document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1105923>Question 1105923</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 3.2 non-work-related e-mails per hour. Assume the arrival of these e-mails is approximated by the Poisson distribution.<BR>\n" );
document.write( " \r<BR>\n" );
document.write( "\n" );
document.write( "What is the probability Linda Lahey, company president, received exactly 2 non-work-related e-mails between 4 P.M. and 5 P.M. yesterday? (Round your probability to 4 decimal places.)\r<BR>\n" );
document.write( "\n" );
document.write( " \r<BR>\n" );
document.write( "\n" );
document.write( " Probability \r<BR>\n" );
document.write( "\n" );
document.write( " \r<BR>\n" );
document.write( "\n" );
document.write( "What is the probability she received 6 or more non-work-related e-mails during the same period? (Round your probability to 4 decimal places.)\r<BR>\n" );
document.write( "\n" );
document.write( " <BR>\n" );
document.write( "Probability <BR>\n" );
document.write( " \r<BR>\n" );
document.write( "\n" );
document.write( " \r<BR>\n" );
document.write( "\n" );
document.write( "What is the probability she received two or less non-work-related e-mails during the period? (Round your probability to 4 decimal places.)\r<BR>\n" );
document.write( "\n" );
document.write( " <BR>\n" );
document.write( "Probability <BR>\n" );
document.write( "  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #721087 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=721087\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> P(2)=e^(-3.2)*(3.2)^2/2! from the Poisson formula, <BR>\n" );
document.write( " P(x)= e^(-lambda)*lambda^x/x!<BR>\n" );
document.write( "=0.2087<BR>\n" );
document.write( "=========================<BR>\n" );
document.write( "For 6, e^(-3.2)*3.2^6/6!=0.0608<BR>\n" );
document.write( "For 7, it is 0.0278<BR>\n" );
document.write( "For 8, it is 0.0111<BR>\n" );
document.write( "For 9, it is 0.0040<BR>\n" );
document.write( "For 10, it is 0.0013<BR>\n" );
document.write( "For 11, it is 0.0004<BR>\n" );
document.write( "For 12, it is 0.0001<BR>\n" );
document.write( "0.1055 is the probability.<BR>\n" );
document.write( "It would have been a little faster to do 5, 4,3,1,and 0 and have subtracted that sum from 1.<BR>\n" );
document.write( "=========================<BR>\n" );
document.write( "0 is e^-3.2 or 0.0408<BR>\n" );
document.write( "1 is 0.1304<BR>\n" );
document.write( "2 is 0.2087 from above<BR>\n" );
document.write( "Two or fewer emails would have probability of 0.3799. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );