Independent solution

How to solve this Poisson Distribution question

Answer in brief

A one-year zero-count probability of 0.90 implies a Poisson mean of -ln(0.90). Adding 15 independent yearly counts gives a Poisson total whose variance is 15[-ln(0.90)]=1.5804, so choice E is correct.

Setup

Setup

Let lambda be the mean count for one year and convert the positive-count probability to a zero-count probability.

Pr⁡(N=0)=1−0.10=0.90\Pr(N=0)=1-0.10=0.90

Model

Model

For a Poisson count, the zero probability determines its parameter. Independent Poisson counts add by summing their parameters.

e−λ=0.90e^{-\lambda}=0.90
λ=−ln⁡(0.90)\lambda=-\ln(0.90)
S=∑j=115Nj∼Poisson⁡(15λ)S=\sum_{j=1}^{15}N_j\sim\operatorname{Poisson}(15\lambda)

Compute

Compute

Use equality of the Poisson mean and variance for the aggregated count.

Var⁡(S)=15λ\operatorname{Var}(S)=15\lambda
Var⁡(S)=−15ln⁡(0.90)=1.5804077349…\operatorname{Var}(S)=-15\ln(0.90)=1.5804077349\ldots

Answer

Answer

The variance of the 15-year total rounds to 1.58.

1.58(E)\boxed{1.58\quad\text{(E)}}