Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Let the count have Poisson mean λ. Convert the given positive-count probability into its zero-count complement.

X∼Poisson⁡(λ)X\sim\operatorname{Poisson}(\lambda)
Pr⁡(X=0)=1−0.368=0.632\Pr(X=0)=1-0.368=0.632

Model

Model

For a Poisson variable, the zero-count probability identifies λ, and the second raw moment is the sum of the variance and the squared mean.

e−λ=0.632e^{-\lambda}=0.632
E⁡[X2]=Var⁡(X)+E⁡[X]2=λ+λ2\operatorname{E}[X^2]=\operatorname{Var}(X)+\operatorname{E}[X]^2=\lambda+\lambda^2

Compute

Compute

Solve for the rate and substitute it into the second-moment identity.

λ=−ln⁡(0.632)=0.4588658848…\lambda=-\ln(0.632)=0.4588658848\ldots
E⁡[X2]=0.4588658848…+(0.4588658848…)2\operatorname{E}[X^2]=0.4588658848\ldots+(0.4588658848\ldots)^2
E⁡[X2]=0.6694237851…\operatorname{E}[X^2]=0.6694237851\ldots

Answer

Answer

The second moment rounds to 0.669.

E⁡[X2]≈0.669(D)\boxed{\operatorname{E}[X^2]\approx0.669\quad\text{(D)}}