Statistics Quest

Poisson distribution

The Poisson distribution models the number of events in a fixed interval when events occur independently at a constant average rate.

P(X=k)=e−λλkk!,k=0,1,2,…P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}, \quad k = 0, 1, 2, \dots
pmf
λ\lambda
expected number of events in the interval (rate × length of interval)
kk
observed number of events
E⁡(X)=λ,Var⁡(X)=λ\E(X) = \lambda, \qquad \Var(X) = \lambda
Mean equals variance
XX
a count, X∼Poisson(λ)X \sim \text{Poisson}(\lambda)

When to use

Counts of events per unit of time, space or exposure: calls per hour, defects per metre, claims per month. It also approximates the binomial when nn is large and pp small (λ=np\lambda = np).

Assumptions

Events occur independently, one at a time, at a constant rate; the count in one interval does not affect another. Rates scale with exposure: 3 per hour means λ=6\lambda = 6 for two hours.

How to interpret

  • Overdispersion check: for Poisson data the variance should be close to the mean. A sample variance much larger than the mean suggests clustering or a varying rate, and Poisson-based intervals and control limits will be too narrow.
  • The distribution is right-skewed for small λ\lambda and close to normal for large λ\lambda.

Common mistakes

  • Forgetting to rescale λ\lambda to the interval in the question.
  • Using Poisson for counts with a fixed maximum and a high success rate (use the binomial).
  • "More than 4" is sf(4)=P(X>4)\text{sf}(4) = P(X > 4); "at least 4" is sf(3)\text{sf}(3).

Example

Calls arrive at λ=3\lambda = 3 per 10 minutes. P(X=2)=0.2240P(X = 2) = 0.2240; P(X≤2)=0.4232P(X \le 2) = 0.4232; P(X>4)=sf(4)=0.1847P(X > 4) = \text{sf}(4) = 0.1847.