Independent solution

How to solve this Normal Distribution question

Setup

Setup

Let the first normal profit have mean μ and standard deviation σ. Translate its positive probability to a z-score.

Pr⁡(X>0)=Φ(μσ)=0.67\Pr(X>0)=\Phi\left(\frac{\mu}{\sigma}\right)=0.67
μσ=Φ−1(0.67)=0.439913166…\frac{\mu}{\sigma}=\Phi^{-1}(0.67)=0.439913166\ldots

Model

Model

The second profit has mean 3μ and standard deviation 3σ. Independence makes the variances add.

E[X+Y]=4μE[X+Y]=4\mu
Var⁡(X+Y)=σ2+(3σ)2=10σ2\operatorname{Var}(X+Y)=\sigma^2+(3\sigma)^2=10\sigma^2

Compute

Compute

Standardize zero under the normal distribution of the total.

Pr⁡(X+Y>0)=Φ(4μ10σ)\Pr(X+Y>0)=\Phi\left(\frac{4\mu}{\sqrt{10}\sigma}\right)
=Φ(4(0.439913166)10)=0.711048719…=\Phi\left(\frac{4(0.439913166)}{\sqrt{10}}\right)=0.711048719\ldots

Answer

Answer

The total-positive probability rounds to 0.71.

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