Worked solution · Exam P

SOA Exam P Sample Question #556 Solution

This Exam P sample reference tests Normal Distribution. The two supplied normal percentiles determine mean 569.34 and standard deviation 1699.89. Applying the 95th-percentile z-score gives 3365.40, which rounds to 3400 and choice D.

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

Let mu and sigma be the normal mean and standard deviation, and record the relevant standard-normal quantiles.

z_{0.60}=0.2533471031,\qquad z_{0.80}=0.8416212336

1000=\mu+z_{0.60}\sigma,\qquad 2000=\mu+z_{0.80}\sigma

Model

Subtract the percentile equations to identify the standard deviation, then recover the mean.

\sigma=\frac{2000-1000}{z_{0.80}-z_{0.60}}

\mu=1000-z_{0.60}\sigma

Compute

Use the 95th-percentile standard-normal value on the calibrated distribution.

\sigma=1699.887770,\qquad \mu=569.338358

q_{0.95}=\mu+1.644853627\sigma=3365.404922

Answer

Rounding the percentile to the nearest hundred gives 3400.

\boxed{3400\quad\text{(D)}}

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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 value 2300 comes from increasing the 80th-percentile amount by 15%, namely 2000(1.15). A 15-percentage-point change is not a 15% change in the quantile value.
BThe value 2400 comes from 2000(0.95/0.80)=2375, treating percentile values as proportional to cumulative probabilities rather than to normal z-scores.
CA linear extrapolation in percentile rank gives 2000+(0.95-0.80)(1000/0.20)=2750, near this choice after downward rounding; normal quantiles are not linear in percentile rank.
EAssuming the mean is zero and calibrating only from the 80th percentile gives sigma=2000/0.84162 and q_0.95=3908.75, about 3900; the 60th percentile rules out a zero mean.

Original practice · fully worked

Original variant: tail probability from symmetric benchmarks

A laboratory score is normally distributed. Its 10th percentile is 42 and its 90th percentile is 58. Calculate the probability that a score exceeds 62.

A
0.0374
B
0.0273
C
0.0668
D
0.1000
E
0.9727

Answer and calculation

Symmetry places the mean at 50, and eight score points equal z_0.90 standard deviations. A score of 62 has z=1.5z_0.90=1.9223, whose upper tail is 0.0273, so choice B is correct.

Setup

Symmetric normal percentiles lie equally far from the mean.

\mu=\frac{42+58}{2}=50

58-50=8=z_{0.90}\sigma

Model

Express the new threshold's standardized distance using the calibrated eight-point benchmark.

z_{62}=\frac{62-50}{\sigma}=\frac{12}{8}z_{0.90}

Compute

Insert the standard-normal 90th percentile and evaluate the upper tail.

z_{62}=1.5(1.281551566)=1.922327348

\Pr(X>62)=1-\Phi(1.922327348)=0.0272822904

Answer

The probability above 62 is approximately 0.0273.

\boxed{0.0273\quad\text{(B)}}

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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: tail probability from symmetric benchmarks

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables