The Poisson Distribution is a theoretical discrete probability distribution that is very useful in situations where the discrete events occur in a continuous manner. This has a huge application in many practical scenarios like determining the number of calls received per minute at a call centre or the number of unbaked cookies in a batch at a bakery, and much more.

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In such cases, only the average number of successes (like the no. of calls received) for a sample space (like the duration of 1 hour) is known. It can sometimes range from 0 to the maximum value, but one doesn’t know how many times it’s going to be a failure (like you can’t know how many calls will not come at a call centre at a given hour).

In such cases, the Poisson Distribution is utilized to determine the probability of exactly *x _{0}* number of successes taking place in unit time. Let us now mention some conditions for a model to qualify for a Poisson Model before looking at the formulae for the distribution itself.

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## The Poisson Model

As we have hinted in the introduction, the calls received per minute at a call centre, forms a basic Poisson Model. Relating to this real-life example, we’ll now define some general properties of a model to qualify as a Poisson Distribution.

Let us first divide the time into *n* number of small intervals, such that *n *→ ∞. We now define a success as a call received in a unit interval of time. Let *p *denote the probability of such a success. As you must have noted, we have already divided the time into infinitesimally small intervals; thus p* *→ 0.

However, this must result in the condition that *np = λ (a finite constant); *because there definitely is a mean number of calls being received per unit time. This way we can understand the following properties of a Poisson Model –

- The event or a success is something that can be counted in whole numbers.
- The probability of having a success in a time interval is independent of any of its previous occurrence.
- The average frequency of successes in a unit time interval is known. (λ in the above case)
- The probability of more than one success in a unit time is very low. (since
*n*→ ∞ and p

The resulting Poisson Distribution is – $$ f(x) = P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!} \text{for x = 0,1,2…. ∞}$$

$$ = 0 \text{ otherwise}$$ In shorthand notation, it is represented as – $$ X \sim P(\lambda) $$ λ is known as the parameter of the Poisson Distribution.

## Characteristics of the Poisson Distribution

⇒ It is uni-parametric in nature. As we can see, only one parameter λ is sufficient to define the distribution.

⇒ The mean of \( X \sim P(\lambda) \) is equal to λ.

⇒ The variance of \( X \sim P(\lambda) \) is also equal to λ. The standard deviation, therefore, is equal to +√λ.

⇒ Depending on the value of the parameter λ, it may be unimodal or bimodal. Look at some cases given below for example –

This illustrates that a Poisson Distribution typically rises, then falls. If λ is an integer, it peaks at x = λ and at x = λ – 1. Otherwise, it simply peaks at the integer part of λ.

⇒ Poisson Approximation to a Binomial Distribution: A typical binomial distribution \( X \sim B(n,p) \) is equivalent to \( X \sim P(\lambda) \) for the condition that *n *→ ∞ and *p → 0*, and *np = λ.*

⇒ Additive Property: If two Poisson Distributions \( X_1 \sim P(\lambda_1) \) and \( X_2 \sim P(\lambda_2) \) are added to give another random variable Y, then Y also obeys a Poisson Distribution given by \( Y = X_1 + X_2 \sim P(\lambda_1 + \lambda_2) \).

Note that no such simple result exists for the difference between two random variables having a Poisson Distribution.

## Fitting a Poisson Distribution to Given Data

For a given frequency distribution of a quantity, if the range of that quantity starts from 0 and proceeds to a positive integer, then a Poisson Probability Distribution can be fitted to that data using the parameter = the observed mean frequency of that quantity.

Note that the distribution may not exactly predict the observed frequency because the Poisson Distribution holds only for random variables assuming discrete values and whose occurrence is completely independent of any of its previous occurrences. Now go through the solved examples below to improve your understanding of the topic!

## Solved Examples for You

### Problem 1

Two types of emails arrive at a company account independently and at random: External emails at a mean rate of one every five minutes and Internal emails at a rate of two every five minutes. Calculate the probability of receiving two or more emails in a two-minute interval.

