document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1000130>Question 1000130</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me solve this problem\r<BR>\n" );
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document.write( "Suppose the number of mechanical failures occurring in an industrial plant follows a Poisson distribution with an average of 1.4 failures per week.<BR>\n" );
document.write( "(a) What is the probability of no mechanical failures in a given week?<BR>\n" );
document.write( "(b) What is the probability that four or more mechanical failures will occur during a three-week period?\r<BR>\n" );
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document.write( "Thanks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #617673 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=617673\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> For (a) f(x)=exp^(-1.4)1.4^0/0!<BR>\n" );
document.write( "that is equal to exp(-1.4)=0.2466. That is the probability of no mechanical failures in a given week.<BR>\n" );
document.write( "For four or more, I need 1,2,3<BR>\n" );
document.write( "exp(-1.4)1.4/1=0.3452<BR>\n" );
document.write( "exp(-1.4)(1.4)^2/2=0.2417<BR>\n" );
document.write( "for 3 exp(-1.4)1.4^3/6=0.1128<BR>\n" );
document.write( "These four sum to 0.9463.<BR>\n" );
document.write( "The probability of more than 4 is 1-0.9463=0.0537<BR>\n" );
document.write( "exp(-lambda)lambda^x/x!, where lambda is the average of failures or whatever is being counted. </SPAN>\n" );
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