Suppose a mean of 20 customers arrive at the drive-through windows at a bank each hour. If each bank teller can handle 6 customers per hour, how many tellers are needed so that the probability all customers can be served in the hour is at least 0.95?
it should be P(x=0)>=0.95, but i do not understand what will be the average (mean) 20 or 20-6n?
I saw an answer to this question recently and my answer differs slightly, so please be critical of my response. Suppose that $X$ customers enter the drive through window in a given hour, and let $n$ denote the number of bank tellers present. Then $X\sim Poisson(20)$ and the event $\{X\leq6n\}$ corresponds to the event that all customers can be served in the given hour. We need to find the smallest value of $n$ so that $$P(X\leq6n)=\sum_{k=0}^{6n}e^{-20}\Bigg(\frac{20^k}{k!}\Bigg)\geq0.95$$ You can verify here that the smallest value of $n$ is 5. The alternate approach I read seems valid but produces an alarmingly small value of $n$. Here it is.
If we take $Y$ to represent the number of customers that cannot be served in a given hour, then $Y\sim Poisson(20-6n)$ and we should to find the smallest value of $n$ so that $P(Y=0)\geq 0.95$. It follows that $$P(Y=0)=e^{6n-20}\geq 0.95 \iff n\geq \left \lceil{\frac{\ln(0.95)+20}{6}}\right \rceil=4.$$ This seems a bit to small for my liking since $X\geq 25$ implies at least one customer doesn't get service and occurs with probability $\approx 16$%.