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# Poisson Distribution

Poisson distribution is one of the important topics of statistics. In statistics, Poisson distribution is one of the discrete probability distribution. This distribution is used for calculating the possibilities for an event with the given average rate of value ie $\lambda$. A Poisson random variable, x, refers to the number of success in a Poisson experiment. Poisson distribution is a limiting process of binomial distribution. Poisson distribution occurs when there are events which do not occur as outcomes of a definite number of outcomes.

Poisson distribution is used under the following conditions:

• Number of trials 'n' tends to infinity
• Probability of success 'p' tends to zero and
• np = 1 is finite.

 Related Calculators Calculating Poisson Distribution Binomial Distribution Calculator Calculator for Distributive Property Cumulative Normal Distribution Calculator

## Poisson Distribution Formula

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A Poisson distribution is the probability distribution that results from a Poisson experiment. A Poisson experiment is a statistical experiment that can be classified as successes or failures.

f(x) = $\frac{e^{-\lambda}\lambda^x }{x!}$

where,
λ is an average rate of value.
x is a Poisson random variable.
e is the base of logarithm.

### Poisson Distribution Function

Probability distribution function for X is

f(x) = $\frac{e^{-\lambda}\lambda^x }{x!}$, where x = 0, 1, 2, 3, .................

Let X be discrete random variable that can take on the values 0, 1, 2,....... such that the probability function of X is given by

f(x) = P(X = x) = $\frac{\lambda^x e^{-1}}{x!}$, x = 0, 1, 2,..................

Where $\lambda$ is given positive constant.

### Continuous Poisson Distribution

If the observed random variable is continuous, the number of the possible values is infinitely large.The Poisson distribution for continuous event numbers X. This can be achieved by replacing the factorial by the gamma-function.

$P(X)$ = $\frac{\lambda^x e^{-1}}{x!}$ = $\frac{\lambda^x e^{-1}}{\gamma x}$

## Poisson Distribution as a Limiting Form

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We shall now deduce the Poisson distribution from the binomial distribution by assuming that n ® ¥ and p ® 0 such that the product np always remains finite, say l.      We shall now use a very important result of limits in Calculus. We state this result without proof: where, e is a constant lying between the number 2 and 3 and ex is defined by with x a real number. From equation (1), we observe that each of (r - 1) factors,  that Thus,   where, l is a finite number and is equal to np.

The sum of the probabilities P(X = r) or simply P(r) for r = 0, 1, 2, … is 1. This can be seen by putting r = 0, 1, 2, … in (4) and adding all the probabilities.   Also, each of the probabilities is a non-negative fraction. This leads to the distribution defined below:

A random variable X taking values 0, 1, 2, … is said to have a Poisson distribution with parameter l (finite), if its probability distribution is given by There are many daily life situations where n is very large and p is very small. In such situations, the Poisson distribution can be more conveniently used as an approximation to binomial distribution which may prove cumbersome for large values of n. This is called Poisson approximation to binomial distribution.

The probability mass function of the Poisson distribution given by Note:   = e-1 e1

= 1

## Poisson Distribution Expected Value

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A random variable is said to have a Poisson distribution with parameter $\lambda$, where $\lambda$ is expected value of the Poisson distribution.

Expected value of the Poisson distribution

E(X) = $\mu$ = $\frac{d(e^{\lambda(t-1)})}{dt}$ at t = 1

= $e^{\lambda(t-1)}$ $\lambda$ at t = 1

= $\lambda$

The expected value (mean) and variance of the Poisson distribution is equal to $\lambda$.

## Compound Poisson Distribution

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The compound Poisson distributions are object of study in probability, it has a number of useful properties. Formulae for the exponential and based on the aggregate claims simplify because mean and variance both are same and P(t) = exp($\lambda(e^t - 1))$. For large $\lambda$ the distribution of the compound Poisson can be approximated by a normal distribution with $\lambda$ E(x) and variance $\lambda E(x^2)$. And the quantile premium is given by:

=> P(a) = $\lambda$ E(x) + $\pi ^{-1}(1 - a)\sqrt{\lambda E(x^2)}$, a lies between (0, 1).

## Poisson Distribution Table

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As with the binomial distribution there is a table that we can use under certain conditions that will make calculating probabilities a little easier when using the Poisson Distribution. Table is showing tabulated values of f(x) = P(X $\geq$ x), where X has a Poisson distribution with parameter $\lambda$. The table displays values of Poisson distribution. ## Cumulative Poisson Distribution

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Cumulative Poisson Distribution is defined as the measure of the probability in which the average number of success estimated within a range of experiments. A cumulative Poisson distribution is used to calculate the probability of getting atleast n successes in a Poisson experiment.

Formula for cumulative Poisson distribution

## Calculate Poisson Distribution

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Poisson distribution can be calculated by using Poisson distribution formula. Let us see with the help of example how to calculate Poisson distribution.

### Solved Example

Question: If the variance of the Poisson distribution is 2, find the probabilities for r = 1, 2, 3, 4 and 5 from the recurrence relation of the Poisson distribution.

Solution:

The variance of the Poisson distribution = l = 2 Recurrence relation is given by               ## Poisson Distribution Examples

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Given below are examples on Poisson distribution for your better understanding:

### Solved Examples

Question 1: A random variable X has a Poisson distribution with parameter l such that P (X = 1) = (0.2) P (X = 2). Find P (X = 0).
Solution:

For the Poisson distribution, the probability function is given by Given P (x = 1) = (0.2) P (X = 2)   Question 2:

Telephone calls arrive at an exchange according to Poisson process with rate $\lambda$ = 2/min. Calculate the probability that exactly 2 calls will be received during each of the first 5 min of the hour.

Solution:

Step 1:

Let N be the number of calls received during a 1 min period.

=> P(N = 2) = $\frac{e^{-2}2^2}{2!}$

= 2$e^{-2}$

Step 2:

Let M be the number of minute, among the 5 min considered, during which exactly 2 calls will be received. M follows a binomial distribution with parameters n = 5 and p = 2$e{-2}$.

P(M = 5) = C(5, 5)$(2e^{-2})^5$$(1 - 2e^{-2})^{5 - 5}$

= 32 * $e^{-10}$

≈  0.00145.

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