Independent solution

How to solve this Normal Distribution question

Setup

Setup

Let one period's result have mean μ and standard deviation σ. Convert its positive-probability statement to a standard-normal quantile.

Pr⁡(X>0)=Φ(μσ)=0.80\Pr(X>0)=\Phi\left(\frac{\mu}{\sigma}\right)=0.80
μσ=Φ−1(0.80)=0.841621234…\frac{\mu}{\sigma}=\Phi^{-1}(0.80)=0.841621234\ldots

Model

Model

The sum of four independent normal variables is normal; its mean is multiplied by four and its standard deviation by two.

S=∑i=14XiS=\sum_{i=1}^{4}X_i
E[S]=4μ,SD⁡(S)=2σE[S]=4\mu,\qquad \operatorname{SD}(S)=2\sigma

Compute

Compute

Evaluate the positive-tail probability using the standardized annual mean.

Pr⁡(S>0)=Φ(4μ2σ)\Pr(S>0)=\Phi\left(\frac{4\mu}{2\sigma}\right)
Pr⁡(S>0)=Φ(2×0.841621234)=0.953835919…\Pr(S>0)=\Phi(2\times0.841621234)=0.953835919\ldots

Answer

Answer

The annual probability rounds to 0.954.

0.954(D)\boxed{0.954\quad\text{(D)}}