Poisson distribution

The Poisson distribution counts occurrences in a fixed window of time or space, on the assumption that events arrive independently at a constant average rate. One parameter sets the whole thing: $\lambda$ (lambda), the mean number of occurrences per window. The probabilities are $P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$ for $k = 0, 1, 2$ and upward with no ceiling, and both the mean and the variance equal $\lambda$.

Take a help desk averaging 3 calls an hour, so $\lambda = 3$. Then $e^{-3} \approx 0.0498$, and $P(X = 2) = \frac{3^2 e^{-3}}{2!} = \frac{9(0.0498)}{2} \approx 0.2240$. The count 3 is exactly as likely, since $P(X = 3) = \frac{27(0.0498)}{6} \approx 0.2240$ too. A quiet hour of at most one call has probability $0.0498 + 3(0.0498) \approx 0.1991$.

"The mean and the variance are both 3, so the standard deviation is 3" is the slip that equality invites. The variance is 3; the standard deviation is $\sqrt{3} \approx 1.732$. For a positive rate the two agree at $\lambda = 1$ and nowhere else, so the coincidence you are remembering is between the mean and the variance, never between the mean and the spread.

The window is part of the parameter rather than part of the background. Three calls an hour means $\lambda = 1.5$ over half an hour and $\lambda = 24$ over an eight-hour day, so carrying 3 into a half-hour question doubles where the answer is centered. The constant-rate assumption is the first thing to break in practice: calls that bunch up after an outage vary more than a Poisson allows, and the tell is a sample variance sitting well above the sample mean.

No topic in the Fall 2026 AP Statistics course covers the Poisson distribution. The count model the course does use is the binomial, topic 2.10 of Unit 2, which fixes the number of trials at $n$ instead of letting the count run without a bound.