Verified 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 August 4, 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)}}

Continue without hunting through PDFs

All 718 Exam P sample solutions in syllabus order

The searchable 3108-page manual includes this complete solution, its error analysis, and one original worked variant for every active reference.

See inside the Exam P study manual PDF

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

Variant answer in brief

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

Turn this solution into a complete study sequence

The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.

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

Cite this solution

ActuaryProof. “SOA Exam P Sample Question #401 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Normal Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: reflect lognormal percentiles across the median

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution