document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=976034>Question 976034</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The average number of phone calls/minute coming into a switch board between 2 p.m. and 4 p.m. is 2.5. Determine the probability that during one particular minute there will be<BR>\n" );
document.write( "i. 4 or fewer<BR>\n" );
document.write( "ii. more than 6 calls </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #597705 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=597705\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We use the Poisson probability distribution to solve<BR>\n" );
document.write( "Probability(P)(X for given lamda) = e^(-lamda)*lamda^X / X! <BR>\n" );
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document.write( "lamda for the problem is 2.5<BR>\n" );
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document.write( "i) P( X < or = 4 ) = P(0) + P(1) + P(2) + P(3) + P(4)<BR>\n" );
document.write( "P(0:2.5) = 0.082084999<BR>\n" );
document.write( "P(1:2.5) = 0.205212497<BR>\n" );
document.write( "P(2:2.5) = 0.256515621<BR>\n" );
document.write( "P(3:2.5) = 0.213763017<BR>\n" );
document.write( "P(4:2.5) = 0.133601886<BR>\n" );
document.write( "P( X < or = 4 ) = 0.89117802 approx 0.89 or 89%<BR>\n" );
document.write( "ii) P ( X > 6 ) = 1 - P( X < or = 6 )<BR>\n" );
document.write( "P(5:2.5) = 0.066800943<BR>\n" );
document.write( "P(6:2.5) = 0.027833726<BR>\n" );
document.write( "P ( X > 6 ) = 1 - 0.985812689 approx 0.01 or 1%<BR>\n" );
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