What is the mean of the Poisson distribution?

The mean equals lambda. The variance also equals lambda. So the standard deviation is sqrt(lambda). This is the only common distribution where mean equals variance.

What is the difference between Poisson and binomial distributions?

The binomial has a fixed number of trials n with success probability p. The Poisson has no upper bound on events and models rare occurrences with lambda = np. As n becomes large and p becomes small (while np = lambda stays fixed), the binomial converges to the Poisson.

What does lambda mean in the Poisson distribution?

Lambda is the average number of events per time (or space) interval. For example, if calls arrive at 3 per hour on average, lambda = 3 for a 1-hour interval, lambda = 1.5 for a 30-minute interval.

Can the Poisson distribution model continuous events?

No. The Poisson distribution is discrete and counts whole-number events (0, 1, 2, ...). For continuous event times, the exponential distribution (the time between successive Poisson events) is used instead.

What is the cumulative Poisson distribution?

The CDF P(X = k) = 1 - P(X <= k-1).

What are examples of Poisson-distributed random variables?

Number of phone calls to a call center per hour, number of cars passing a checkpoint per minute, number of defects per square metre of fabric, radioactive decay counts per second, and number of server requests per second during normal load.

Poisson Distribution Calculator

Q: When should I use the Poisson distribution?

Use the Poisson distribution when counting rare, independent events over a fixed interval of time or space, where the average rate lambda is known and the probability of two events at the exact same moment is negligible.

Q: What are examples of Poisson-distributed random variables?

Number of phone calls to a call center per hour, number of cars passing a checkpoint per minute, number of defects per square metre of fabric, radioactive decay counts per second, and number of server requests per second during normal load.

Find exact and cumulative Poisson probabilities, mean, variance, and a full distribution table for any rate lambda.

📊 What is the Poisson Distribution?

The Poisson distribution is a discrete probability distribution that models the number of events occurring in a fixed interval of time or space, given that these events happen independently of each other at a constant average rate. It is named after the French mathematician Simeon Denis Poisson, who described it in 1837.

The Poisson distribution is widely used across many fields. In telecommunications, it models the number of calls arriving at a switchboard per minute. In healthcare, it describes the number of patients arriving at an emergency room per hour. In manufacturing quality control, it counts the number of defects per unit of product. In physics, it models the number of radioactive decay events per second. In ecology, it estimates the number of organisms found in a sample area. Any process involving rare, independent events with a known average rate is a candidate for Poisson modelling.

The Poisson distribution has a remarkable property: its mean and variance are both equal to the rate parameter lambda. This means that if you know the average event count, you also know the spread of the distribution. A small lambda produces a right-skewed distribution concentrated near zero. A large lambda produces an approximately symmetric, bell-shaped distribution, consistent with the Central Limit Theorem. When lambda exceeds about 10, the normal distribution N(lambda, lambda) gives a good approximation.

A common point of confusion is the relationship between the Poisson and binomial distributions. The Poisson distribution emerges as a limiting case of the binomial when the number of trials n becomes very large and the success probability p becomes very small, while the product np remains constant at lambda. This is why the Poisson distribution is sometimes called the law of rare events. If you have a binomial situation with n above 20 and p below 0.05, the Poisson with lambda = np is an accurate and computationally simpler approximation.

📐 Formula

P(X = k) = (λ^k × e^−λ) ÷ k!

λ (lambda) = average number of events per interval (rate parameter, > 0)

k = number of events (0, 1, 2, ...)

e = Euler's number ≈ 2.71828

k! = factorial of k (e.g., 3! = 6)

Mean = λ

Variance = λ

Std Deviation = √λ

Example: λ = 4, k = 2: P(X = 2) = (4² × e⁻⁴) / 2! = (16 × 0.01832) / 2 = 14.65%

📖 How to Use This Calculator

Steps

Choose your mode - Select Calculate Probability to find P(X=k) and cumulative probabilities for a specific k, or select Distribution Table to generate the full PMF and CDF for all k from 0 to approximately lambda + 4 x sqrt(lambda).

Enter lambda - Set lambda to the average number of events per interval. For example, if a server receives 7.5 requests per second on average, lambda = 7.5. Must be positive. Probability mode supports lambda up to 500; Table mode supports lambda up to 50.

Enter k (Probability mode) - Set k to the specific number of events you want to evaluate. k must be a non-negative integer (0, 1, 2, ...). The upper tail result P(X ≥ k) is especially useful for asking "what is the probability of at least k events?"

Read the results - The calculator shows P(X=k), cumulative P(X ≤ k), upper tail P(X ≥ k), mean (= lambda), variance (= lambda), and standard deviation (= sqrt(lambda)). The Distribution Table lists every k with its PMF and CDF values.

💡 Example Calculations

Example 1 - Call Center Arrivals

A call center receives 4 calls per minute on average. What is the probability of exactly 2 calls in the next minute?

lambda = 4 (average calls per minute), k = 2.

P(X = 2) = (4² x e^-4) / 2! = (16 x 0.01832) / 2 = 0.1465 = 14.65%.

P(X ≤ 2) = P(0) + P(1) + P(2) = 1.83% + 7.33% + 14.65% = 23.81%.

Mean = 4, Std Dev = sqrt(4) = 2.

P(X = 2) = 14.65%, P(X ≤ 2) = 23.81%, Mean = 4

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Example 2 - Website Traffic Spike

A website gets an average of 2.5 errors per hour. What is the probability of getting 5 or more errors in the next hour?

lambda = 2.5, k = 5. We need P(X ≥ 5) from the upper tail result.

