Poisson distribution - question

Question

A company receives 720 messages ordering its product in 8 hours of operation. What is the probability that:

a) In 6 minutes receive at least 4 messages?

b) In 4 minutes do not receive any messages?

Answer 1

Poisson distribution term for $n$ messages $=e^{-\lambda t}\frac{(\lambda t)^n}{n!}$ where $t$ is the time and $\lambda$ is the mean message rate. For you case, the time is in minutes and $\lambda =1.5$.

Set up th Poisson distribution for the given mean. For a) sum up the terms for less than 4 messages for 6 minutes and subtract from 1. For b) Look at the 0 term for 4 minutes. Thus a)$=1-e^{-9}\sum_{n=0}^3\frac{9^n}{n!}$ while b)$=e^{-6}$.