document.write( "Question 1058515: Phone calls arrive at the rate of 48 per hour at the reservation desk for Regional Airways.
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\n" ); document.write( "a. Compute the probability of receiving four calls in a 10-minute interval of time.
\n" ); document.write( "b. Compute the probability of receiving exactly 12 calls in 20 minutes.
\n" ); document.write( "c. Suppose, no calls are currently on hold. If the agent takes 5 minutes to complete the current
\n" ); document.write( " call, how many callers do you expect to be waiting by that time? What is the probability that
\n" ); document.write( " none will be waiting?
\n" ); document.write( "d. If no calls are currently being processed, what is the probability that the agent can take 3 minutes
\n" ); document.write( " for personal time without being interrupted by a call?
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Algebra.Com's Answer #673647 by Boreal(15235)\"\" \"About 
You can put this solution on YOUR website!
Poisson distribution--discrete, proportional to time, theoretically could be infinite, random.
\n" ); document.write( "8 calls in 10 minutes is the proportion. That is the parameter. p(x)=e^(-lambda)*lambda^x/x!
\n" ); document.write( "So for 4, it is e^(-8)*8^4/4!=0.0573
\n" ); document.write( "For 12 calls in 20 minutes, expected is 16
\n" ); document.write( "e^(-16)16*12/12!=0.0661
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\n" ); document.write( "expected calls in five minutes is 1/12 of an hour during which 48 calls are received. I would expect 4 calls.
\n" ); document.write( "probability that none is waiting is e^(-4)*4^0/0!=e^(-4)=0.0183
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\n" ); document.write( "For 3 minutes without a call it would be an expectation of 2.4 (1/20 th of 48)
\n" ); document.write( "e^(-2.4)=0.0907
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