#### The Poisson Distribution

- The Poisson Distribution is used as a model for the number, X, of events in a given interval of space or time. It has the probability formula:

- where λ is equal to the mean number of events in the given interval

- A Poisson distribution with mean λ can be noted as:
- X ~ Po(λ)

__Suitability of a Poisson Distribution__

__Suitability of a Poisson Distribution__

- It occurs randomly in space or time
- It occurs singly — events cannot occur simultaneously
- It occurs independently
- It occurs at a constant rate — mean number of events in a given time interval is proportional to the size of the interval

__Expectation and Variance__

__Expectation and Variance__

- For a Poisson distribution X ~ Po(λ)
- Mean = μ = E(X) = λ
- Variance = σ² = Var(X) = λ
- The mean and the variance of a Poisson distribution are equal

__Addition of Poisson Distributions__

__Addition of Poisson Distributions__

- If X and Y are independent Poisson random variables with parameters λ and μ respectively, then X + Y has a Poisson distribution with parameter λ + μ

__Examples__

__Examples__

- The number of emissions per minute from two radioactive objects A and B are independent Poisson variables with mean 0.65 and 0.45 respectively. Find the probabilities that:
- in a period of three minutes there are at least three emissions from A
- Write the distribution using the correct notation:
- A ~ Po(0.65 × 3) = A ~ Po(1.95)

- Use the limits given in the question to find the probability:
- P(A ≥ 3) = 1 – P(A < 3)

- = 1 – 0.690
- = 0.310

- P(A ≥ 3) = 1 – P(A < 3)

- Write the distribution using the correct notation:
- in a period of two minutes there is a total of less than four emissions from A and B together
- Write the distribution using the correct notation;
- (A + B) ~ Po(2(0.65 + 0.45)) = (A + B) ~ Po(2.2)

- Use the limits given in the equation to find the probability:

- = 0.819

- Write the distribution using the correct notation;

- in a period of three minutes there are at least three emissions from A

__Relationship of Inequalities__

__Relationship of Inequalities__

- P(X < r) = P(X ≤ r – 1 )
- P(X = r) = P(X ≤ r) – P(X ≤ r – 1)
- P(X > r) = 1 – P(X ≤ r)
- P(X ≥ r) = 1 – P(X ≤ r – 1)

__Poisson Approximation of a Binomial Distribution__

__Poisson Approximation of a Binomial Distribution__

- To approximate a binomial distribution given by X ~ B(n, p)
- If n > 50 and np > 5
- Then we can use a Poisson distribution given by X ~ Po(np)

__Examples__

__Examples__

- A randomly chosen doctor in general practice sees, on average, one case of a broken nose per year and each case is independent of the other similar cases
- Regarding a month as a twelfth part of a year,
- Show that the probability that, between them, three such doctors see no cases of a broken nose in a period of one month is 0.779
- Write down the information we know and need:
- 1 doctor = 1 nose per year =
^{1}/_{12}noses per month - 3 doctors =
^{3}/_{12}=^{1}/_{4}nose per month

- 1 doctor = 1 nose per year =
- Write the distribution using the correct notation:
- X ~ Po(0.25)

- Use the limits given in the equation to find the probability:

- = 0.779

- Write down the information we know and need:
- Find the variance of the number of cases seen by three such doctors in a period of six months
- Var(X) = μ = λ
- Calculate λ in this scenario;
- λ = 6 × μ(in one month)
- = 6 × 0.25
- = 1.5
- ∴ Var(X) = 1.5

- Show that the probability that, between them, three such doctors see no cases of a broken nose in a period of one month is 0.779
- Find the probability that, between them, three such doctors see at lest three cases in one year
- Calculate λ in this scenario:
- λ = 12 × μ(in one month)
- = 12 × 0.25
- = 3

- Use the limits given in the equation to find the probability:
- P(X ≥ 3) = 1 – P(X ≤ 2)

- = 1 – 0.423
- = 0.577

- P(X ≥ 3) = 1 – P(X ≤ 2)

- Calculate λ in this scenario:
- Find the probability that, of three such doctors, one sees three cases and the other two see no cases in one year
- We will need two different λs in this scenario:
- λ for one doctor in one year = 1
- λ for the other two doctors in one year = 2 × 1 = 2

- For the first doctor:

- For the other two doctors:

- Considering that any of the three could be first

- = 0.025

- We will need two different λs in this scenario:

- Regarding a month as a twelfth part of a year,

__Normal Approximation of a Poisson Distribution__

__Normal Approximation of a Poisson Distribution__

- To approximate a Poisson distribution given by X ~ P(λ)
- If λ > 15
- Then we can use a normal distribution given by X ~ N(λ, λ)
- Apply continuity correction to the limits

__Examples__

__Examples__

- The number of flaws in a length of cloth, l m long has a Poisson distribution with mean 0.04l
- Find the probability that a 10 m length of cloth has fewer than 2 flaws
- Form the parameters of the Poisson distribution
- l = 10
- λ = 0.04l
- ∴ λ = 0.4

- Write down the distribution using the correct notation:
- X ~ Po(0.4)

- Write the probability required by the question:
- P(X < 2)

- From earlier equations:

- = 0.938

- Form the parameters of the Poisson distribution
- Find the approximate value for the probability that a 1000 m length of cloth has at least 46 flaws
- Using the question to form the parameters:
- l = 1000 m
- λ = 0.04l
- ∴ λ = 40, which is greater than 15
- Thus, we can use the normal approximation

- Write down our distribution using the correct notation:
- X ~ Po(40) → Y ~ N(40, 40)

- Write the probability required by the question:
- P(X ≥ 46)

- Apply continuity correction for the normal distribution
- P(Y ≥ 45.5)

- Evaluate the probability:
- P(Y ≥ 45.5) = 1 – Φ(
^{45.5 – 40}/_{√40}) - = 0.192

- P(Y ≥ 45.5) = 1 – Φ(

- Using the question to form the parameters:
- Given that the cost of rectifying X flaws in a 1000 m length of cloth is X² pence, find the expected cost
- Using the variance formula:
- Var(X) = E(X²) – (E(X))²

- For a Poisson distribution:
- E(X) = Var(X) = λ and λ = 40

- Substitute into the equation and solve for the unknown:
- 40 = E(X²) – 40²
- E(X²) = 1640 pence
- E(X²) = £ 16.40
- ∴ The expected cost for rectifying the cloth is £ 16.40

- Using the variance formula:

- Find the probability that a 10 m length of cloth has fewer than 2 flaws