Calculating the probability using Poisson Distribution

Question

During working hours, an office switchboard receives telephone calls at random and at an average of 3 calls per minute.

a) What is the expected number of calls received during a five minute period?

Ans: Let X = number of calls per minute therefore $\Bbb E[X] = 3$. During a five minute period, $5\,\Bbb E[X] = 15$ expected calls.

b) What is the probability that exactly the expected no. from (a) are received in a five minute period?

Ans: Let $Y$ = number of calls during a five minute period therefore from the previous question, $\Bbb E[Y] = 15$.

So, $\Bbb P(Y = 15) = 15^{15}\,\mathsf e^{-15} / 15!$

Are the answers right? Thanks!

Answer 1

Yes, that is correct.   There is not much more to say.

Answer 2

For Poisson Distribution, X=number of occurrences in a time span, $\lambda$=average number of occurrences per unit of time: $$\mathsf P(X{=}x)~=~\mathsf e^{-\lambda t}(\lambda t)^x/x! \qquad\big[x\in \Bbb N\big]$$ So I believe your answers are correct.