document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1170266>Question 1170266</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In a food processing and packing plant, there are on average two packing machine breakdowns per week.\r<BR>\n" );
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document.write( "a) (2 marks) Determine the mean and the standard deviation of the distribution.\r<BR>\n" );
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document.write( "b) (2 marks) What is the probability that there are no machine breakdowns during a given week?\r<BR>\n" );
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document.write( "c) (4 marks) What is the probability that there are at most two machine breakdowns during a given week?\r<BR>\n" );
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document.write( "d) (3 marks) What is the probability that there are exactly three machine breakdowns during the next two weeks?\r<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 #795336 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=795336\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> would assume a Poisson both because likely to be a large number of such machines and also because there is no way to calculate the probability without being given the sd, unless there is a binomial, Poisson distribution, or a proportion where the two are dependent. \r<BR>\n" );
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document.write( "The mean and variance are both 2, so the sd is sqrt(V)=sqrt(2) or 1.414\r<BR>\n" );
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document.write( "P(0)=e^(-2)2^0/0!=0.1353<BR>\n" );
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document.write( "at most 2 is 0,1,2<BR>\n" );
document.write( "probability of 1 is e^(-2)*2^1/!=0.2707<BR>\n" );
document.write( "for 2 it is e^-2)*2^2/2!=0.2707<BR>\n" );
document.write( "the probability is 0.6767<BR>\n" );
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document.write( "for 3 in 2 weeks it is equivalent of parameter 4 for that time period, since the distribution is proportional to time. e^(-4)^4^3/3!=0.1954\r<BR>\n" );
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