Poisson Distribution Calculator

StatisticsLast updated: August 22, 2026

The Poisson distribution models the number of rare events that occur independently of one another within a fixed interval of time, area, or volume. The number of calls arriving in an hour, the number of breakdowns in a day, the number of defects per square meter, and the number of patients of a given type presenting at an emergency room in a week are all classic Poisson examples. The distribution has a single parameter: λ, the mean number of events expected in one unit interval.

For the λ you enter (and, if you like, an observation coefficient t), this calculator returns the exact probability P(X = k), the cumulative probabilities P(X ≤ k), P(X < k), P(X ≥ k), and P(X > k), the probability of the interval [a, b] you choose, and the mean, variance, standard deviation, and mode of the distribution. The defining feature of the Poisson is that its mean equals its variance; the calculator shows this identity explicitly and points out when λ has grown large enough for the normal approximation to apply.

If 4 calls arrive per hour, enter 4. Either a point or a comma may be used as the decimal separator.
How many unit intervals you are observing. Enter 2 for a two-hour observation; the effective parameter becomes λ·t.
The event count whose probability you want. It must be a non-negative integer.
The lower bound for P(a ≤ X ≤ b). If left empty it is taken as 0.
The upper bound for P(a ≤ X ≤ b). If left empty the upper bound is taken as infinite.

Poisson Distribution Formulas

Mass function:     P(X = k) = e⁻λ · λᵏ / k!,  k = 0, 1, 2, …
Scaling:           λ_eff = λ · t   (t = number of unit intervals)
Cumulative:        P(X ≤ k) = Σᵢ₌₀ᵏ e⁻λ · λⁱ / i!
Upper tail:        P(X ≥ k) = 1 − P(X ≤ k − 1)
Interval:          P(a ≤ X ≤ b) = P(X ≤ b) − P(X ≤ a − 1)
Mean:              μ = λ
Variance:          σ² = λ        (mean = variance)
Standard dev.:     σ = √λ
Mode:              ⌊λ⌋  (λ − 1 and λ when λ is an integer)
Binomial approx.:  for n ≥ 20 and p ≤ 0.05, Binomial(n, p) ≈ Poisson(n·p)

The cumulative probability is evaluated through the regularized incomplete gamma relation P(X ≤ k) = 1 − P(k + 1, λ); because no summation loop is used, precision is preserved even for large λ. Once λ ≥ 10 the distribution can be approximated by N(λ, λ).

How to Calculate

  1. Enter the mean number of events expected in one unit interval (λ): if 4 calls arrive per hour, type 4.
  2. If your observation window differs from the unit interval, enter the coefficient (t); type 2 for a two-hour observation and the calculator works with λ·t.
  3. Enter the number of events (k) whose probability you are interested in.
  4. If you wish, set the interval bounds (a and b); leaving the upper bound empty computes P(X ≥ a).
  5. Read the exact, cumulative, and tail probabilities, then check the mean = variance = λ identity and the mode.
  6. Look at the mass plot to see how skewed the distribution is; as λ grows the distribution becomes more symmetric and approaches the normal.

Worked Examples

Four calls an hour: exactly 5 calls

With λ = 4, the probability of exactly 5 calls is P(X = 5) = 0.1563. The probability of at most 5 calls is 0.7851 and the probability of at least 5 calls is 0.3712. P(2 ≤ X ≤ 6) = 0.7977; the mean and the variance are both 4, the standard deviation is 2, and the mode is 3 and 4.

Exactly 5 events — P(X = 5): 0.1563 · At most 5 events — P(X ≤ 5): 0.7851 · Fewer than 5 — P(X < 5): 0.6288

A two-hour window: at least 8 events

In a process averaging 3 events per hour, a two-hour observation has an effective parameter of λ·t = 6. The probability of exactly 8 events is P(X = 8) = 0.1033 and the probability of at least 8 events is P(X ≥ 8) = 0.2560. The probability of at most 8 events is 0.8472; the mean and the variance are both 6 and the standard deviation is 2.4495.

Exactly 8 events — P(X = 8): 0.1033 · At most 8 events — P(X ≤ 8): 0.8472 · Fewer than 8 — P(X < 8): 0.7440

Rare events: no events at all when λ = 0.5

For λ = 0.5 the probability of seeing no events at all is P(X = 0) = e⁻⁰·⁵ = 0.6065. The probability of at most 1 event is 0.9098 and the probability of at least 1 event is 0.3935. Because λ < 10, the distribution is clearly right-skewed and the normal approximation cannot be used.

Exactly 0 events — P(X = 0): 0.6065 · At most 0 events — P(X ≤ 0): 0.6065 · Fewer than 0 — P(X < 0): 0

Frequently Asked Questions

When is the Poisson distribution used?
It is used for the number of events that occur independently within a given unit of time, area, or volume and that are individually rare. Calls arriving at a call center, machine breakdowns, defects in a bolt of fabric, and accidents at an intersection are typical examples. The events must be independent and the event rate must stay constant throughout the observation.
Why are the mean and the variance equal in a Poisson distribution?
The Poisson distribution has a single parameter: λ fixes both the expected number of events and the variability. That is why μ = σ² = λ and the standard deviation is √λ. If real data shows a variance clearly larger than the mean (overdispersion), the Poisson model is inadequate and the negative binomial distribution should be preferred.
How do you adjust λ for a period other than the unit interval?
In a Poisson process the parameter is directly proportional to the length of the observation: for t units you use λ·t instead of λ. In a process with 3 events per hour, a two-hour observation has a parameter of 6 and a half-hour observation has a parameter of 1.5. The observation coefficient field does this conversion for you.
How do you compute Poisson probabilities in Excel, SPSS, and R?
In Excel, use =POISSON.DIST(k, λ, FALSE) for the exact probability and =POISSON.DIST(k, λ, TRUE) for the cumulative one. In R, dpois(k, lambda) and ppois(k, lambda) give the same results, as does scipy.stats.poisson.pmf/cdf in Python. SPSS provides the functions PDF.POISSON(k, λ) and CDF.POISSON(k, λ).
When can the binomial distribution be replaced by the Poisson?
When the number of trials is large and the success probability is small (the common criteria are n ≥ 20 with p ≤ 0.05, or n ≥ 100 with n·p ≤ 10), the binomial distribution is approximated very well by a Poisson with λ = n·p. This makes hand calculation far easier: instead of a combination you need only a single exponential term. The error of the approximation shrinks as n grows.
Can the normal approximation be used when λ is large?
Yes. Once λ ≥ 10 the Poisson distribution is reasonably represented by the normal distribution N(λ, λ); some sources set the boundary at 20 instead. Because you are replacing a discrete distribution with a continuous one, the continuity correction (±0.5) should be applied. This calculator returns the exact value without needing an approximation, so the approximation only matters for hand calculation.
Why do the Poisson probabilities not add up to 1?
The support of the Poisson distribution is infinite: k = 0, 1, 2, … with no upper limit. The table and the chart show only a window that covers almost all of the probability; the tail probability outside that window is stated separately in the notes. Mathematically, summing over all values of k gives exactly 1.