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MasterMath

Poisson Distribution Calculator

The probability that exactly k events happen in an interval when you know the average, plus the cumulative figures. It fits calls per hour, customers per minute or breakdowns per month.

Exactly k events

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Exactly k events—
At most k—
At least k—
Mean—
Standard deviation—
None in the interval—
kExactly kUp to k

How this was worked out

    The formula

    P(X = k) = e^(−λ) · λᵏ / k!

    What it means

    The Poisson counts rare events across a continuous interval of time or space: there is no fixed number of trials, only an average rate. Its defining feature is that the mean and the variance are the same number, and that is also its check: if your data show a variance far above the mean, the events are not independent and the Poisson does not describe what is happening.

    How to work it out by hand

    1. Note the average number of events per interval, which is lambda
    2. Raise lambda to the power k and divide by the factorial of k
    3. Multiply by e to the power minus lambda
    4. For the cumulative figures, add the cases from zero up to k

    What is worth knowing

    Lambda has to match the interval you are asking about: if the rate is twelve calls an hour and you ask about fifteen minutes, lambda is three, not twelve. That is the commonest mistake with this distribution. And independence matters: calls to a support line are not independent when there is a general outage, and on those days the Poisson badly underestimates the chance of a spike.

    Frequently asked questions

    How does it differ from the binomial?

    The binomial counts successes in a fixed number of trials; the Poisson counts events in a continuous interval, with no trial count at all.

    What is lambda?

    The average number of events per interval. It has to refer to the same interval you are asking about.

    Why does the variance equal the mean?

    It is a property of the distribution. If your data do not show it, the events are probably not independent.

    Can lambda be a fraction?

    Yes, and it usually is: 2.7 breakdowns a month is a perfectly reasonable average.

    Does it work for queues and waiting times?

    For the number of arrivals, yes. For the time between one and the next, the right distribution is the exponential.