Worked 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 July 31, 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

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)}}

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.

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

Answer and calculation

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)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

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

How to solve this Normal Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: percentile rank of a shared performance threshold

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables