Enter the average number of events per interval. Add a count to get the probability of exactly, at most and at least that many events, or leave it blank for the mean, variance and standard deviation alone.
Each bar is P(X = k). The darker bars lie within one standard deviation (σ = 1.789) of the mean.
Show the working, step by step
A Poisson distribution has a single parameter, the rate λ — the average number of events per interval.
λ = 3.2
Its mean and its variance are both equal to λ. That equality is the distribution's signature.
μ = λ = 3.2 σ² = λ = 3.2
So the standard deviation is the square root of the rate.
σ = √3.2 = 1.78885
The probability of exactly 2 events comes from the mass function.
P(X = k) = λᵏ e^(−λ) ÷ k! P(X = 2) = 3.2^2 × e^(−3.2) ÷ 2! = 0.208702
Cumulative probabilities add the mass function from 0 up to k.
P(X ≤ 2) = Σ P(X = i), i = 0 … 2 = 0.379904
The Poisson model assumes events occur independently at a constant average rate. If the rate drifts or events cluster, the real spread is larger than √λ.
What the Poisson distribution models
The Poisson distribution describes counts: how many times something happens in a fixed stretch of time or space when the events occur independently at a constant average rate. Calls reaching a switchboard in an hour, defects along a metre of cable, goals in a football match, typos on a page, particles registered by a Geiger counter in a minute — each is a count with no fixed ceiling, driven by a rate rather than by a number of trials.
That single rate, λ, is the only parameter. Give the calculator λ and it returns the whole distribution; give it a count k as well and it returns the probability of that count and of the counts either side of it.
The formula
P(X = k) = λᵏ · e^(−λ) ÷ k! mean: μ = λ variance: σ² = λ std dev: σ = √λ
Here e is 2.71828 and k! is k factorial. The e^(−λ) term is the probability of no events at all, and each successive k multiplies it by λ and divides by k, which is why the mass function rises to a peak near λ and then tails off.
A worked example
A help desk receives 3.2 tickets an hour on average. What is the chance of exactly 2 in the next hour? With λ = 3.2 and k = 2:
- λᵏ = 3.2² = 10.24
- e^(−λ) = e^(−3.2) = 0.04076
- k! = 2! = 2
- P(X = 2) = 10.24 × 0.04076 ÷ 2 = 0.2087
So roughly a one-in-five chance. Adding P(X = 0) = 0.0408 and P(X = 1) = 0.1304 gives P(X ≤ 2) = 0.3799: the desk sees two tickets or fewer in about 38% of hours, and three or more in the remaining 62%. The standard deviation is √3.2 = 1.789, so a quiet hour with none and a busy hour with six are both within two standard deviations of the mean and neither should surprise anyone.
Mean, variance and standard deviation
The mean and the variance of a Poisson distribution are the same number. That is not a coincidence of the example; it is the distribution's defining property, and it means the standard deviation is always the square root of the rate. Nothing else needs to be known.
It has a consequence worth noticing. The absolute spread √λ grows with the rate, but the spread relative to the mean shrinks: √λ ÷ λ = 1 ÷ √λ. At λ = 1 the standard deviation equals the mean, a 100% relative spread. At λ = 4 it is 50%, at λ = 25 it is 20%, and at λ = 100 it is 10%. That is why a count of 100 events is far more informative than a count of 1 — the estimate of the rate gets proportionally tighter as more events are observed. The same ratio, expressed as a percentage, is the relative standard deviation.
The distribution at λ = 3.2
| k | P(X = k) | P(X ≤ k) |
|---|---|---|
| 0 | 0.0408 | 0.0408 |
| 1 | 0.1304 | 0.1712 |
| 2 | 0.2087 | 0.3799 |
| 3 | 0.2226 | 0.6025 |
| 4 | 0.1781 | 0.7806 |
| 5 | 0.1140 | 0.8946 |
| 6 | 0.0608 | 0.9554 |
The mode is 3, the largest whole number below λ, and the distribution is skewed to the right: it stops dead at zero on the left but carries on indefinitely on the right, with about 4.5% of the probability still sitting at seven events or more. The chart above draws the same mass function, and you can watch the skew fade as you raise λ.
