Poisson Distribution Calculator

Poisson Distribution Guide

What should I know about The Poisson Distribution?

P(X = k) = (λᵏ × e⁻ˡ) / k!. Where λ = mean number of events per interval, k = number of events, e ≈ 2.71828. The Poisson distribution models events that occur randomly and independently at a constant average rate. Properties: mean = variance = λ. Standard deviation = √λ. The distribution is right-skewed for small λ and approaches the normal distribution as λ increases (above 15-20).

When to Use the Poisson Distribution?

Conditions: events occur independently. Events cannot occur simultaneously. The average rate (λ) is constant over the interval. Common applications: number of calls to a call centre per hour (λ = 45 calls/hour). Number of accidents at a junction per month (λ = 2.3). Radioactive decay events per second. Number of bacteria on a microscope slide. Number of printing errors per page. Number of goals in a football match (roughly Poisson, λ ≈ 2.7 in Premier League). Website visitors per minute.

What do I need to know about Poisson as Binomial Approximation?

The Poisson distribution approximates the binomial when n is large and p is small (rare events). Use Poisson when n ≥ 50 and p ≤ 0.05. λ = np. Example: 1000 items, 0.3% defect rate. n=1000, p=0.003. λ = 3. P(X ≥ 5) using Poisson ≈ P(X ≥ 5) using binomial — much simpler calculation. The approximation is because n!/(r!(n-r)!) becomes unwieldy for large n, while the Poisson formula uses only λ.

What should I know about Fitting a Poisson Distribution?

Test whether data follows a Poisson distribution: calculate the sample mean and variance — if the data is Poisson, these should be approximately equal (mean = variance). Large variance relative to mean (overdispersion) suggests the Poisson model is inappropriate — a negative binomial distribution may fit better. Chi-squared test: compare observed frequencies to expected Poisson frequencies. If χ² < critical value, the Poisson model fits. Application: insurance companies use Poisson models for claims frequency, helping them price policies based on the expected number of claims in a given period rather than just historical averages.

What should I know about the Poisson Formula and a Worked Example?

The Poisson distribution gives the probability of observing exactly k events in a fixed interval, given that events occur with a known average rate λ (lambda) and independently of each other. The formula is P(X = k) = (λᵏ × e^(−λ)) / k!, where e ≈ 2.71828 is Euler's number and k! is k factorial. As a worked example, if a call centre receives an average of λ = 4 calls per minute, the probability of receiving exactly 2 calls in a given minute is P(X=2) = (4² × e^(−4)) / 2! = (16 × 0.0183) / 2 ≈ 0.2927 × 0.5 — working through it carefully: 4² = 16, e^(−4) ≈ 0.01832, so 16 × 0.01832 ≈ 0.2931, divided by 2! = 2, giving P(X=2) ≈ 0.1465, or about 14.7%. This formula lets businesses, hospitals, and engineers calculate the likelihood of rare or random discrete events — from customer arrivals to equipment failures — occurring a specific number of times within a set period.

What's the key thing to understand about the Mean and Variance of a Poisson Distribution?

A distinctive mathematical property of the Poisson distribution is that its mean and its variance are always equal, both equal to λ, the same rate parameter used in the probability formula. This is unlike most other distributions, where the mean and variance are independent and unrelated quantities that must be specified separately. For example, a Poisson-distributed process with an average of 9 events per interval also has a variance of 9, and therefore a standard deviation of √9 = 3. This equal-mean-variance property gives statisticians a useful diagnostic check: if real-world count data has a sample variance noticeably larger than its sample mean (called 'overdispersion'), the Poisson model is probably not a good fit, and a different distribution (such as the negative binomial) that allows variance and mean to differ should be used instead. This single property is one of the quickest ways to test whether count data genuinely behaves like a Poisson process before relying on Poisson-based predictions.

What's the key thing to understand about the Assumptions Behind the Poisson Model?

The Poisson distribution models the count of independent events happening at a constant average rate within a fixed interval of time or space, but it relies on a few assumptions that are worth checking before trusting it. Events must occur independently of each other (one customer arriving doesn't make the next arrival more or less likely), the average rate λ must stay constant across the interval being modeled (no rush hours or seasonal spikes within the window), and two events essentially can't occur at exactly the same instant. When these assumptions break down — for example, customer arrivals that cluster because of a scheduled bus drop-off, or server requests that surge during a sale — the real variance in the data tends to exceed the Poisson's predicted variance (which must always equal its mean), a pattern called "overdispersion" that signals a different model, like the negative binomial distribution, may fit better.

Poisson Probability Table for λ = 3

Here is the full probability mass function P(X = k) = e^(-λ)λ^k / k! evaluated at λ = 3, which models, for example, the number of emails arriving per minute when the long-run average is 3 per minute.

Notice that k = 2 and k = 3 are tied for the most likely single outcome, which is expected whenever λ is a whole number (the mode is both λ and λ − 1). By k = 6, cumulative probability already exceeds 96.6%, showing how quickly the Poisson tail thins out once k is roughly double the mean.

Worked Example: Staffing a Call Center with the Poisson Distribution

A call center receives an average of λ = 4 complaints per hour. The manager wants to know (a) the probability of getting exactly 6 complaints in an hour, and (b) the probability of getting 2 or fewer, which would mean staff are overstaffed for that hour. (a) P(X = 6) = e^(-4) × 4^6 / 6! = 0.018316 × 4096 / 720 = 75.02 / 720 = 0.1042, so there is about a 10.4% chance of exactly 6 complaints in a given hour. (b) P(X ≤ 2) = P(0) + P(1) + P(2) = e^(-4)[1 + 4 + 8] = 0.018316 × 13 = 0.2381, so there's roughly a 23.8% chance of two or fewer complaints — nearly a 1-in-4 hour where a single staff member could easily have handled the load. Because the Poisson variance equals its mean (σ² = λ = 4), the standard deviation is √4 = 2 complaints per hour, so a load anywhere from 2 to 6 complaints (mean ± 1 SD) is unremarkable; it's only when counts stray much beyond that band — say 9 or more — that a manager should investigate whether something unusual (an outage, a promotion, a product bug) is driving an above-average arrival rate.

Poisson vs. Normal Approximation for Large λ

When λ grows large (commonly λ ≥ 20 is used as a rule of thumb), the Poisson distribution becomes closely approximated by a normal distribution with mean μ = λ and standard deviation σ = √λ, which is far easier to compute by hand or with a basic calculator. For example, with λ = 25 (average daily website sign-ups), estimate P(X ≤ 30) using the normal approximation with a continuity correction: z = (30.5 − 25) / √25 = 5.5/5 = 1.10. Looking up Φ(1.10) in a standard normal table gives approximately 0.8643, so there's roughly an 86.4% chance of 30 or fewer sign-ups on a given day. The exact Poisson calculation would require summing 31 individual factorial-based terms, which is exactly why the normal approximation is so useful for quick estimates once λ is reasonably large — the two methods typically agree to within about 1 percentage point once λ exceeds roughly 20, but diverge more noticeably for small λ (under about 10), where the Poisson's right-skew is still pronounced and a symmetric normal curve is a poor fit. As a rule of thumb, many textbooks also suggest the square-root transformation √X as a way to stabilize the Poisson's variance before applying normal-based methods, since the variance of √X is approximately constant (about 0.25) regardless of λ, which isn't true of X itself.