A dispatcher for an airport shuttle averages sending a van to the airport 2 times every 60 minutes.The distribution is Poisson,and the driver must take a 15-minute lunch break.The probability that he can finish lunch before getting a call is:
A) 0.1350.
B) 0.6065.
C) 0.3935.
D) 1.6490.
E) 0.8650.

Question

Quizplus · Accepted Answer

The rate of sending a van to the airport is 2 vans per 60 minutes, which can be written as λ = 2/60 = 1/30 vans per minute.

We need to find the probability that no calls come in during the 15-minute lunch break. We can use the Poisson distribution formula:

P(X = 0) = (e^(-λ)*λ^0)/0! = e^(-λ) = e^(-(1/30)*15) = e^(-1/2) = 0.6065

Therefore, the probability that the driver can finish lunch before getting a call is 0.6065 or 60.65%. The best choice is C.

A Dispatcher for an Airport Shuttle Averages Sending a Van