Poisson distribution question?

The Poisson discrete probability distribution is given by

P(k, λ) = λ^k.e^(-λ)/k!

where k is the number of occurrences of the property or event under examination, and λ the mean value under the test conditions. It would be necessary to avoid or ignore centres of population where the density of telephone boxes could be atypically high, but if the overall density is 300 boxes in 500_km² it doesn't sound as if there are many of them.

One of the advantages of the Poisson distribution is that it is characterised by a single parameter λ which is related proportionally to the sample area. So the overall density of 300 boxes in 500 km² may be used to calculate the mean for 1 km² sized sample areas as λ = 300/500 = 0.6.

So the student could carry out the project in two stages: first, survey the number of telephone boxes found in each of the 1 km² squares in the sample area of 50 km², and establish how well the results fitted the Poisson distribution above with mean λ = 0.6. Second, assuming that the results were in reasonable agreement with those calculated, other statistical correlations could be studied with a very simple division of the sample areas into those with no telephone boxes or those with one or more, since for λ = 0.6

P(0, λ) = 0.549, P(1, λ) = 0.329, P(2, λ) = 0.099, P(3, λ) = 0.020 ...

can be replaced by P(k=0, λ) = 0.549 and P(k>0, λ) = 1 - P(k=0, λ) = 0.451, with a considerable reduction in the amount of work required.

HTH