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Poisson Distribution Formula

Poisson distribution is actually an important type of probability distribution formula.  As in the binomial distribution, we will not know the number of trials, or the probability of success on a certain trail. The average number of successes will be given for a certain time interval. The average number of successes is called “Lambda” and denoted by the symbol \(\lambda\). In this article, we will discuss the Poisson distribution formula with examples. Let us begin learning!

poisson distribution formula

                                                                                                                                                                  Source: en.wikipedia.org

Poisson Distribution Formula

Concept of Poisson distribution

The French mathematician Siméon-Denis Poisson developed this function in 1830. This is used to describe the number of times a gambler may win a rarely won game of chance out of a large number of tries.

The Poisson random variable follows the following conditions:

  1. The number of successes in two disjoint time intervals is independent.
  2. The probability of success during a given small time interval is proportional to the entire length of the time interval.

Besides the disjoint time intervals, the Poisson random variable also applies to disjoint regions of space.

Some Applications of Poisson distribution are as following:

  • The number of deaths by horse kicking in the army of Prussian.
  • Birth defects and genetic mutations.
  • Rare diseases like Leukemia, because it is very infectious and so not independent mainly in legal cases.
  • Car accident prediction on roads.
  • Traffic flow and the ideal gap distance between vehicles.
  • The number of typing errors found on a page in a book.
  • Hairs found in McDonald’s hamburgers.
  • The spread of an endangered animal in Africa.
  • Failure of a machine in one month.

Formula for Poisson Distribution

The probability distribution of a Poisson random variable let us assume X. It is representing the number of successes occurring in a given time interval is given by the formula:

\(\displaystyle{ P }{\left({ X }\right )}=\frac{{{ e }^{-\mu}\mu^{ x }}}{{{ x }!}} \)

where

\(\displaystyle{x}={0},{1},{2},{3},…\)

\(\displaystyle{e}={2.71828}\)

\(\mu\)= mean number of successes in the given time interval or region of space.

Mean and Variance of Poisson distribution:

If \(\mu\) is the average number of successes occurring in a given time interval or region in the Poisson distribution. Then the mean and the variance of the Poisson distribution are both equal to \(\mu\).

Thus,

E(X) = \(\mu\)

and

V(X) = \(\sigma^2 = \mu\)

Remember that, in a Poisson distribution, only one parameter, \(\mu\) is needed to determine the probability of any given event.

Some Solved Examples for You

Example-1: Some vehicles pass through a junction on a busy road at an average rate of 300 per hour.

  1. Find out the probability that none passes in a given minute.
  2. What is the expected number of passing in two minutes?
  3. Find the probability that this expected number found above actually pass through in a given two-minute period.

Solution: First we will compute,

The average number of cars per minute is:

\(\displaystyle\mu = \frac{300}{{60}}\)

\(\displaystyle\mu\) = 5

(a)Applying the formula:

\(\displaystyle{P}{\left({ X }\right )}=\frac{{{ e }^{-\mu }\mu^{x}}}{{{x}!}} \)

\(\displaystyle{ P }{\left({ x }_{{ 0 }}\right)}=\frac{{{e}^{ -{{5}}}{5}^{0}}}{{{0}!}}={ 6.7379 }\times{10}^{ -{{3}}} \)

(b) Expected number each 2 minutes = E(X) = 5 × 2 = 10

(c) Now, with \(\mu\) = 10, we have:

\(\displaystyle{ P }{\left({ x }_{{ 10 }}\right)}=\frac{{{e}^{ -{{10}}}{10}^{10}}}{{{10}!}}={ 0.12511 }\)

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