Independent solution

How to solve this Poisson Distribution question

Answer in brief

The weekly mean gives a daily Poisson mean of 9 and standard deviation 3. A count at least two standard deviations below the mean is at most 3, whose Poisson cumulative probability is 0.02122649, so choice B is correct.

Setup

Setup

Convert the weekly expected count to a daily Poisson mean.

λ=637=9\lambda=\frac{63}{7}=9
DPoisson(9)D\sim\operatorname{Poisson}(9)

Model

Model

A Poisson count has variance equal to its mean, so determine the requested lower cutoff.

SD(D)=9=3\operatorname{SD}(D)=\sqrt9=3
92(3)=39-2(3)=3

Compute

Compute

Sum the Poisson masses through the integer cutoff.

Pr(D3)=e9k=039kk!\Pr(D\le3)=e^{-9}\sum_{k=0}^{3}\frac{9^k}{k!}
Pr(D3)=0.0212264863\Pr(D\le3)=0.0212264863\ldots

Answer

Answer

The probability rounds to 0.021.

0.021(B)\boxed{0.021\quad\text{(B)}}