How to use
- Enter the average count λ for the interval you care about.
- Under Find, choose exactly, at most, at least k and so on.
- Enter k and read the result, chart and table.
Options explained
- Formula
- P(X = k) = λᵏ · e⁻λ / k!. The mean and the variance are both λ.
- When it applies
- Events happen independently, at a constant average rate, and two can't happen at exactly the same moment.
- Match λ to the interval
- With 3.5 per hour on average, a half-hour question uses λ = 1.75.
Worked example
With λ = 3.5 (say 3.5 customers an hour), the default asks for at most 3:
| Find | Result |
|---|---|
| P(X ≤ 3) | 0.5366 |
| P(X = 2) | 0.1850 |
| P(X ≤ 5) | 0.8576 |
| P(X > 5) | 0.1424 |
| Mean = variance | 3.5 |
Two more:
| Question | λ | Find | Probability |
|---|---|---|---|
| 2 breakdowns a day on average, none today | 2 | P(X = 0) | 0.1353 |
| 4 complaints a day on average, more than 9 today | 4 | P(X > 9) | 0.0081 |
Frequently asked questions
How does it relate to the binomial distribution?
A binomial with large n and small p is close to a Poisson with λ = np. 1000 items with a 0.2% defect rate give roughly Poisson λ = 2.
How does this map to Excel?
Exactly k is POISSON.DIST(k, λ, FALSE); at most k is POISSON.DIST(k, λ, TRUE).
My data's variance is much bigger than the mean.
Then the data are overdispersed and the Poisson model doesn't fit well; a negative binomial model is the usual alternative.
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