Answer to Question #136595 in Statistics and Probability for Anarfi Peter

Question #136595
If X has a Poisson distribution with parameter λ, and if
P(X=0)=0.2.
Evaluate
P(X>2).
1
Expert's answer
2020-10-05T17:41:47-0400

Poisson distribution has following PDF:

P(X=k)=λkk!eλP(X=k) = \frac{\lambda^k}{k!} e^{-\lambda}

We can use given value P(X=0)=0.2P(X=0)=0.2 to find the value of λ\lambda.

P(X=k)=11eλ=0.2P(X=k) = \frac{1}{1} e^{-\lambda} = 0.2

λ=ln0.2=1.609\lambda = - \ln{0.2=1.609}, and, since the coefficient eλe^{-\lambda} is the same for every X=k, we can use directly eλ=0.2e^{-\lambda}=0.2.


P(X>2)=1P(X2)=1P(X=0)P(X=1)P(X=2)P(X>2) = 1 - P(X \leq 2) = 1 - P(X=0) - P(X=1) - P(X=2)

P(X=0)=0.2P(X=0)=0.2

P(X=1)=1.60910.2=0.322P(X=1) = \frac{1.609}{1} 0.2 =0.322

P(X=2)=1.6092210.2=0.259P(X=2) = \frac{1.609^2}{2 \cdot 1} 0.2=0.259

So, P(X>2)=10.20.3220.259=0.219P(X>2) = 1 -0.2-0.322-0.259 = 0.219

Answer: P(X>2)=0.219P(X>2)=0.219


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