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The Poisson distribution questions
Count events that arrive at random in a fixed window and one number describes everything. That number is the mean rate. It also happens to be the variance, which makes the model easy to test and easy to break.
6 original questions · 24 marks · the the poisson distribution notes · Further Statistics 1
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State the conditions under which a count can be modelled by a Poisson distribution, and name its one parameter.
Worked answer
Events must occur singly, at random, independently of one another, and at a constant average rate over the window in question. The only parameter is that mean rate λ, which is also the variance. B1 B1 for the four conditions, B1 for naming λ. All four conditions are needed, and 'independently' and 'at a constant rate' are the two most often left out.X ~ Po(4). Find P(X = 3) and P(X ≤ 2), each to 4 decimal places.
Worked answer
P(X = 3) = e⁻⁴ × 4³/3! = 0.1954. P(X ≤ 2) = e⁻⁴(1 + 4 + 8) = 0.2381. M1 for the Poisson formula, A1 for 0.1954, A1 for 0.2381. The cumulative value comes fastest from a calculator's own Poisson function.Calls arrive at a switchboard at a mean rate of 2.5 per minute, at random. Find the probability of no calls in a two-minute period, and state the variance of the count over that period.
Worked answer
Scale λ with the window. Over two minutes the count is Po(2.5 × 2) = Po(5), so P(X = 0) = e⁻⁵ = 0.0067 and the variance is 5, equal to the mean. M1 for scaling λ, A1 for Po(5), A1 for 0.0067, B1 for the variance. Rescaling λ to the interval actually asked about is the step that decides the whole answer, and leaving it at 2.5 gives 0.0821.X ~ Po(6). Find P(X ≥ 8) to 4 decimal places.
Worked answer
P(X ≥ 8) = 1 − P(X ≤ 7) = 1 − 0.7440 = 0.2560. M1 for the one minus form, A1 for 0.2560. Converting an 'at least' to one minus an 'at most', with the boundary dropped by one, is the standard manoeuvre.A component fails with probability 0.01, independently. In a batch of 200, use a Poisson approximation to estimate the probability of at most one failure, and compare with the exact binomial value.
Worked answer
n is large and p small, so B(200, 0.01) ≈ Po(2) with λ = np = 2. Then P(X ≤ 1) = e⁻²(1 + 2) = 0.4060. The exact binomial gives 0.4046, so the approximation is out by 0.0014: close enough for any practical purpose. B1 for λ = 2, M1 for the Poisson probability, A1 for 0.4060, B1 for the binomial value, A1 for the comparison.Two independent sources of faults occur at rates following Po(1.5) and Po(2) per hour. State the distribution of the total number of faults in a three-hour shift, explain why its mean and variance are equal, and find the probability that the shift produces at most 8 faults.
Worked answer
Independent Poisson variables add, so the two sources pool to Po(1.5 + 2) = Po(3.5) per hour. Scaling the window to three hours multiplies the rate by 3, giving Po(10.5) for the shift.
Both the mean and the variance are 10.5. A Poisson distribution has only one parameter, and it serves as both, so a sample whose variance is far from its mean is evidence against the model.
P(X ≤ 8) = 0.279 to three decimal places, from the cumulative Poisson function with λ = 10.5.
M1 for pooling the rates, M1 for scaling to three hours, A1 for Po(10.5), B1 for the mean and variance, B1 for why they are equal, M1 for the cumulative probability, A1 for 0.279.
Two steps have to happen in the right order. Pool the rates first and then scale, or scale each and then pool, but never scale only one of them. The additive property needs the two sources to be independent, and that should be stated.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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