# What is the variance of the time of the arrival of the 3rd customer?

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.

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 $\lambda$.
From the incorrect speculation in the question that the answer is 5+5+5, I think there may be confusion about the distribution of the waiting time for the first customer to arrive. That distribution is exponential with rate $\lambda$, hence mean $1/\lambda$ and variance $1/\lambda^2$. So the variance of the waiting time for the first customer is 1/25; and then the variance of the waiting time for the third customer is 1/25 + 1/25 + 1/25.
Notes: (1) Eventually, in doing more with queues, it is essential to know that the name of the distribution of the waiting time for the $n$th customer is the 'gamma' (here also 'Erlang') distribution, but that information is not necessary to solve the elementary problem asked. (2) A Poisson random variable has mean numerically equal to variance; an exponential random variable has mean equal to standard deviation.