### Poisson Distribution

Suppose that the number of cars that goes past the gates of Heathfield Secondary school in London, between 8.00 am and 8.15 am on a week day is 15. If this rate remains the same during the
morning peak hours - - 8 am till 9 am - one can safely assume that the number of cars goes past the gate of this particular school between 8.15 am and 8.30 am as 15 as well. Moreover, the number of cars that goes past between,
8.15 am and 8.45 am can be assumed as 2 x 15 = 30 and between 8.15 am and 9.00 am as 3 x 15 = 45. In a situation of this kind, the parameter ** the number of cars going past the school in 15
minutes,** can be a basis for a statistics model.

It is *Poisson Distribution*.

As you can see the **parameter** depends on the **time interval** - the greater the time interval , the greater the parameter.
Similarly, the parameter can be confined to
**space** as well. For instance, the number of bacteria in 5ml of water in a bucket, is a case in point - the greater the volume, the greater the number of bacteria.

In situations of this nature, the outcomes can be modelled by **Poisson Distribution.**

Poisson Distribution can be defined as follows:

#### X ~ Po(λ)

P( X = r ) = e^{-λ}λ^{r} / r!

Is Poisson a true distribution?

Let's say the values of success - r - are 0, 1, 2, 3, 4, ...........n

So, the corresponding probabilities - e^{-λ}λ^{r} / r! - are e^{-λ}λ^{0} / 0!, e^{-λ}λ^{1} / 1!,
e^{-λ}λ^{2} / 2!, e^{-λ}λ^{3} / 3!
etc.

The following table summarizes the distribution:

**t** | P(X = r) |

0 | e^{-λ}λ^{0} / 0! |

1 | e^{-λ}λ^{1} / 1! |

2 | e^{-λ}λ^{2} / 2! |

3 | e^{-λ}λ^{3} / 3! |

n | e^{-λ}λ^{n} / n! |

So, Σ (P X = r) = Σ e^{-λ}λ^{r} / r! = e^{-λ}[(λ^{0}/0!) + (λ^{1}/1!) + .... + (λ^{n}/n!)]
= e^{-λ}[e^{λ}] = e^{0} = 1

[the purple part comes from the expansion of the exponential function, e^{x}].

Since the sum of probabilities adds up to 1, this is a true probability distribution.

Check list for Poisson Distribution

- The probability of a certain event is constant in an interval based on
space or time.
- The Poisson parameter is proportional to the length of the interval.

Situations where Poisson Distribution model does not work:

- Modelling the number of cars going past a particular school during 8.30 am and 9.00 am on a week day and the at weekends - the parameter is not the same.
- Modelling the number of bacteria in a bucket of water, when it is under observation with a certain anti-biotic - the parameter changes by the minute.

You can use the following calculator in the calculations of Poisson Distribution.

#### Poisson Calculator

**E.g.1**

The number of bacteria found in 5ml of water in a bucket is 8. Calculate the following probabilities:

- Having not a single bacteria
- Having exactly two bacteria
- Having two or fewer bacteria
- Having more than 3 bacteria

λ = 8

- P(r = 0) = e
^{-8}8^{0} / 0! = 0.0003
- P(r = 2) = e
^{-8}8^{2} / 2! = 0.0107
- P(r < = 2) = e
^{-8}8^{0} / 0! + e^{-8}8^{1} / 1! + e^{-8}8^{2} / 2! = 0.0003 + 0.0027 + 0.0107 = 0.0137
- P(r > 3) = 1- P(r < = 2) = 1- 0137 = 0.9863

**E.g.2**

When Angelina types in an article in Word, she makes, on average, 3 spelling mistakes in a single page. She has to type in an article consisting of 3 pages. Find the following probabilities:

- Making no spelling mistakes in the whole article
- Making just two mistakes
- Making 2 or fewer mistakes
- Making more than 3 mistakes

Since we are considering three pages - the new space - λ = 3 * 3= 9

- P(r = 0) = e
^{-9}9^{0} / 0! = 0.0001
- P(r = 2) = e
^{-9}9^{2} / 2! = 0.005
- P(r < = 2) = e
^{-9}9^{0} / 0! + e^{-9}9^{1} / 1! + e^{-9}9^{2} / 2! = 0.0001 + 0.0011 + 0.0107 = 0.0119
- P(r > 3) = 1- P(r < = 2) = 1- 0119 = 0.9881

#### Expectation and Variance of Poisson Distribution

If X ~ Po(λ)

E(X) = λ; Var(X) = λ

You can practise the Poisson Distribution interactively here:

You can choose the number of trials(n), probability(p) and the range of values for which you need probabilities. Click the selector, [, [], or ], then movable slider will appear below the chart.

**Please answer the following questions.**

- If X ~ Po(5), find the following:
- Jason is an electrician in a school. He normally replaces 3 bulbs in a week at his school. Find the following:
- Replacing no bulb in a week
- Replacing 4 bulbs in a week
- Replacing 3 or fewer bulbs in a week

- The average demand for a skip hiring company on a week day is 3. Find the following probabilities:
- Hire no skip on Monday
- Hiring a skip on Tuesday and Thursday
- Hiring 5 skips during the whole week

- Tim plays for his local football club. On average, his manager lets him play 1 out of 3 matches. This season they are going to play 24 matches in all. Find the following
probabilities:
- He plays no match
- He plays exactly two matches
- He plays more than 4 matches

- On average, every week , two members of Randall family go abroad. Find the probability of 3 members of the family being abroad in three-week
period
- The number of defects found in a 20 m pipe is 5. Find the probability of finding 27 defects in a pipe of 100m long.
- The number of telephone calls received by a receptionist between 8.00 am and 8.05 am is 4. Find the probability of not receiving a single call, during the next three minutes.
- A plane spotter says on average he sees three aircrafts near a certain airfield in an hour. He says he is prepared to watch the sky for three hours. Find the probability of him seeing more than 5 aircrafts
during this period.
- A network engineer has been told about 4 potential cracks on a certain stretch of a railway track in every 10 km. He decided to inspect a track of 50 km. Find the probability of him finding more than
15 cracks.
- In a certain village of a certain county, 5 people are suffering from a certain birth defect, that has been attributed high background radiation in the area. It has been reported that this is the case
throughout the county. In a health survey, 8 villages were chosen for an experiment. Find the probability of finding exactly 20 people suffering from the defect.