Worked solution · Exam P

SOA Exam P Sample Question #164 Solution

This Exam P sample reference tests Normal Distribution. A variance of 4 gives standard deviation 2. The 12th-percentile standard-normal score is about -1.175, so transforming back gives 7.650, choice B.

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.

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Independent solution

How to solve this Normal Distribution question

Setup

Convert the stated variance to a standard deviation before standardizing.

\mu=10,\qquad \sigma=\sqrt4=2

Model

Let q be the desired percentile and match its standardized value to the lower-tail normal quantile.

\Pr(X\le q)=0.12

\frac{q-10}{2}=\Phi^{-1}(0.12)

Compute

Evaluate the standard-normal quantile and transform it to the original units.

\Phi^{-1}(0.12)=-1.1749868

q=10+2(-1.1749868)=7.6500264

Answer

The 12th percentile is approximately 7.65 years.

\boxed{7.65\quad\text{(B)}}

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.

See the cover and sample pages

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 5.30 uses 4 as the standard deviation: 10+4(-1.175)=5.30. The given 4 is a variance, so the standard deviation is 2.
CThe value 8.41 corresponds to a standardized score near -0.795, not the 12th-percentile score -1.175.
DThe value 12.35 uses the correct magnitude but the wrong sign, placing a lower-tail percentile above the mean.
EThe value 14.70 combines both errors: it treats the variance as a standard deviation and reverses the lower-tail sign.

Original practice · fully worked

Original variant: recover a normal scale from symmetric percentiles

A calibrated sensor reading is normally distributed. Its 20th percentile is 25.792 and its 80th percentile is 34.208. Using these values, calculate the probability that a reading exceeds 38.

A
0.0228
B
0.0548
C
0.1000
D
0.9452
E
0.9772

Answer and calculation

Symmetry gives mean 30, while the 80th-percentile offset identifies a standard deviation essentially equal to 5. The threshold 38 has z-score 1.6, whose upper tail is 0.0548, choice B.

Setup

Use symmetry of the two supplied percentiles to identify the mean.

\mu=\frac{25.792+34.208}{2}=30

Model

The 80th-percentile standard-normal score calibrates the scale.

z_{0.80}=0.8416212

\sigma=\frac{34.208-30}{0.8416212}=4.9998739

Compute

Standardize 38 and evaluate the corresponding upper tail.

z=\frac{38-30}{4.9998739}=1.6000404

\Pr(X>38)=1-\Phi(1.6000404)=0.0547948

Answer

The exceedance probability is approximately 0.0548.

\boxed{0.0548\quad\text{(B)}}

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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: recover a normal scale from symmetric percentiles

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables