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