Time until the tenth customer arrives

Question

Customers arrive as a Poisson process at rate 4 per hour, so the gaps between arrivals are independent exponentials with rate $\lambda = 4$ per hour. Let $T$ be the time (in hours) until the 10th customer arrives.

What are the mean and variance of $T$ ?

Accepted Answer

A Poisson process has exponential inter-arrival times, and the time to the 10th arrival is the sum of 10 independent exponential gaps, an Erlang. Each gap has mean 1/\lambda = 1/4 hour and variance 1/\lambda^2 = 1/16. Adding 10 of them: \mathbb{E}[T] = \frac{10}{4} = 2.5 	ext{ hours}, \qquad \operatorname{Var}(T) = \frac{10}{16} = 0.625. Standard deviation \sqrt{0.625} \approx 0.79 hour. This connects two distributions: Poisson counts and exponential (Erlang) waiting times are two views of the same random arrival stream. Takeaway. With Poisson rate 4/hour, the wait for the 10th arrival has m…

Time until the tenth customer arrives

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