Four signups in a quarter hour

Question

New signups arrive on a freshly launched website as a Poisson process at rate \lambda = 20 per hour. What is the probability that exactly 4 users sign up in the next 15 minutes?

Accepted Answer

Fifteen minutes is a quarter of an hour, so the mean number of signups in that window is \mu = \lambda t = 20 	imes 	frac14 = 5. We expect about five signups in fifteen minutes. The count N is Poisson with mean 5, and the chance of exactly k = 4 is P(N = 4) = e^{-5}\,\frac{5^4}{4!} = e^{-5}\,\frac{625}{24} \approx 26.04 	imes 0.006738 \approx 0.175. So about a 17.5\% chance of exactly four signups. Notice that 4 is close to the mean of 5, which is why this probability is relatively large: counts near the mean are the most likely outcomes. Takeaway. Values of k near the mean \mu carry the m…

Four signups in a quarter hour

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