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I am studying probability theory and one of the questions that I have faced is this. The problem is that I either don't know where to go about with this question or even if I do do something about it, I have no way of knowing if I'm on the right path. Is it possible to please help explain this?

Q. In Germany, every year 125 people out of 100 000 report the occurrence of Parkinson's Disease. a) Which kind of distribution would you use to model the occurrence of new cases per year?

b) Garching has a population of 17500, compute the probability that no more than 2 cases of the disease are encountered per year.(don’t calculate numeric value, but show the pdf or cdf computation that you do)

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In Germany, every year 125 people out of 100 000 report the occurrence of Parkinson's Disease. a) Which kind of distribution would you use to model the occurrence of new cases per year?

This can be modeled as Poisson distribution with parameter $\lambda$ which is the average number of events per interval. Poisson distribution expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant rate and independently of the time since the last event.

Garching has a population of 17500, compute the probability that no more than 2 cases of the disease are encountered per year.

to answer this one can make a simplifying assumption that the $125$ cases in Germany are uniformly distributed over Germany and the population is also uniformly distributed in Germany. Then you can find out the average number of cases in Garching, which should be $125*(17500/100,000)$. Then you can model it similarly as above and can find out the relevant probabilities.

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This scenario seems best to follow a Poisson Distribution. The Poisson distribution is popular for modelling the number of times an event occurs in an interval of time or space.

a) A $poisson$ distribution with $\lambda$= $125$ would best model the phenomenon. If $k$ is the number of cases reported in a year,

$P(k ) = \frac{\lambda^k e^{-\lambda}}{k!}$

b) Let $k$ be the number of cases reported in Garching in a year.

We have $\lambda= \frac{(125) (17500)}{100000}$ and

$P(k \leq 2) = P(k=0) + P(k=1) + P(k=2)$.

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  • $\begingroup$ Could you tell me how you came up with the lambda value? shouldn't it be (125)(17500)/100000 ?? $\endgroup$ – Aurangzeb Rathore Feb 26 at 5:02
  • $\begingroup$ Yes. Edited it. Was a typo. This is the solution to your problem. $\endgroup$ – rgk Feb 26 at 16:03
  • $\begingroup$ The lambda that you calculate is not consistent throughout. $\lambda$ for Garching is $17500$ times more than that for Germany according to your calculation which is awkward. $\lambda$ for garching is correctly calculated while for Germany it is not. Please note: parameter $\lambda$ is the average number of events per interval. $\endgroup$ – naive Feb 26 at 17:26
  • $\begingroup$ Edited it, now. $\endgroup$ – rgk Feb 26 at 17:46

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