document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1202769>Question 1202769</A>: <I><SPAN STYLE=\"word-break: break-all;\"> An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of 3.5 non-work-related e-mails per hour. Assume the arrival of these e-mails is approximated by the Poisson distribution.\r<BR>\n" );
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document.write( "a. What is the probability Linda Lahey, company president, received exactly 4 non-work-related e-mails between 4 p.m. and 5 p.m. yesterday? <BR>\n" );
document.write( "b. What is the probability she received 6 or more non-work-related e-mails during the same period?<BR>\n" );
document.write( "c. What is the probability she received four or less non-work-related e-mails during the period?<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 #838218 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=838218\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> for 4 it is e^(-3.5)*3.5^4/4!=0.1888<BR>\n" );
document.write( "for more than 6 it is 1-poisson cdf(3.5,6)=0.0653<BR>\n" );
document.write( "for 4 or fewer it is <BR>\n" );
document.write( "0.1888 for 4<BR>\n" );
document.write( "0.2158 for 3<BR>\n" );
document.write( "0.1850 for 2<BR>\n" );
document.write( "0.1057 for 1<BR>\n" );
document.write( "0.0302 for 0<BR>\n" );
document.write( "0.7254 is the answer </SPAN>\n" );
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