Independent solution

How to solve this Normal Distribution question

Setup

Setup

Translate the historical lower-tail probability into its standard-normal quantile.

Φ−1(0.1056)≈−1.25\Phi^{-1}(0.1056)\approx-1.25
400−500σ=−1.25\frac{400-500}{\sigma}=-1.25

Model

Model

Solve for the historical spread and apply the future standard-deviation multiplier.

σ=80\sigma=80
σfuture=1.25(80)=100\sigma_{\mathrm{future}}=1.25(80)=100

Compute

Compute

Both future endpoints are equally distant from the future mean.

370−550100=−1.8,730−550100=1.8\frac{370-550}{100}=-1.8,\qquad \frac{730-550}{100}=1.8
Pr⁡(370<N<730)=Φ(1.8)−Φ(−1.8)=0.928139362…\Pr(370<N<730)=\Phi(1.8)-\Phi(-1.8)=0.928139362\ldots

Answer

Answer

The interval probability rounds to 0.928.

0.928(B)\boxed{0.928\quad\text{(B)}}