Independent solution

How to solve this Poisson Distribution question

Answer in brief

The aggregate count has mean and variance 5000. Standardizing 5100 under its normal approximation gives z=sqrt(2), whose upper-tail probability is 0.078650, so choice A is correct.

Setup

Setup

Let T denote the aggregate count. Independent Poisson counts add to another Poisson count, with parameter equal to the sum of the component parameters.

T∼Poisson⁡(100⋅50)=Poisson⁡(5000)T\sim\operatorname{Poisson}(100\cdot 50)=\operatorname{Poisson}(5000)
E⁡[T]=Var⁡(T)=5000\operatorname{E}[T]=\operatorname{Var}(T)=5000

Model

Model

Because the aggregate mean is large, approximate T by a normal random variable with the same mean and variance.

T ∼˙ N(5000,5000)T\ \dot\sim\ N(5000,5000)
Z=T−50005000Z=\frac{T-5000}{\sqrt{5000}}

Compute

Compute

Standardize the stated threshold and evaluate the standard normal upper tail.

z=5100−50005000=2=1.4142136…z=\frac{5100-5000}{\sqrt{5000}}=\sqrt{2}=1.4142136\ldots
Pr⁡(T>5100)≈1−Φ(2)=0.0786496…\Pr(T>5100)\approx 1-\Phi(\sqrt{2})=0.0786496\ldots

Answer

Answer

The normal approximation gives an upper-tail probability of about 0.08.

Pr⁡(T>5100)≈0.08(A)\boxed{\Pr(T>5100)\approx 0.08\quad\text{(A)}}