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Poisson process; intersection accidents

Short answer

A Poisson process has independent events at constant average rate; count has Poisson distribution with mean lambda and standard deviation sqrt(lambda). For lambda=5 per month, probabilities are: no accidents 0.0067, exactly one 0.0337, at most one 0.0404.

a) Explain the Poisson process. What are mean and standard deviation of Poisson distribution? Give one example of Poisson distribution. b) Suppose that we are investigating the safety of a dangerous intersection. Past police records indicate a mean of 5 accidents per month at this intersection. Calculate the probability, in any month, of: i) No accidents occurring ii) Exactly one accident occurring iii) At most one accident occurring

a) Poisson process, mean, SD and example

A Poisson process is a random process in which events occur one at a time, at a constant average rate, independently of each other, in a continuous interval of time, length or area. Its properties:

  1. The number of events in non-overlapping intervals is independent.
  2. The probability of exactly one event in a very short interval is proportional to the length of the interval.
  3. The probability of two or more events in a very short interval is negligible.
  4. The average rate is constant over the interval.

The number of events X in an interval with average λ follows the Poisson distribution P(X = x) = e⁻ˡ λˣ / x!, x = 0, 1, 2, …

  • Mean = λ
  • Variance = λ, so standard deviation = √λ

Example: Number of telephone calls arriving at an exchange per minute, or the number of printing errors per page.

b) Accidents at the intersection

λ = 5 per month, e⁻⁵ = 0.006738.

i) No accident: P(0) = e⁻⁵ = 0.0067

ii) Exactly one accident: P(1) = e⁻⁵ × 5 = 0.033690 ≈ 0.0337

iii) At most one accident: P(X ≤ 1) = 0.006738 + 0.033690 = 0.040428 ≈ 0.0404

Answer: (i) 0.0067, (ii) 0.0337, (iii) 0.0404

✓ Answer checked by Claude Sonnet 5.5

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