SOA Exam P Sample Question #591. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #591 — Verified Solution

Independent solution

How to solve this Normal Distribution question

Answer in brief

This is a symmetric normal-quantile calculation. A central probability of 0.25 leaves cumulative probability 0.625 at the upper endpoint, whose z-score is 0.31864; multiplying by the standard deviation 4 gives a half-width of 1.2746 and the interval (18.7,21.3), so choice E is correct.

Setup

The variance is 16, so the standard deviation is 4. Write the centered interval as the mean plus or minus a multiple of this standard deviation.

\sigma=\sqrt{16}=4

I=(20-4z,\,20+4z)

Model

Symmetry splits the remaining probability equally between the two tails. Therefore the upper standardized endpoint has cumulative probability 0.5+0.25/2.

\Pr(-z<Z<z)=0.25

\Phi(z)=0.625

Compute

Find the standard-normal quantile, rescale it, and add and subtract the resulting half-width from the mean.

z=\Phi^{-1}(0.625)=0.318639364\ldots

4z=1.274557456\ldots

I=(18.7254425\ldots,\,21.2745575\ldots)

Answer

The listed endpoints that match this centered interval are 18.7 and 21.3.

\boxed{(18.7,\,21.3)\quad\text{(E)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis uses roughly two standard deviations on each side, producing a central probability near 95% instead of 25%.
BThis uses about 1.28 standard deviations on each side, which corresponds to a much larger central probability near 80%.
CThis uses one standard deviation on each side and therefore invokes the familiar central probability near 68%.
DThis uses about 0.67 standard deviations on each side, corresponding to a central probability near 50% rather than 25%.

Original practice · fully worked

Original variant: paired sensor threshold

Independent readings X and Y are normally distributed with means 50 and 44 and variances 9 and 16, respectively. Determine c such that the probability that X-Y exceeds c is 0.10.

A
-0.4078
B
6.0000
C
9.8447
D
11.1262
E
12.4078

Variant answer in brief

The difference X-Y is normal with mean 6 and standard deviation 5. A 10% upper tail uses the 90th-percentile z-score 1.28155, giving c=6+5(1.28155)=12.4078 and choice E.

Setup

Form the distribution of the difference. Independence makes the two component variances add.

D=X-Y

E[D]=50-44=6

\operatorname{Var}(D)=9+16=25

Model

An upper-tail probability of 0.10 places the threshold at the 90th percentile of the difference distribution.

\Pr(D>c)=0.10

\Pr\!\left(Z\le \frac{c-6}{5}\right)=0.90

Compute

Use the standard-normal 90th percentile and transform back to the original scale.

\Phi^{-1}(0.90)=1.281551566\ldots

c=6+5(1.281551566\ldots)=12.40775783\ldots

Answer

The required threshold is approximately 12.4078.

\boxed{c=12.4078\quad\text{(E)}}

Continue by topic

Build a complete proof notebook

The Probability Proof Manual reorganizes verified solutions by syllabus skill and adds formula proofs, error patterns, and original practice.

Explore study products Get the free formula sheet

Cite this solution

ActuaryProof. “SOA Exam P Sample Question #591 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.