Verified solution · Exam P

SOA Exam P Sample Question #523 Solution

This Exam P sample reference tests Normal Distribution. This is a normal-quantile matching calculation with a common center and a 1.5 standard-deviation ratio. Multiplying z₀.₈₁ by 1.5 and returning through the normal CDF gives p=90.605%, so choice D is correct.

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.

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Independent solution

How to solve this Normal Distribution question

Setup

Write the two quantiles as their common mean plus a standard deviation times a standard-normal quantile.

q_I(p)=8000+kz_p

q_{II}(0.81)=8000+1.5kz_{0.81}

Model

Equating the two values cancels the common mean and the positive scale k.

8000+kz_p=8000+1.5kz_{0.81}

z_p=1.5z_{0.81}

Compute

Find the 81st-percentile standard score, scale it, and apply the standard-normal CDF.

z_{0.81}=0.8778962951

z_p=1.5(0.8778962951)=1.3168444426

p=\Phi(1.3168444426)=0.9060546138

Answer

The percentile is approximately 90.6.

\boxed{p\approx 90.6\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.

AA percentile of 83.2 corresponds to z≈0.9621, only about 1.096 times z_0.81. Substitution leaves the first deviation 0.355k below the required 1.3168k.
BThis is near Φ(√(1.5)z_0.81)=0.8585, which results from treating 1.5 as a variance ratio and taking an unnecessary square root.
CThis is tempting after replacing z_0.81 by the percentile decimal 0.8: Φ(1.5 × 0.8)=Φ(1.2)≈0.884, but a probability is not its normal z-score.
EThis reports Φ(1.5)=0.9332 by using the standard-deviation multiplier itself as the final z-score and discarding z_0.81.

Original practice · fully worked

Original variant: percentile rank of a shared performance threshold

Instrument A produces normally distributed scores with mean 50 and standard deviation 6. Instrument B produces normally distributed scores with mean 56 and standard deviation 4. A cutoff is set at the 90th percentile of Instrument A. What proportion of Instrument B scores fall below that same numerical cutoff?

A
0.3364
B
0.5000
C
0.6636
D
0.9000
E
0.9727

Variant answer in brief

The cutoff is 50+6z₀.₉₀=57.6893. Relative to Instrument B this is 0.422327 standard deviations above its mean, giving probability 0.663607 and choice C.

Setup

Convert the stated percentile of Instrument A into a numerical cutoff.

z_{0.90}=1.2815515655

t=50+6z_{0.90}=57.6893093933

Model

Standardize that same cutoff using Instrument B's center and scale.

z_B=\frac{t-56}{4}

Compute

Evaluate the standardized location and its normal cumulative probability.

z_B=\frac{57.6893093933-56}{4}=0.4223273483

\Pr(B\le t)=\Phi(0.4223273483)=0.6636069505

Answer

About 66.36% of Instrument B scores lie below the cutoff.

\boxed{0.6636\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #523 — 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: percentile rank of a shared performance threshold

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution