Independent solution

How to solve this Normal Distribution question

Setup

Setup

Let X and Y denote the two independent normal contributions and define the combined result S.

S=X+YS=X+Y
E⁡[S]=400+(−100)=300\operatorname{E}[S]=400+(-100)=300

Model

Model

An independent sum of normal variables is normal. Independence also makes the component variances additive.

Var⁡(S)=2002+5002=290000\operatorname{Var}(S)=200^2+500^2=290000
S∼N ⁣(300,290000)S\sim N\!\left(300,290000\right)

Compute

Compute

Standardize zero relative to the aggregate distribution and take the probability above that boundary.

SD⁡(S)=290000=538.5164807135…\operatorname{SD}(S)=\sqrt{290000}=538.5164807135\ldots
Pr⁡(S>0)=Φ ⁣(300538.5164807135)\Pr(S>0)=\Phi\!\left(\frac{300}{538.5164807135}\right)
Pr⁡(S>0)=Φ(0.5570860145…)=0.7112656688…\Pr(S>0)=\Phi(0.5570860145\ldots)=0.7112656688\ldots

Answer

Answer

The requested probability rounds to 0.71.

0.71(E)\boxed{0.71\quad\text{(E)}}