Independent solution

How to solve this Normal Distribution question

Setup

Setup

Let the independent annual outcomes be X and Y, each with mean mu and standard deviations s1 and s2. Convert the two positive-tail probabilities into standard normal quantiles.

Pr⁡(X>0)=Φ(μs1)=0.8531\Pr(X>0)=\Phi\left(\frac{\mu}{s_1}\right)=0.8531
Pr⁡(Y>0)=Φ(μs2)=0.9192\Pr(Y>0)=\Phi\left(\frac{\mu}{s_2}\right)=0.9192
z1=Φ−1(0.8531)=1.0498219,z2=Φ−1(0.9192)=1.3997106z_1=\Phi^{-1}(0.8531)=1.0498219,\qquad z_2=\Phi^{-1}(0.9192)=1.3997106

Model

Model

Solve the two standard deviations in units of the common mean and aggregate the independent normal outcomes.

s1=μz1,s2=μz2s_1=\frac{\mu}{z_1},\qquad s_2=\frac{\mu}{z_2}
S=X+Y∼N(2μ,μ2(1z12+1z22))S=X+Y\sim N\left(2\mu,\mu^2\left(\frac1{z_1^2}+\frac1{z_2^2}\right)\right)

Compute

Compute

Standardize zero under the distribution of the two-year sum.

Pr⁡(S>0)=Φ(2μμz1−2+z2−2)\Pr(S>0)=\Phi\left(\frac{2\mu}{\mu\sqrt{z_1^{-2}+z_2^{-2}}}\right)
2z1−2+z2−2=1.6796926\frac{2}{\sqrt{z_1^{-2}+z_2^{-2}}}=1.6796926
Pr⁡(S>0)=Φ(1.6796926)=0.9534914310\Pr(S>0)=\Phi(1.6796926)=0.9534914310

Answer

Answer

The two-year positive-total probability rounds to 0.9535.

0.9535(D)\boxed{0.9535\quad\text{(D)}}