Worked solution · Exam P

SOA Exam P Sample Question #623 Solution

This Exam P sample reference tests Normal Distribution. This is a normal-quantile calibration problem. The supplied center and lower quantile imply a standard deviation of about 15, making the target z-score 0.63337 and its cumulative probability 0.73675, so choice B 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.

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Independent solution

How to solve this Normal Distribution question

Setup

A normal distribution is symmetric and unimodal, so its mode identifies its mean. Let sigma denote its unknown standard deviation.

\mu=56

\Phi^{-1}(0.40)=-0.2533471031\ldots

Model

Standardize the known lower quantile to determine the scale.

\frac{52.20-56}{\sigma}=-0.2533471031\ldots

\sigma=14.9991847271\ldots

Compute

Convert the target score to a z-score and evaluate the standard-normal cumulative distribution.

z=\frac{65.50-56}{14.9991847271}=0.6333677578\ldots

\Phi(z)=0.7367532425\ldots

Answer

A cumulative probability of about 0.737 corresponds to the 74th percentile.

\boxed{\text{74th percentile}\quad\text{(B)}}

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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.

AThe 71st percentile corresponds to a z-score near 0.55. It results from inflating the inferred standard deviation instead of using the standard-normal 40th-percentile quantile accurately.
CThe 78th percentile requires a z-score near 0.77, so it reflects a standard deviation that is too small. The known quantile must first be converted with Phi inverse, not read as a raw distance.
DThe 81st percentile similarly overstates the target z-score by compressing the scale too much. Normal percentiles are not linearly spaced in the original score units.
ETreating 0.40 itself as the magnitude of the known z-score gives sigma=9.5 and a target z-score of 1, which points near the 84th or 85th percentile.

Original practice · fully worked

Original variant: paired calibration gap

A calibration facility records independent readings A and B. Reading A is normal with mean 80 and standard deviation 6, while B is normal with mean 72 and standard deviation 8. A gap of 18 is observed for A-B. Approximately what percentile of the distribution of A-B does this gap represent?

A
16th
B
50th
C
76th
D
84th
E
97th

Answer and calculation

The difference A-B is normal with mean 8 and standard deviation 10. The observed gap has z=(18-8)/10=1, whose cumulative probability is 0.84134, so it is approximately the 84th percentile.

Setup

Let D be the difference between the two independent calibration readings.

D=A-B

E[D]=80-72=8

Model

A linear combination of independent normal variables is normal, and the variances add even though one variable is subtracted.

\operatorname{Var}(D)=6^2+8^2=100

D\sim N(8,10^2)

Compute

Standardize the observed gap and evaluate the normal cumulative probability.

z=\frac{18-8}{10}=1

\Pr(D\le18)=\Phi(1)=0.8413447461\ldots

Answer

A cumulative probability near 0.84 corresponds to the 84th percentile.

\boxed{\text{84th percentile}\quad\text{(D)}}

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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: paired calibration gap

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables