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#### 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
• 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
• 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
1. 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:
1. 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
2. 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
1. 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
1. Regarding a month as a twelfth part of a year,
1. 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
2. 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
2. 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
3. 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
1. The number of flaws in a length of cloth, l m long has a Poisson distribution with mean 0.04l
1. 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
2. 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
3. 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