Independent solution

How to solve this Poisson Distribution question

Answer in brief

For a Poisson count with mean 3, the cumulative probabilities through 3 and 4 are 0.6472 and 0.8153. Four is the smallest reimbursement limit meeting 75 percent, so choice C.

Setup

Setup

Let X be the Poisson count with rate 3. A limit of m reimburses every item precisely when X does not exceed m.

X∼Poisson⁡(3)X\sim\operatorname{Poisson}(3)
Pr⁡(all reimbursed under limit m)=Pr⁡(X≤m)\Pr(\text{all reimbursed under limit }m)=\Pr(X\le m)

Model

Model

Find the smallest listed limit whose Poisson cumulative probability is at least 0.75.

F(m)=e−3∑k=0m3kk!F(m)=e^{-3}\sum_{k=0}^{m}\frac{3^k}{k!}

Compute

Compute

The cumulative probability remains below the target through three and first exceeds it at four.

F(3)=e−3(1+3+322!+333!)=0.6472318888…F(3)=e^{-3}\left(1+3+\frac{3^2}{2!}+\frac{3^3}{3!}\right)=0.6472318888\ldots
F(4)=F(3)+e−3344!=0.8152632445…F(4)=F(3)+e^{-3}\frac{3^4}{4!}=0.8152632445\ldots

Answer

Answer

The plan with a maximum of four is the first one that meets the probability requirement.

m=4(C)\boxed{m=4\quad\text{(C)}}