Independent solution

How to solve this Normal Distribution question

Answer in brief

The probability condition gives a mean-to-standard-deviation ratio of Phi inverse 0.92, or 1.405072. Combining this ratio with E[X^2]=mu^2+sigma^2=74 yields variance 24.8804, so choice A is correct.

Setup

Setup

Write the normal variable in terms of its mean and standard deviation and translate the positive-outcome probability into a standard-normal quantile.

X∼N(μ,σ2)X\sim N(\mu,\sigma^2)
Pr⁡(X>0)=Φ ⁣(μσ)=0.92\Pr(X>0)=\Phi\!\left(\frac{\mu}{\sigma}\right)=0.92

Model

Model

The quantile fixes the ratio of the mean to the standard deviation, while the supplied second raw moment fixes their squared sum.

μσ=Φ−1(0.92)=1.4050715603…\frac{\mu}{\sigma}=\Phi^{-1}(0.92)=1.4050715603\ldots
E[X2]=μ2+σ2=74E[X^2]=\mu^2+\sigma^2=74

Compute

Compute

Substitute the quantile ratio into the second-moment equation and solve for the variance.

(1.40507156032+1)σ2=74\left(1.4050715603^2+1\right)\sigma^2=74
σ2=24.8804219219…\sigma^2=24.8804219219\ldots

Answer

Answer

The variance rounds to 24.88.

24.88(A)\boxed{24.88\quad\text{(A)}}