Independent solution

How to solve this Central Limit Theorem question

Setup

Setup

Let X_i be the count on day i and let T be the aggregate across the full observation window.

T=∑i=1303XiT=\sum_{i=1}^{303}X_i
E⁡[T]=303(3300)=999900\operatorname{E}[T]=303(3300)=999900

Model

Model

Independence makes the variances additive. With many observations, the central limit theorem gives a normal approximation for T.

Var⁡(T)=303(1302)\operatorname{Var}(T)=303(130^2)
T ∼˙ N ⁣(999900, 303(1302))T\ \dot\sim\ N\!\left(999900,\,303(130^2)\right)

Compute

Compute

Standardize the specified aggregate cutoff and evaluate the upper standard-normal tail.

z=1000000−999900130303=0.0441911530…z=\frac{1000000-999900}{130\sqrt{303}}=0.0441911530\ldots
Pr⁡(T>1000000)≈1−Φ(0.0441911530)=0.4823760170…\Pr(T>1000000)\approx 1-\Phi(0.0441911530)=0.4823760170\ldots

Answer

Answer

The approximate probability rounds to 0.48.

0.48(B)\boxed{0.48\quad\text{(B)}}