P(X ≥ 5) = 1 - P(X ≤ 4) = 1 - (P(0)+P(1)+P(2)+P(3)+P(4)).

P(X ≤ 4) = 0.0821 + 0.2052 + 0.2565 + 0.2138 + 0.1336 = 0.8912 = 89.12%.

P(X ≥ 5) = 1 - 0.8912 = 10.88%. Mean = 2.5, Std Dev = sqrt(2.5) = 1.581.

P(X ≥ 5) = 10.88%, Mean = 2.5, Std Dev = 1.581

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Example 3 - Manufacturing Defects

A fabric producer averages 1.2 defects per square metre. What is the probability of zero defects in a 1 m2 sample?

lambda = 1.2, k = 0.

P(X = 0) = (1.2⁰ x e^-1.2) / 0! = 1 x e^-1.2 / 1 = e^-1.2 = 0.3012 = 30.12%.

There is a 30.12% chance of a defect-free square metre even at an average of 1.2 defects per m2.

Mean = 1.2, Std Dev = sqrt(1.2) = 1.095.

P(X = 0) = 30.12%, Mean = 1.2, Std Dev = 1.095

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❓ Frequently Asked Questions

What is the Poisson distribution formula?+

The formula is P(X = k) = (lambda^k x e^(-lambda)) / k!, where lambda is the average event rate, k is the number of events, and e is Euler's number (2.71828). For example, with lambda = 3 and k = 1: P(X = 1) = (3 x e^(-3)) / 1 = 3 x 0.0498 = 14.94%.

What is the mean and variance of the Poisson distribution?+

Both the mean and variance equal lambda. This is the defining property of the Poisson distribution. The standard deviation is sqrt(lambda). If you observe that the sample mean and sample variance of your data are approximately equal, the Poisson distribution is likely a good model.

When should I use the Poisson distribution instead of the binomial?+

Use the Poisson distribution when the number of trials n is very large and success probability p is very small, making np = lambda manageable. As a rule of thumb, use Poisson when n > 20 and p < 0.05. Use the binomial when n is small or p is not close to zero. Both give accurate results when lambda = np is around the same value.

What are the assumptions of the Poisson distribution?+

Four assumptions must hold: (1) events occur one at a time (no simultaneous events); (2) events are independent of each other; (3) the rate lambda is constant over the interval; and (4) the probability of an event in a very small sub-interval is proportional to the sub-interval size. If any assumption is violated (e.g., clustering or time-varying rate), a modified model like the negative binomial may be more appropriate.

How does the shape of the Poisson distribution change with lambda?+

For small lambda (less than 1), the distribution has most of its weight at k = 0 and is strongly right-skewed. As lambda increases, the mode shifts right and the distribution becomes more symmetric. For lambda above 10, the Poisson distribution is approximately bell-shaped and well approximated by a normal distribution N(lambda, lambda).

What is the relationship between the Poisson and exponential distributions?+

If events follow a Poisson process with rate lambda (events per unit time), then the time between consecutive events follows an exponential distribution with rate lambda. Conversely, counting events from an exponential inter-arrival process over a fixed time interval gives a Poisson distribution. They are two views of the same random process.

How do I scale the Poisson distribution to a different time interval?+

Multiply lambda by the new interval size. If events arrive at 5 per hour on average, then over 2 hours the rate is lambda = 10. Over 15 minutes (0.25 hours) the rate is lambda = 1.25. The same formula applies with the rescaled lambda. This linearity is one of the most useful properties of the Poisson process.

What is the cumulative Poisson probability P(X ≤ k)?+

P(X ≤ k) is the sum of P(X = 0) + P(X = 1) + ... + P(X = k). It gives the probability of observing k or fewer events. There is no closed-form expression, so it must be summed numerically (which this calculator does automatically). For large lambda, the normal CDF with mean lambda and variance lambda is a convenient approximation.

Can the Poisson distribution take non-integer lambda?+

Yes. Lambda can be any positive real number, including non-integers such as 2.7 or 0.35. Only k (the number of events) must be a non-negative integer. Non-integer lambda values are common in practice: a call center might receive 4.2 calls per minute on average, giving lambda = 4.2.

What is the normal approximation to the Poisson distribution?+

For lambda above 10, Poisson(lambda) is well approximated by N(lambda, lambda) (normal with mean and variance both equal to lambda). Apply a continuity correction for better accuracy: P(X = k) is approximated by P(k - 0.5 < Z < k + 0.5) where Z is the standard normal. This simplifies calculations when exact Poisson computation is inconvenient.

What is overdispersion and when does the Poisson model fail?+

Overdispersion occurs when the observed variance of your data is greater than the mean, violating the Poisson assumption that mean = variance. This often happens with count data that exhibits clustering or extra variability (e.g., accident counts where some days have inherently higher risk). In such cases, the negative binomial distribution is a better model, as it allows variance to exceed the mean.

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📌 Quick Tips

💡The Poisson distribution has the unique property that mean equals variance: both equal lambda.

💡For large lambda (above 10), the normal distribution N(lambda, lambda) is a good approximation.

💡Use the complement rule: P(X >= k) = 1 - P(X <= k-1) for upper tail calculations.

💡When lambda is small (below 5), the distribution is strongly right-skewed. As lambda grows it becomes symmetric.

💡The Poisson distribution approximates the binomial when n is large and p is small, with lambda = np.