Poisson Distribution Calculator
The Poisson distribution models the number of events in a fixed interval when events happen at a constant average rate λ, independently. Enter λ and a target k to get the point probability, the cumulative probabilities and the (equal) mean and variance.
One parameter, and everything follows
Some things are counted rather than measured: calls arriving at a switchboard in an hour, flaws along a length of wire, decays from a sample in a second. What these have in common is an average rate and no natural upper bound — there is no maximum number of calls the way a coin toss has a maximum of one head.
The Poisson distribution is the model for exactly that situation, and it needs only the average rate. This calculator takes that rate and a target count, and reports the probability of hitting it exactly along with the cumulative probabilities either side.
Most distributions need two numbers, a centre and a spread. Poisson needs one: the rate fixes both, because the variance equals the mean. That is not a convenience but a structural property, and it is the sharpest test of whether the model fits — data whose spread is much wider than its average is telling you something Poisson cannot capture.
The tool reports the mean, the variance and the standard deviation alongside the probabilities for that reason. Comparing them against a sample's own figures is the quickest check on whether the distribution is the right one before its probabilities are trusted.
How to use this calculator
- Enter the rate The average number of events per interval. It must be non-negative, need not be a whole number, and carries the units of whatever interval you have chosen.
- Enter the target count A whole number of zero or more. Counts are discrete, so a fractional target is rejected rather than rounded.
- Pick the right probability line Four are reported: exactly the target, at most, strictly below, and at least. The distinctions matter here in a way they never do for a continuous distribution.
- Check the mean against the variance Both equal the rate, by construction. Comparing them against your data's actual mean and variance is the fit test worth running first.
The formula, and where it comes from
P(X = k) = e^(−λ)·λᵏ / k! P(X ≤ k) = Σᵢ₌₀ᵏ P(X = i) μ = λ, σ² = λ, σ = √λ
The point probability multiplies three pieces: the exponential factor, which shrinks as the rate rises and keeps the total probability at one; the rate raised to the target count; and the factorial in the denominator, which eventually overwhelms everything else and drives the tail to zero.
The cumulative probability is a genuine summation rather than a closed form. Every point probability from zero up to the target is computed and added, which is exact but means the work grows with the target rather than being a single evaluation.
The upper tail is obtained as one minus the probability of falling strictly below the target, not one minus the probability of being at or below it. That distinction is what makes at least the target include the target itself, and it is the single most common source of an off-by-one error in Poisson work.
The mean and variance both equalling the rate has a practical consequence worth internalising: the standard deviation is the square root of the rate, so relative variability shrinks as the rate grows. A rate of 4 has a standard deviation of 2 — half the mean — while a rate of 100 has one of 10, a tenth of it.
What each input means
- λ Rate — form field “Rate λ”
- The average number of events per interval. Non-negative and not necessarily whole — an average of 2.5 calls an hour is ordinary even though 2.5 calls never occur.
- k Target count — form field “Target k”
- The number of events asked about. A non-negative whole number, since the quantity being modelled is a count.
- μ, σ² Mean and variance
- Both equal to the rate. Their equality is a defining property, and a useful diagnostic when checking the model against data.
Worked examples
Every number below is produced by the same calculation engine the tool above runs. Nothing here is typed by hand, so the walkthrough cannot drift from what you get when you enter the same values yourself.
A modest rate below the mean
A rate of 3 with a target of 2. The target sits just below the average, which is where the distribution is at its densest.
Inputs Rate λ = 3, Target k = 2
- Distribution X ~ Poisson(λ = 3)
- P(X = k) e^(−λ)·λ^k/k! = e^(−3)·3^2/2! = 0.224042
- P(X ≤ k) 0.42319
- P(X < k) 0.199148
- P(X ≥ k) 1 − P(X < k) = 0.800852
- Mean μ = λ = 3
- Variance σ² = λ = 3
- Standard deviation σ = √λ = 1.73205
Result P(X = 2) = 0.224042, P(X ≤ 2) = 0.42319, P(X ≥ 2) = 0.800852
The point probability is around a fifth, which is close to the maximum for this rate — the most likely single count is 2 or 3, and they are nearly tied. A Poisson distribution peaks at the whole number just below its rate.
