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Online Poisson Distribution Calculator

Get probabilities, cumulative values and percentiles for events per interval with mean rate λ.

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Explanation

The Poisson distribution models how many times an event occurs in a fixed interval of time, space or any other unit when the events are independent of each other and occur at a stable mean rate \( \lambda \). It is a discrete distribution with support on the non-negative integers \( \{0, 1, 2, \ldots\} \). Its most distinctive property is that the mean and the variance are equal: both equal \( \lambda \). If in your data the variance clearly exceeds the mean (overdispersion), the negative binomial may be a more suitable model.

Use it when the events can occur any number of times in the interval, there is no fixed upper limit on occurrences (unlike the binomial, which has \( n \) trials), the events are rare relative to the potential opportunities, and the rate is stable across the interval. Typical examples: phone calls at a call center, customers in a store, errors on a printed page, radioactive emissions, or server failures in production. The Poisson distribution is also the limit of the binomial when \( n \) is large, \( p \) is small and \( \lambda = np \) is moderate.

Formula

\( P(X=k)=\frac{\lambda^k e^{-\lambda}}{k!} \)

Parameters

  • λ (lambda): expected number of events in the interval.
  • k: observed event count.

Worked example

Situation: A telephone switchboard receives on average \(\lambda = 3\) calls per minute. The number of calls in any given minute follows a Poisson distribution with \(\lambda = 3\).

Question 1: What is the probability that no call arrives in a given minute, \(P(X = 0)\)?

Solution: We apply the Poisson mass function with \(k = 0\): \[ P(X = 0) = \frac{e^{-3} \cdot 3^0}{0!} = e^{-3} \approx 0.0498 \] There is approximately a 4.98% probability that a minute passes without any call.

Question 2: What is the probability of receiving at most 2 calls in a minute, \(P(X \leq 2)\)?

Solution: We add \(P(X = 0) + P(X = 1) + P(X = 2)\): \[ P(X=1) = e^{-3}\cdot 3 \approx 0.1494, \quad P(X=2) = e^{-3}\cdot\frac{9}{2} \approx 0.2240 \] \[ P(X \leq 2) = 0.0498 + 0.1494 + 0.2240 \approx 0.4232 \] There is approximately a 42.32% probability of receiving 2 or fewer calls in a minute.

Interpretation: With a mean of 3 calls per minute, there is more than a 57% probability of receiving 3 or more calls. For a Poisson with integer \(\lambda\), the values \(\lambda-1 = 2\) and \(\lambda = 3\) are both modes (bimodal): \(P(X=2) = P(X=3) \approx 0.224\). These calculations allow you to size the service capacity needed for different service levels.

How to interpret the result

The calculator offers three distinct outputs. The PMF, \( P(X = k) \), gives the probability that exactly \( k \) events occur in the analyzed interval. If \( P(X = 8) = 0.103 \) with \( \lambda = 6 \), it means that approximately 10% of equivalent intervals would have exactly 8 events. In the chart, each bar represents the probability of a specific count; the bars in the middle (near \( \lambda \)) are the tallest because they are the most typical values.

The CDF, \( P(X \leq k) \), accumulates the probability of observing \( k \) or fewer events. A value \( P(X \leq 10) = 0.957 \) indicates that 95.7% of intervals produce 10 events or fewer; the right tail \( P(X > 10) = 0.043 \) quantifies the rarity of receiving more than 10. The shaded green area in the chart visualizes that cumulative probability for the selected range; the horizontal axis shows the possible non-negative integer counts.

The percentile result returns the smallest integer \( k \) such that \( P(X \leq k) \geq p \). For example, if the 99th percentile is \( k = 13 \) for \( \lambda = 6 \), only 1% of intervals exceed 13 events; that value can be used as an alert threshold in monitoring systems or as capacity sizing in service queues.

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Frequently asked questions

  • What does the parameter λ represent? It is the average number of events per unit of time or space (calls per hour, defects per metre). In the Poisson distribution, λ is both the mean and the variance.
  • When should I use Poisson instead of the binomial? When you count rare events over a continuous interval with no fixed number of trials, or as an approximation to the binomial when n is large and p is small (taking λ = np).
  • How is it related to the exponential distribution? In a Poisson process the number of events per interval follows a Poisson distribution, and the time between consecutive events follows an exponential with rate λ: they are two views of the same process.

Reference: Poisson distribution — Wikipedia