Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Write the total charge incurred when the count equals n.

T(n)=1+2++n=n(n+1)2T(n)=1+2+\cdots+n=\frac{n(n+1)}2
T=N(N+1)2T=\frac{N(N+1)}2

Model

Model

Use the raw second moment of a Poisson variable with mean four.

E[N]=4,Var(N)=4E[N]=4,\qquad \operatorname{Var}(N)=4
E[N2]=Var(N)+E[N]2=4+16=20E[N^2]=\operatorname{Var}(N)+E[N]^2=4+16=20

Compute

Compute

Apply linearity of expectation to the quadratic charge.

E[T]=12(E[N2]+E[N])E[T]=\frac12\left(E[N^2]+E[N]\right)
E[T]=12(20+4)=12E[T]=\frac12(20+4)=12

Answer

Answer

The expected total charge is 12.

12(D)\boxed{12\quad\text{(D)}}