• 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

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