Worked solution · Exam P

SOA Exam P Sample Question #401 Solution

This Exam P sample reference tests Normal Distribution. This problem uses the fact that a normal median equals its mean. The 90th percentile lies 1.28155 standard deviations above the median, so the difference is 500(1.28155)=640.776 and choice C.

Independent analysis by ActuaryProof · reviewed July 31, 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

Recover the standard deviation from the stated variance.

\sigma=\sqrt{250000}=500

Model

For a normal variable, the median is the location parameter. Express the 90th percentile using the standard-normal quantile.

q_{0.50}=\mu

q_{0.90}=\mu+\sigma z_{0.90}

z_{0.90}=\Phi^{-1}(0.90)=1.2815515655

Compute

Subtracting the median cancels the unknown location.

q_{0.90}-q_{0.50}=500(1.2815515655)

=640.7757828

Answer

The requested difference rounds to 641.

\boxed{641\quad\text{(C)}}

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.

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Error diagnosis

Why the other letters tempt

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

AA difference of 241 corresponds to a standardized displacement 241/500=0.482, whose cumulative probability is only about 0.685. It results from an incorrect standard-normal lookup for the 90th percentile.
BThe calculation 500(1.2816-0.40)=440.8 subtracts the magnitude of the supplied observation's z-score. The median has z-score zero, so that location information should not enter the percentile-minus-median gap.
DThe value 500(1.64485)=822.4 uses the 95th-percentile standard-normal quantile instead of the 90th-percentile quantile.
EThe value 500(1.96)=980 uses the 97.5th-percentile critical value associated with a central 95% interval, not the 90th percentile.

Original practice · fully worked

Original variant: reflect lognormal percentiles across the median

A service duration is lognormally distributed. Its median is 20 minutes and its 90th percentile is 50 minutes. Calculate its 10th percentile, in minutes.

A
0.4
B
2.5
C
8.0
D
20.0
E
30.0

Answer and calculation

Normal symmetry applies after taking logarithms. Complementary lognormal percentiles satisfy q_0.10 q_0.90=median², so q_0.10=20²⁄⁵⁰=8 minutes and choice C.

Setup

Let Y be the logarithm of the duration. Its normal mean is the logarithm of the lognormal median.

Y=\ln X\sim N(\mu,\sigma^2)

\mu=\ln20

Model

The 10th and 90th standard-normal quantiles have equal magnitude and opposite signs.

\ln q_{0.90}=\mu+z_{0.90}\sigma

\ln q_{0.10}=\mu-z_{0.90}\sigma

Compute

Add the two log-quantile equations and solve for the lower percentile.

\ln q_{0.10}+\ln q_{0.90}=2\mu=2\ln20

q_{0.10}=\frac{20^2}{50}=8

Answer

The 10th percentile is 8 minutes.

\boxed{8.0\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Normal Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: reflect lognormal percentiles across the median

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables