Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Let X_i denote the count on day i and let N be the total across the seven-day horizon.

Xi∼Poisson⁡(0.20)X_i\sim\operatorname{Poisson}(0.20)
N=∑i=17XiN=\sum_{i=1}^{7}X_i

Model

Model

Independent Poisson variables add to another Poisson variable whose mean is the sum of the component means.

N∼Poisson⁡(7⋅0.20)=Poisson⁡(1.40)N\sim\operatorname{Poisson}(7\cdot0.20)=\operatorname{Poisson}(1.40)
Pr⁡(N=k)=e−1.401.40kk!\Pr(N=k)=e^{-1.40}\frac{1.40^k}{k!}

Compute

Compute

Evaluate the aggregate Poisson mass at the requested count.

Pr⁡(N=3)=e−1.401.4033!\Pr(N=3)=e^{-1.40}\frac{1.40^3}{3!}
Pr⁡(N=3)=0.1127770115…\Pr(N=3)=0.1127770115\ldots

Answer

Answer

The requested probability rounds to 0.113.

0.113(C)\boxed{0.113\quad\text{(C)}}