Worked solution · Exam P

SOA Exam P Sample Question #598 Solution

This Exam P sample reference tests Normal Distribution. The probability condition gives a mean-to-standard-deviation ratio of Phi inverse 0.92, or 1.405072. Combining this ratio with E[X²]=mu²+sigma²=74 yields variance 24.8804, so choice A is correct.

Independent analysis by ActuaryProof · reviewed July 30, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

Independent solution

How to solve this Normal Distribution question

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\sim N(\mu,\sigma^2)

\Pr(X>0)=\Phi\!\left(\frac{\mu}{\sigma}\right)=0.92

Model

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

\frac{\mu}{\sigma}=\Phi^{-1}(0.92)=1.4050715603\ldots

E[X^2]=\mu^2+\sigma^2=74

Compute

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

\left(1.4050715603^2+1\right)\sigma^2=74

\sigma^2=24.8804219219\ldots

Answer

The variance rounds to 24.88.

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

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

See the cover and sample pages

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

BThis uses mu squared equal to z times sigma squared instead of z squared times sigma squared, producing 74/(1+1.40507)=30.77.
CThis writes the second-moment identity as mu squared plus sigma rather than mu squared plus sigma squared, leading to a reported variance near 34.50.
DThis corresponds to using an incorrect standardized distance of about 0.82 in place of the 92nd-percentile value 1.40507.
EThis writes the second moment as mu plus sigma squared, omitting the square on the mean and producing variance about 62.86.

Original practice · fully worked

Original variant: centered gauge readings

A precision gauge reading X is normally distributed with mean 50. The probability that X lies between 44 and 56 is 0.6826894921. Calculate the second raw moment E[X²].

A
36
B
2500
C
2506
D
2536
E
3136

Answer and calculation

The interval is centered at the mean and contains the standard-normal probability between -1 and 1, so the gauge standard deviation is 6. The second raw moment is 50 squared plus 6 squared, or 2536, which is choice D.

Setup

Standardize the two endpoints around the known mean.

\Pr(44<X<56)=\Pr\!\left(-\frac6\sigma<Z<\frac6\sigma\right)

Model

The stated central probability equals the standard-normal probability between minus one and one.

\Phi(1)-\Phi(-1)=0.6826894921\ldots

\frac6\sigma=1

Compute

Recover the standard deviation and apply the second-moment identity.

\sigma=6

E[X^2]=\operatorname{Var}(X)+E[X]^2=6^2+50^2=2536

Answer

The second raw moment is 2536.

\boxed{2536\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Normal Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: centered gauge readings

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables