When does the fifth call arrive?

Question

Calls arrive at a help desk as a Poisson process at rate \lambda = 2 calls per minute. What is the expected time until the fifth call arrives, and what is the standard deviation of that time?

Accepted Answer

In a Poisson process, the gaps between consecutive arrivals are independent exponential random variables, each with rate \lambda = 2. So the time until the fifth call is the sum of five such gaps: T5 = G1 + G2 + G3 + G4 + G5, each Gi \sim 	ext{Exponential}(2) with mean \frac{1}{\lambda} = 0.5 minute and variance \frac{1}{\lambda^2} = 0.25. (This sum has an Erlang, or Gamma(5, \lambda), distribution.) Mean. By linearity, add the five gap means: E[T5] = \frac{5}{\lambda} = \frac{5}{2} = 2.5 	ext{ minutes}. Variance and SD. The gaps are independent, so variances add: \operatorname{Var}(T5) = \…

When does the fifth call arrive?

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