Customers arrive in single server line to be served according to Poisson process with intensity 5 customers an hour. (there is a single server who services each customer in the order they arrive while all the customers wait in line). The customers begin to arrive at 8am.
What is the variance of the time of the arrival of the 3rd customer?
Is it simply 5+5+5 = 15? Because variance for Poisson r.v. is the rate if I recall correctly.
Best Answer
The r.v. for the number of arrivals in a given time has a Poisson distribution but here the r.v. we want is the waiting time for the third arrival. This r.v., call it $S_3$, is known to have a Gamma distribution with parameters $n$ and $\lambda$ ($\lambda=5$ being this distribution's "rate" parameter and $n=3$). Then,
$$Var(S_3) = \dfrac{n}{\lambda^2} = \dfrac{3}{25}.$$
Note: $S_3$ is the sum of three independent inter-arrival time r.v.s, each of which has the exponential distribution with parameter $\lambda$.