Independent solution

How to solve this Normal Distribution question

Setup

Setup

Let the positive mean, measured in billions, be mu and translate the stated variance into a standard deviation.

E⁡[X]=μ,Var⁡(X)=μ3\operatorname{E}[X]=\mu,\qquad \operatorname{Var}(X)=\mu^3
σ=μ3/2\sigma=\mu^{3/2}

Model

Model

Standardize the zero-profit threshold using the 5th percentile of a standard normal variable.

Pr⁡(X<0)=0.05\Pr(X<0)=0.05
0−μμ3/2=z0.05=−1.6448536\frac{0-\mu}{\mu^{3/2}}=z_{0.05}=-1.6448536

Compute

Compute

Simplify the power of mu and solve for the mean.

μ−1/2=1.6448536\mu^{-1/2}=1.6448536
μ=11.64485362=0.3696115 billion\mu=\frac{1}{1.6448536^2}=0.3696115\ \text{billion}

Answer

Answer

The expected profit is approximately 370 million.

370 million(A)\boxed{370\ \text{million}\quad\text{(A)}}