**Solution: **Let us first define the random variables that we need –

X – the number of External emails received per a two-minute interval

Y – the number of Internal emails received per a two-minute interval

Since on an average, we expect the mean of X = 1 for every five minute-interval and the mean of Y = 2 for every five-minute interval, we can calculate the parameters of the Poisson Distribution satisfied by X and Y in the following way –

λ_{X} = \(\frac{1}{5}\times 2 = 0.4\)

λ_{Y} = \(\frac{2}{5}\times 2 = 0.8\)

Thus, we have –

\(X \sim P(0.4)\)

\(Y \sim P(0.8)\)

Since here we have to consider both types of emails received together, let us now define a third random variable Z = X + Y. It can be equivalently expressed as –

\(Z = X + Y = P(0.4) + P(0.8)\)

\(Z = P(1.2)\)

Now in this problem, we simply have to calculate Pr(Z ≥ 2). Using simple probability logic and the Poisson distribution, it can be calculated as –

Pr(Z ≥ 2) = 1 – Pr(Z < 2) = 1 – Pr(Z = 0) – Pr(Z = 1)

Since Z can only take values 0 and 1 before it attains a value = 2.

Now, \(Pr(Z = 0) = e^{-1.2} = 0.301\)

\(Pr(Z = 1) = e^{-1.2}\frac{1.2}{1} = 0.361\)

Hence,

Pr(Z ≥ 2) = 1 – 0.301 – 0.361

= 0.338

= almost 34%

### Problem 2

The table below is the data collected from a group of 500 students. *x *represents the number of different foreign countries visited, and corresponding to it, the number of students is given.** **Fit the given data to a Poisson distribution and see the deviation from it.

x | Frequency |

0 | 56 |

1 | 156 |

2 | 132 |

3 | 92 |

4 | 37 |

5 | 22 |

6 | 4 |

7 | 0 |

8 | 1 |

**Solution: **We first need to find the mean for this data – $$\mu = \frac{\Sigma x_if_i}{\Sigma f_i}$$ $$ = \frac{0.56 + 1.156 + 2.132 + 3.92 + 4.37 + 5.22 + 6.4 + 7.0 + 8.1}{56 + 156 + 132 + 92 + 37 + 22 + 4 + 0 + 1}$$ $$ = \frac{986}{500} = 1.972$$

Thus the variable X, if indeed it is random, can be expressed as –

\(X \sim P(1.972)\)

Now according to the Poisson Distribution, let us find the values that Pr(X = x) will assume can be calculated as –

x | Pr (X = x) | Expected Frequency = Pr(X = x) ×N | Observed Frequency |

0 | \(Pr(X = 0) = e^{-1.972}\frac{1.972^0}{0!} = 0.139178\) | 69.589 | 56 |

1 | \(Pr(X = 1) = e^{-1.972}\frac{1.972^1}{1!} = 0.274459\) | 137.2295 | 156 |

2 | \(Pr(X = 2) = e^{-1.972}\frac{1.972^2}{2!} = 0.270617\) | 135.3085 | 132 |

3 | \(Pr(X = 3) = e^{-1.972}\frac{1.972^3}{3!} = 0.177885\) | 88.9425 | 92 |

4 | \(Pr(X = 4) = e^{-1.972}\frac{1.972^4}{4!} = 0.087697\) | 43.8485 | 37 |

5 | \(Pr(X = 5) = e^{-1.972}\frac{1.972^5}{5!} = 0.034588\) | 17.294 | 22 |

6 | \(Pr(X = 6) = e^{-1.972}\frac{1.972^6}{6!} = 0.011368\) | 5.684 | 4 |

7 | \(Pr(X = 7) = e^{-1.972}\frac{1.972^7}{7!} = 0.003202\) | 1.601 | 0 |

8 | \(Pr(X = 8) = e^{-1.972}\frac{1.972^8}{8!} = 0.000789\) | 0.3945 | 1 |

Total = 499.89 ≅ 500 | Total = 500 |

Thus you see that there indeed are some departures in the data from the ones calculated by the Poisson Distribution. Still, it is definitely a good approximation in certain situations.

In problem two, the sum of standard deviations should be 4.4 not 3.16.