document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1150679>Question 1150679</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The customer service department of HSBCC Bank receives the complaints following a Poisson distribution with the mean rate of two complaints per day.<BR>\n" );
document.write( "(i) What is the probability of at most two complaints per day?<BR>\n" );
document.write( "(ii) Assume that a week has five working days. What is the probability that a randomly selected week has three days with at most two complaints per day? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #772318 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=772318\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> at most 2 complaints is 0, 1, 2<BR>\n" );
document.write( "for 0 prob is e^(-2)*2^0/0!<BR>\n" );
document.write( "this is e^(-2) or 0.1353\r<BR>\n" );
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document.write( "for 1 p(1) is e^(-2)*2^1/1! or twice the above or 0.2707\r<BR>\n" );
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document.write( "for 2, it would be e^(-2)*2^2/2!<BR>\n" );
document.write( "or the same as for p(1) or 0.2707<BR>\n" );
document.write( "sum is 0.6767\r<BR>\n" );
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document.write( "For 3 days with at most 2 complaints per day is a binomial distribution<BR>\n" );
document.write( "5C3*0.6767^3*0.3233^2=0.3239, the 5C3 or 10 are the number of ways it can occur in 5 days. </SPAN>\n" );
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