SOA Exam P Sample Question #593. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #593 — Verified Solution

Independent solution

How to solve this Normal Distribution question

Answer in brief

The 28th-percentile z-score is -0.582842, so the 18-unit gap below the mean determines standard deviation 30.8832. A value of 3 is 1.165683 standard deviations below the mean, with cumulative probability 0.121871, so choice E is correct.

Setup

Translate the known percentile into a standard-normal equation.

\frac{21-39}{\sigma}=\Phi^{-1}(0.28)=-0.5828415073\ldots

Model

Solve for the normal standard deviation.

\sigma=\frac{-18}{-0.5828415073}=30.8831814060\ldots

Compute

Standardize the target value and evaluate its cumulative probability.

z=\frac{3-39}{30.8831814060}=-1.1656830145\ldots

\Pr(X\le3)=\Phi(z)=0.1218713150\ldots

Answer

A cumulative probability near 12.2% corresponds to the listed 12th percentile.

\boxed{\text{12th percentile}\quad\text{(E)}}

Error diagnosis

Why the other letters tempt

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

AThis substitutes the 20th-percentile z-score, about -0.842, for the stated 28th percentile; doubling it gives about -1.68 and a lower tail near 4.6%.
BThis treats probability as linear in distance from the mean: 0.50-2(0.50-0.28)=0.06, even though normal CDF probabilities do not scale linearly.
CThis squares the known percentile probability, 0.28^2=0.0784, and rounds it to the 8th percentile instead of doubling the standardized distance.
DThis subtracts the 18-unit drop from 21 to 3 directly from the percentile number, 28-18=10, incorrectly treating profit units as percentile points.

Original practice · fully worked

Original variant: spread from paired quantiles

A normally distributed sensor reading has 10th percentile 40 and 90th percentile 60. Calculate the variance of the reading.

A
7.80
B
10.00
C
60.89
D
100.00
E
400.00

Variant answer in brief

Normal symmetry places the mean at 50. The half-width 10 equals Phi inverse 0.90 times sigma, giving sigma=7.80304 and variance 60.88746, so choice C is correct.

Setup

Use symmetry of the paired normal percentiles to find the center.

\mu=\frac{40+60}{2}=50

Model

Relate the upper percentile's distance from the mean to its standard-normal quantile.

\frac{60-50}{\sigma}=\Phi^{-1}(0.90)=1.2815515655\ldots

Compute

Solve for the standard deviation and square it.

\sigma=\frac{10}{1.2815515655}=7.8030414607\ldots

\sigma^2=60.8874560378\ldots

Answer

The variance rounds to 60.89.

\boxed{60.89\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #593 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.