## Inverse Poisson Distribution...

I'm wondering if there is an available command that can evaluate the number of terms required to produce a desired outcome.

Specifically, I am interested in determining the probability of a Poisson distribution, given the parameter (mean) value and the probability outcome. I can obtain the desired result using trial and error / brute force, but I am curious to know if there is a more efficient way.

Suppose that, lambda = 2.6 and the cumulative sum of the probabilities is 95%. I know that I must add the first 6 terms for P(x) in the series (x=0,1, ..,5) to sum to 0.95. Each term ...  P(x=0)= 0.07, P(x=1)=0.19, and so on.

However, how can we know that desired 95% outcome can be determined from the first 5 terms without trial & error?

## Poisson Distribution...

How to do this with Maple? Is it possible?

http://www.mapleprimes.com/ViewTemp.ashx?f=5794_1319135560/poissonHRI-1.jpg

Gracias

## confidence limit of Poisson Distribution...

restart;

P := proc (x, lambda) options operator, arrow; piecewise(x < 0, 0, lambda^x*exp(-lambda)/factorial(x)) end proc;

The upper confidence limit. When the confidence limit is 95%, e is 0.05

Lu := proc (a) options operator, arrow; sum(P(x, lambda), x = 0 .. a) = 0.5e-1 end proc

Lu(10)

exp(-lambda)+lambda*exp(-lambda)+(1/2)*lambda^2*exp(-lambda)+(1/6)*lambda^3*exp(-lambda)+(1/24)*lambda^4*exp(-lambda)+(1/120)*lambda^5*exp(-lambda)+(1/720...

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