supervisor of a factory, Poisson distribution

Apr 2010
Any help offered in solving the below question will be much appreciated.

The supervisor of a factory, on average, two complaints of poor service from its customers per week.

(i) Given; that the supervisor receives 3 complaints in a month, find the probability that he receives all of them in the first three weeks of the month. (Assume that there are 4 weeks in a month) ( .422)

(ii) Find the probability, that, in a random sample of six weeks, the supervisor receives no complaint in one week, one complaint each in three weeks and more than complaint in each of the remaining two weeks. (.0568)
May 2010
If the number of complaints in 1 week is poisson with mean 2, then the number of complaints in 3 weeks is poisson with mean 6.

(NB: this is a special property of the poisson distribution, it is not true for all other distributions).

Now, the paramter of a poisson distribution is equal to the mean; so the number of claims in 3 weeks is \(\displaystyle Poi(6)\)

You can find the chance that x=3 using the PF of a poisson distribution
\(\displaystyle P(X=x) = \frac{e^{-6}6^x}{x!}\)

You can use similar arguments for part b.

Note: You may have been taught a different formulation of the poisson distribution where the parameter is (1/mean). In this case, the answer is the same but the method needs adjusting somewhat.
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