Suppose the average number of calls received by an operator in one minute is 2. What is the probability of receiving 10 calls in 5 minutes ?
This is a Poisson distribution with the mean of 2∗5=10.2*5=10.2∗5=10.
P(X=10)=e−10101010!=0.1251.P(X=10)=e^{-10}\frac{10^{10}}{10!}=0.1251.P(X=10)=e−1010!1010=0.1251.
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