Hello all!
I'm having some difficulty understanding how to solve this problem. I rewrote it a bit for better understanding as the real problem has over 50 people and also to practice myself after I know how it should be done.

There are 10 people waiting in the dentist office. The average waiting time is 5 minutes. What is the chance that the total waiting time is more than 60 minutes?
It is exponentially distributed.

This is how far I got:
I understand that it is not the chance of everyone waiting 6 minutes or more, someone waiting 40 minutes while the rest is waiting just 5 will have the same result. So I can't multiply all the chances of someone waiting more than 6 minutes.
On the internet I found a formula that says to substract the time someone waited from the 60 minutes, but this leaves me with over 50 variables in my problem. This can hardly be the solution?
Can someone give me a hint?

2. Re: Adding exponential waiting time

The sum of $n$ i.i.d. exponentially distributed rvs has an Erlang distribution.

let $X_k \sim \lambda e^{-\lambda t},~k=1,~\dots,~n$

$W = \sum \limits_{k=1}^n~X_k \sim \dfrac{\lambda^n t^{n-1} e^{-\lambda t}}{(n-1)!}$

and thus, with $n=10,~\lambda = \dfrac 1 5$ we have

$P[W > 60] = 1 - \displaystyle \int_0^{60}~\dfrac{5^{-10} t^9 e^{-t/5}}{9!}~dt \approx 0.2424$