Verified solution · Exam P

SOA Exam P Sample Question #521 Solution

This Exam P sample reference tests Normal Distribution. This is a probability for the sum of independent normal variables. The sum has mean 35 and variance 41, so standardizing 45 gives probability 0.940825 and 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

Let S be the combined annual amount. Add the two means and, because the components are independent, add their variances.

S=X+Y

E(S)=15+20=35

\operatorname{Var}(S)=4^2+5^2=41

Model

A sum of independent normal variables is normal.

S\sim\mathcal N(35,41)

\Pr(S<45)=\Phi\left(\frac{45-35}{\sqrt{41}}\right)

Compute

Standardize the cutoff and evaluate the standard-normal distribution function.

z=\frac{10}{\sqrt{41}}=1.5617376189

\Phi(1.5617376189)=0.9408250936

Answer

The requested probability rounds to 0.94.

\boxed{0.94\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.

AThis reports the two-sided Chebyshev lower bound 1-41/10²=0.59, rounded to 0.60, instead of using the exact normal distribution.
BThis reports the one-sided Cantelli lower bound 10²/(41+10²)=0.7092, rounded to 0.71, as though that bound were the normal probability.
CThis adds standard deviations, using 4+5=9 and obtaining Φ(10/9)=0.8667, rather than adding independent variances.
EThis replaces Y by its mean and calculates P(X<25)=Φ(2.5)=0.9938, thereby ignoring the variability contributed by Y.

Original practice · fully worked

Original variant: interval probability for a calibration contrast

Independent calibration readings A and B are normally distributed with means 10 and 4 and standard deviations 3 and 2, respectively. A lab forms the contrast R=2A-B. Calculate the probability that R is between 13 and 25.

A
0.0261
B
0.5159
C
0.6050
D
0.7113
E
0.9226

Variant answer in brief

The contrast is normal with mean 16 and variance 40. Standardizing both interval endpoints gives probability 0.605007, so choice C is correct.

Setup

Calculate the center and spread of the stated linear contrast.

E(R)=2(10)-4=16

\operatorname{Var}(R)=2^2(3^2)+(-1)^2(2^2)=40

Model

The independent normal inputs make R normal, so both interval endpoints can be standardized.

R\sim\mathcal N(16,40)

\Pr(13<R<25)=\Phi\left(\frac{25-16}{\sqrt{40}}\right)-\Phi\left(\frac{13-16}{\sqrt{40}}\right)

Compute

Evaluate the two standardized endpoints and subtract their cumulative probabilities.

z_L=-0.4743416490,\qquad z_U=1.4230249471

\Phi(z_U)-\Phi(z_L)=0.6050073903

Answer

The interval probability rounds to 0.6050.

\boxed{0.6050\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #521 — 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: interval probability for a calibration contrast

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution