Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Write the cumulative probabilities through counts one and two.

F(1)=e−λ(1+λ)F(1)=e^{-\lambda}(1+\lambda)
F(2)=e−λ(1+λ+λ22)F(2)=e^{-\lambda}\left(1+\lambda+\frac{\lambda^2}{2}\right)

Model

Model

Insert the reported ratio and cancel the common factor.

1+λ+λ2/21+λ=2.6\frac{1+\lambda+\lambda^2/2}{1+\lambda}=2.6

Compute

Compute

Clear denominators and solve the resulting quadratic, rejecting the negative root.

5λ2−16λ−16=05\lambda^2-16\lambda-16=0
(5λ+4)(λ−4)=0,λ=4(5\lambda+4)(\lambda-4)=0,\qquad \lambda=4

Answer

Answer

The Poisson expected value is 4.0.

4.0(B)\boxed{4.0\quad\text{(B)}}