The probability of at most 2 is well under a half, so more than 2 events is the likelier outcome despite 2 being among the most probable individual counts. Point probabilities and cumulative ones answer different questions.
A higher rate with a target below it
A rate of 10 with a target of 7 — the shape of a busy help line where the question is whether a quiet hour is unusual.
Inputs Rate λ = 10, Target k = 7
- Distribution X ~ Poisson(λ = 10)
- P(X = k) e^(−λ)·λ^k/k! = e^(−10)·10^7/7! = 0.0900792
- P(X ≤ k) 0.220221
- P(X < k) 0.130141
- P(X ≥ k) 1 − P(X < k) = 0.869859
- Mean μ = λ = 10
- Variance σ² = λ = 10
- Standard deviation σ = √λ = 3.16228
Result P(X = 7) = 0.0900792, P(X ≤ 7) = 0.220221, P(X ≥ 7) = 0.869859
The standard deviation is a little over 3, so a count of 7 is around one standard deviation below the rate. That is unremarkable, and the cumulative probability confirms it: well over a fifth of hours come in at or below 7.
As the rate grows the distribution becomes increasingly symmetric and bell-shaped. At a rate of 10 the normal approximation is already reasonable, which is not true at a rate of 2.
Reading the result
The four probability lines are genuinely different
For a discrete distribution, at most k and strictly below k differ by the whole probability of exactly k, which can be substantial. Matching the wording of the question to the right line is not pedantry here.
Relative variability falls as the rate rises
Because the standard deviation is the square root of the rate, a rate of 100 varies by about ten per cent while a rate of 4 varies by fifty. Rare-event counts are proportionally far noisier than common ones.
When you would use this
Staffing and capacity questions
Given an average arrival rate, the cumulative probability says how often demand will exceed a given capacity — the basis for deciding how much headroom a service needs.
Quality and defect counting
Flaws per unit length or per batch follow this model when they occur independently at a steady rate. An observed count far into the tail is evidence the rate itself has changed.
Assumptions and limitations
What this calculator assumes
- Events occur independently of one another, at a constant average rate over the interval.
- Two events cannot occur at exactly the same instant.
- The rate is non-negative; the target is a non-negative whole number.
- The cumulative probability is computed by summing point probabilities from zero to the target.
Where it stops being the right tool
- One target at a time: the probability of a range requires two runs and a subtraction.
- The inverse direction is not offered, so finding the count corresponding to a given probability needs repeated trials.
- The model is assumed rather than tested; overdispersed data needs a different distribution.
Common mistakes
Confusing at least with more than
Why it happens. In ordinary speech the two are almost interchangeable, and for a continuous distribution they genuinely are. For a count they differ by the entire probability of hitting the target exactly.
How to avoid it. At least the target is one minus the probability of being strictly below it, which is the line reported. More than the target is one minus the at-most line instead.
Mismatching the rate and the interval
Why it happens. The rate is quoted per hour and the question asks about a day, or vice versa. Nothing in the calculation knows what interval was intended.
How to avoid it. Scale the rate to the interval in the question before entering it. Twelve per hour over a quarter of an hour is a rate of three, not twelve.
Treating a count near the rate as confirmation of the model
Why it happens. Observations cluster around the rate under almost any count distribution, so agreement of the mean says little on its own.
How to avoid it. Compare the variance too. Poisson requires it to equal the mean, and data whose spread is much wider is not Poisson however well the average matches.
Frequently asked questions
When does Poisson apply?
When events occur independently at a constant average rate and what is modelled is the count over a fixed window — calls per hour, defects per metre, decays per second. It needs no upper bound on the count, which is what distinguishes it from a binomial setting.
Why are the mean and variance both λ?
It is a defining property of the distribution rather than a coincidence. The practical consequence is that the standard deviation is √λ, so variability grows only as the square root of the rate and shrinks in relative terms as the rate rises.
How does Poisson relate to the binomial distribution?
It is the limiting case as the number of trials grows without bound and the success probability shrinks, with their product held fixed at λ. That is why it models rare events over many opportunities so well.
Can the rate be a fraction?
Yes. The rate is a long-run average, so 2.5 events per interval is perfectly meaningful even though no single interval contains half an event. Only the target count must be a whole number.