Poisson Distribution Calculator

Background.

Some things happen at a steady average rate but at moments nobody can predict: calls reaching a help desk, typing errors on a page, clicks of a Geiger counter, goals in a football match. When the events are independent of one another and the average rate holds steady, the number that turns up in a fixed stretch of time or space follows the Poisson distribution, named after the French mathematician Siméon Denis Poisson. One number describes it completely, λ, the expected count for the interval, and from λ alone you can say how likely any particular count is.

That makes it the natural way to measure surprise. A help desk averaging 4 calls an hour gets exactly 3 in a given hour about 19.5% of the time, and 3 or fewer about 43.3% of the time. An hour with ten or more calls happens only about 0.81% of the time at that rate, so if it keeps happening, the rate has probably changed. The same reasoning sets staffing levels, spare-part stocks and quality-control limits, and it is the starting point for many football-score models: a side averaging 1.5 goals a game will, if goals really arrive independently, fail to score in about 22.3% of its matches and score three or more in about 19.1%.

The distribution has a signature: its variance equals its mean. Real counts often break that rule. Calls bunch after an outage, defects cluster in a bad batch, and many records hold more zeros than the model allows. When observed counts spread much wider than their average, Poisson figures understate the chance of extreme values, and a model built for overdispersion, such as the negative binomial, fits better.

Here you supply λ and a count k and get back the probability of exactly k events and of k or fewer. Both inputs stop at 10,000. Long before that, the distribution is so close to a normal curve with mean λ and standard deviation √λ that the normal approximation is the usual shortcut.

What is poisson distribution calculator?

The Poisson distribution gives the probability of observing exactly k events in a fixed interval when events occur independently at a constant average rate, λ being the expected number in that interval. Its probabilities are P(X = k) = e^−λ λ^k ÷ k! for k = 0, 1, 2 and upwards, with no upper limit on k. The mean and the variance are both λ, so the standard deviation is √λ. The cumulative probability P(X ≤ k) adds up the exact probabilities from zero to k. One consequence regularly surprises people. An event that averages once per interval, λ = 1, fails to happen at all in e^−1 ≈ 36.8% of intervals, so a flood or breakdown that comes "once a year on average" arrives at least once in only about 63.2% of years, and some years bring two or three. The model rests on three assumptions: events happen one at a time, they do not influence each other, and the rate is the same throughout the interval. It can be derived as the limit of a binomial count when the interval is cut into ever more, ever smaller pieces, each with a tiny chance p of holding an event, while the product np stays fixed at λ.

How to use this calculator.

1. 01Enter the Mean count λ for the same interval your question is about. If a junction averages 6 accidents a year and you are asking about one quarter, λ is 1.5, not 6.
2. 02When the intervals you care about differ in length, set λ for each one as rate × exposure. A machine that faults 0.5 times per running hour has λ = 5 for a 10-hour shift and λ = 3 for a 6-hour one; using one average across both would misstate each.
3. 03Enter the Event count k as a whole number from 0 to 10,000. Fractional counts are refused, since you cannot observe two and a half events.
4. 04Read P(X = k) for exactly that many events and P(X ≤ k) for that many or fewer. For at least k, enter k − 1 and subtract the cumulative result from 1.
5. 05If λ comes from your own records, compare the variance of the counts with their mean before trusting the tail probabilities; a variance well above the mean means the Poisson model is too narrow.

The formula.

P(X = k) = e^−λ × λ^k ÷ k!, and P(X ≤ k) is the sum of those terms for every count from 0 up to k. Working out λ^k and k! separately would soon produce enormous numbers (171! already exceeds what ordinary floating-point arithmetic can hold), so the calculator builds each term from the one before. It starts at P(0) = e^−λ, the chance of no events at all, then applies P(i) = P(i − 1) × λ ÷ i, adding every term to a running total. The last term is the exact probability and the total is the cumulative one. Each step is one multiplication and one division, so the work grows in step with k, which is why λ and k are both capped at 10,000. The ratio λ ÷ i also explains the shape: terms grow while i is below λ and shrink once it passes, so the most likely count sits at or just below λ. Results are given to 12 significant digits, and the cumulative value is never reported above 1. For large means the familiar shortcut is the normal curve with a continuity correction, P(X ≤ k) ≈ Φ((k + 0.5 − λ) ÷ √λ). At λ = 100 and k = 110 it gives 0.8531 against the exact 0.8529, close enough for most purposes. At λ = 4 and k = 3 it gives 0.4013 against the exact 0.4335, because a distribution with so small a mean is lopsided, bunched against zero with a long right tail. Exact terms matter most in that small-mean region, which is also where most real questions about rare events live. Events that arrive in bursts break the one-at-a-time assumption: a single pile-up can generate several accident reports at once, so count the incidents rather than the reports, or the variance will run well above the mean.

A worked example.

Frequently asked questions.

References& sources.

1. [1]NIST/SEMATECH e-Handbook, Poisson Distribution.
2. [2]OpenStax, Introductory Statistics 2e, section 4.6, Poisson Distribution.
3. [3]BIPM, The International System of Units (SI Brochure), 9th ed., version 3.01, coherent derived units and quantity equations.
4. [4]NIST Special Publication 811, Guide for the Use of the International System of Units, 2008 edition.
5. [5]Penn State STAT 414, Poisson Distribution.