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
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
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
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
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
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
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
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
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
Write the distribution using the correct notation:
X ~ Po(0.25)
Use the limits given in the equation to find the probability:
= 0.779
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
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
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
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
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
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
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
