Independent solution

How to solve this Poisson Distribution question

Answer in brief

Adding the 25 independent Poisson counts gives a manuscript total with mean 5. Summing its masses from zero through three produces 0.265026, which rounds to choice E.

Setup

Setup

Let X_j be the count on item j and let T be the count across all 25 items.

Xj∼Poisson⁡(0.20),j=1,…,25X_j\sim\operatorname{Poisson}(0.20),\qquad j=1,\ldots,25
T=∑j=125XjT=\sum_{j=1}^{25}X_j

Model

Model

A sum of independent Poisson variables is Poisson with mean equal to the sum of the component means.

T∼Poisson⁡(25×0.20)=Poisson⁡(5)T\sim\operatorname{Poisson}(25\times 0.20)=\operatorname{Poisson}(5)

Compute

Compute

Three or fewer includes each total from zero through three, so add all four corresponding masses.

Pr⁡(T≤3)=e−5∑k=035kk!\Pr(T\le 3)=e^{-5}\sum_{k=0}^{3}\frac{5^k}{k!}
Pr⁡(T≤3)=e−5(1+5+252+1256)=0.2650259153…\Pr(T\le 3)=e^{-5}\left(1+5+\frac{25}{2}+\frac{125}{6}\right)=0.2650259153\ldots

Answer

Answer

The cumulative probability rounds to 0.265.

Pr⁡(T≤3)≈0.265(E)\boxed{\Pr(T\le 3)\approx 0.265\quad\text{(E)}}