Independent solution

How to solve this Poisson Distribution question

Answer in brief

The zero-count probability is 0.632, so the Poisson mean is minus the natural logarithm of 0.632. Using the Poisson identity for the second moment gives 0.6694, which selects choice D.

Setup

Setup

Let the count have Poisson mean lambda. 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 lambda, 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)}}