Independent solution

How to solve this Poisson Moments question

Setup

Setup

Let the two Poisson means differ by eight as stated. For a Poisson variable, the second raw moment is the sum of its mean and the square of its mean.

a=b−8,E[X2]=a+a2,E[Y2]=b+b2a=b-8,\quad E[X^2]=a+a^2,\quad E[Y^2]=b+b^2

Model

Model

Substituting the difference relation into the relationship between second moments produces two algebraic candidates, 4 and 35.

b−8+(b−8)2=0.6(b+b2)b-8+(b-8)^2=0.6(b+b^2)

Compute

Compute

The candidate 4 would force the other Poisson mean to be negative. Therefore the admissible mean is 35, which is also the requested Poisson variance.

0.4b2−15.6b+56=0,b∈{4,35},b=350.4b^2-15.6b+56=0,\quad b\in\{4,35\},\quad b=35

Answer

Answer

The Poisson variance of the afternoon count equals its mean, 35.

35(E)\boxed{35\quad\text{(E)}}