Poisson versus binomial
The binomial distribution counts successes in n trials with probability p each. When n is large and p is small, the binomial becomes awkward to compute and the Poisson with λ = np is an excellent stand-in. With n = 1,000 and p = 0.0032, the exact binomial gives P(X = 2) = 0.2088; the Poisson with λ = 3.2 gives 0.2087. The usual rule of thumb is n ≥ 20 and p ≤ 0.05, or more comfortably n ≥ 100 and np ≤ 10.
The conceptual difference is the ceiling. A binomial count can never exceed n, and its standard deviation √(npq) is always a little smaller than √(np) because the q = 1 − p factor trims it. A Poisson count has no ceiling, and as p shrinks q approaches 1 and the two standard deviations converge. When the events are rare, the distinction stops mattering.
When the normal approximates the Poisson
As λ grows the skew flattens out, and above about λ = 20 the Poisson is close enough to a normal distribution with mean λ and standard deviation √λ to use one in its place. Add a continuity correction of 0.5, because you are replacing a discrete distribution with a continuous one: for P(X ≤ 20) at λ = 25, use z = (20.5 − 25) ÷ 5 = −0.90, which gives 0.1841 against the exact 0.1855. Below λ = 10 the approximation is poor and the exact sum is the right tool.
Overdispersion
The first thing to check with real count data is whether the variance really does equal the mean. Very often it is larger — daily admissions to a hospital, tweets about a product, insurance claims by region — because the rate is not constant and the events are not independent: a bad day is bad for everyone, and one claim tends to bring others. That excess is called overdispersion, and when it is present the Poisson understates the spread and every probability derived from it is too confident. The standard remedy is the negative binomial distribution, which keeps the Poisson's shape but adds a second parameter for the extra variance.
Check the units of λ
λ must be the average for exactly the interval k refers to. A rate of 6.4 calls per hour is 3.2 per half hour and 0.107 per minute, and the probability of 2 events is different in each case — 0.2087 in a half hour but only 0.0051 in a minute. Convert the rate first, then compute. The same applies to space: defects per metre must be scaled to the length you are inspecting before the mass function is applied.
Related calculators
-
Binomial distribution
Counts from a fixed number of trials — the Poisson approximates it when p is small.
-
Normal distribution
The continuous approximation once λ is above about 20.
-
Standard deviation
The general case, computed from raw data rather than a rate.
-
Empirical rule
How much of a distribution sits within 1, 2 and 3 standard deviations.
Common questions
What is the standard deviation of a Poisson distribution?
σ = √λ, the square root of the average rate. A Poisson distribution has one
parameter, and its mean and variance are both equal to it, so the standard deviation
follows from λ alone. At λ = 3.2 the standard deviation is √3.2 = 1.789.
What is the Poisson distribution formula?
P(X = k) = λk e−λ / k!, where λ is the average
number of events per interval, k is the count you are asking about and e is 2.71828.
Cumulative probabilities such as P(X ≤ k) add the terms from k = 0 upwards; there is no
closed form, so the calculator sums them directly.
What is the difference between Poisson and binomial?
The binomial counts successes in a fixed number of trials, each with probability p, so the count can never exceed n. The Poisson counts events in a continuous interval of time or space with no upper limit, from a rate λ. When n is large and p is small the two agree closely with λ = np, which is why the Poisson is used as the binomial's shortcut for rare events.
Can the mean and variance of a Poisson be different?
Not for a true Poisson — the equality is built into the distribution. If your data show a variance noticeably larger than the mean, the events are not arriving independently at a constant rate, and the Poisson will understate the spread. That situation is called overdispersion and is usually handled with a negative binomial model instead.
When can I use the normal distribution instead of the Poisson?
Once λ is above about 20 the Poisson is close to symmetric and a normal with mean λ and standard deviation √λ approximates it well, provided you apply a continuity correction of ±0.5 to the count. At λ = 25, the exact P(X ≤ 20) is 0.1855 and the corrected normal approximation gives 0.1841. Below λ = 10 the skew is too pronounced and you should use the exact probabilities.
What does λ have to be measured in?
The same interval as the count you are asking about. If a switchboard averages 6.4 calls an hour and you want the probability of exactly 2 calls in half an hour, λ must be halved to 3.2 first. Using the hourly rate with a half-hour count is the most common way to get a Poisson probability wrong.