Independent solution

How to solve this Poisson Distribution question

Answer in brief

Independence lets the monthly Poisson variances add across two locations and three months. The total variance is three times the combined monthly mean, or 135, so the answer is B.

Setup

Setup

Represent each location-month count separately. Every component is Poisson, so its variance equals its mean.

Var⁡(Xm)=15,Var⁡(Ym)=30\operatorname{Var}(X_m)=15,\qquad \operatorname{Var}(Y_m)=30
m=1,2,3m=1,2,3

Model

Model

The six counts are mutually independent, so the variance of their sum is the sum of their variances.

T=∑m=13(Xm+Ym)T=\sum_{m=1}^{3}(X_m+Y_m)
Var⁡(T)=∑m=13(Var⁡(Xm)+Var⁡(Ym))\operatorname{Var}(T)=\sum_{m=1}^{3}\left(\operatorname{Var}(X_m)+\operatorname{Var}(Y_m)\right)

Compute

Compute

Each month contributes variance 45, and there are three independent months.

Var⁡(T)=3(15+30)=3(45)=135\operatorname{Var}(T)=3(15+30)=3(45)=135

Answer

Answer

The variance of the full three-month total is 135.

Var⁡(T)=135(B)\boxed{\operatorname{Var}(T)=135\quad\text{(B)}}