Parameters

Worked example: calls to a help desk

A help desk receives 24 calls per hour on average. What is the chance of exactly 6 calls in a 10-minute window, and how unusual would 8 or more be?

  1. Scale the rate to the interval. The rate must match the window you ask about: 24 calls per hour is 4 calls per 10 minutes, so λ = 4.
  2. Exactly 6 calls: P(X = 6) = e−4 × 46 / 6! = 0.0183 × 4096 / 720 = 0.104, about a 10% chance.
  3. 8 or more calls: P(X ≥ 8) = 1 − P(X ≤ 7) = 0.051. A burst like that happens in about one 10-minute window in 20, so it is unusual but not alarming.
  4. A quiet spell: P(X = 0) = e−4 = 0.018, so a completely silent 10 minutes happens about 2% of the time.

Enter λ = 4 and k = 6 or k = 8 above to check these values. A quick sanity check on any Poisson model: the mean and variance should be roughly equal in your data. If the counts vary much more than their mean, events are clustered and a negative binomial model fits better.

Guide: The Poisson Distribution

The Poisson distribution models the number of events that occur in a fixed interval of time or space, when those events happen independently at a constant average rate λ. Classic examples: calls arriving at a switchboard, decay events from a radioactive source, or typos per page.

Use the binomial when you have a fixed number of trials n each with probability p. Use the Poisson when events occur continuously and you only know the average rate — there's no fixed n. The Poisson is also the limit of the binomial when n is large and p is small (with λ = np).

P(X = k) is the probability of exactly k events. P(X ≤ k) is the cumulative probability of k or fewer. P(X ≥ k) = 1 − P(X ≤ k−1) is the probability of at least k events — useful for questions like "what's the chance of 10 or more calls this hour?"

  • Events occur independently
  • The average rate λ is constant over the interval
  • Two events cannot occur at exactly the same instant

A key property: the mean and variance are both equal to λ. If your data's variance is much larger than its mean (overdispersion), a negative binomial model may fit better.

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