# Question:confidence limit of Poisson Distribution

## Question:confidence limit of Poisson Distribution

Maple

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)*lambda^6*exp(-lambda)+(1/5040)*lambda^7*exp(-lambda)+(1/40320)*lambda^8*exp(-lambda)+(1/362880)*lambda^9*exp(-lambda)+(1/3628800)*lambda^10*exp(-lambda) = 0.5e-1

`assuming`([solve(Lu(10), lambda, useassumptions)], [lambda > 0])

No solition. ?

The lower confidence limit

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

Ll(10)

(1/362880)*(362880*exp(lambda)-362880-362880*lambda-181440*lambda^2-60480*lambda^3-15120*lambda^4-3024*lambda^5-504*lambda^6-72*lambda^7-9*lambda^8-lambda^9)/exp(lambda) = 0.5e-1

RealDomain:-solve((1/362880)*(362880*RealDomain:-exp(lambda)-362880-362880*lambda-181440*RealDomain:-`^`(lambda, 2)-60480*RealDomain:-`^`(lambda, 3)-15120*RealDomain:-`^`(lambda, 4)-3024*RealDomain:-`^`(lambda, 5)-504*RealDomain:-`^`(lambda, 6)-72*RealDomain:-`^`(lambda, 7)-9*RealDomain:-`^`(lambda, 8)-RealDomain:-`^`(lambda, 9))/RealDomain:-exp(lambda) = 0.5e-1, lambda);

-2.635610964
Lambda should be positive.

Can be solved using the package Statistcs?

Gracias

Poisson

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