Waiting for three buses to pass

Question

Buses pass a stop with independent exponential gaps averaging 12 minutes apart. Let T be the time until the 3rd bus passes. What are the mean and variance of T?

Accepted Answer

The time until the 3rd bus is the sum of 3 independent exponential gaps, an Erlang. Each gap has mean 1/\lambda = 12 minutes and variance 1/\lambda^2 = 12^2 = 144. Summing 3 of them: \mathbb{E}[T] = 3 	imes 12 = 36 	ext{ min}, \qquad \operatorname{Var}(T) = 3 	imes 144 = 432. Standard deviation \sqrt{432} \approx 20.8 minutes. Even though you expect 36 minutes, the spread is wide because a single long gap can dominate the total. Takeaway. Summing 3 independent exponentials with mean 12 gives mean n/\lambda = 36 and variance n/\lambda^2 = 432.

Waiting for three buses to pass

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