppose that one healthy female rabbit is randomly chosen.\r<BR>\n" );
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document.write( "Fill in “True” or “False”: The random variable X represents the number of litters per year that a healthy female rabbit can have. <BR>\n" );
document.write( "Fill in “True” or “False”: X may not take on whole number values 0, 1, 2, 3, … <BR>\n" );
document.write( "The probability (in % terms) that she has exactly on litter in one year is <BR>\n" );
document.write( " %. (round to two decimals)<BR>\n" );
document.write( "The probability (in % terms) that she has at least two litters in one year is <BR>\n" );
document.write( " %. (round to two decimals)<BR>\n" );
document.write( "The probability (in % terms) that she has exactly three litters in one year is <BR>\n" );
document.write( " %. (round to two decimals)<BR>\n" );
document.write( "The probability (in % terms) that she has between 2 and 4 litters in one year is <BR>\n" );
document.write( " %. (round to two decimals) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #819212 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=819212\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> True<BR>\n" );
document.write( "False--X must take on whole number values in this instance. Mathematically it does not have to, but with litters, there are no fractional ones.<BR>\n" );
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document.write( "e^(-7)*7^1/1!=0.0064, or 0.01<BR>\n" );
document.write( "confirm with Poissonpdf(7,1) ENTER<BR>\n" );
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document.write( "we know that 1 litter has probability 0.0064 and 0 is e^(-7)=0.0009. They add to 0.0073, and the complement or 0.9927, is the answer, at least two litters. Two decimal places is 0.99<BR>\n" );
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document.write( "exactly 3 litters is e^(-7)*7^3/3!=0.0521, or 0.05<BR>\n" );
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document.write( "exactly 4 litters is e^(-7)*7^4/24=0.0912, or 0.09<BR>\n" );
document.write( "so the answer is that plus 0.0521 and the probability of 2 litters (0.0223)=0.1656, or 0.17 </SPAN>\n" );
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