Verified solution · Exam P

SOA Exam P Sample Question #575 Solution

This Exam P sample reference tests Normal Distribution. The four independent quarterly results sum to a normal variable with mean 4(8)=32 and variance 4(24²)=2304, so its standard deviation is 48. Standardizing zero gives -32/48=-0.67, making choice C 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 Y be the four-quarter sum. Add the four identical means.

E[Y]=4(8)=32

Model

Independence makes the quarterly variances additive, and a sum of independent normal variables remains normal.

\operatorname{Var}(Y)=4(24^2)=2304

Y\sim N(32,48^2)

Compute

Standardize the zero boundary for a positive aggregate result.

\Pr(Y>0)=\Pr\!\left(Z>\frac{0-32}{48}\right)

\Pr(Y>0)=\Pr(Z>-0.666666\ldots)

Answer

The standardized probability matches the expression using -0.67.

\boxed{\Pr(Z>-0.67)\quad\text{(C)}}

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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 reverses the one-quarter standardized ratio, using -24/8=-3.00, and also ignores aggregation over four quarters.
BThis first obtains aggregate mean 32 and standard deviation 48 but reverses the z-score ratio, using -48/32=-1.50.
DAdding the four quarterly standard deviations gives 96, so -32/96=-0.33; independent standard deviations do not add directly.
EThis uses one quarterly mean with the four-quarter standard deviation, giving -8/48=-0.17.

Original practice · fully worked

Original variant: demand-supply buffer

Independent daily demand X and supply Y are normally distributed with means 70 and 60 and standard deviations 8 and 6, respectively. Determine the buffer b for which P(X-Y≤b)=0.95.

A
-6.449
B
10.000
C
18.704
D
33.028
E
26.449

Variant answer in brief

The difference X-Y is normal with mean 10 and standard deviation √(8²+6²)=10. Its 95th percentile is 10+1.64485(10)=26.449, so choice E is correct.

Setup

Form the normal distribution of the demand-supply difference.

D=X-Y

E[D]=70-60=10

\operatorname{Var}(D)=8^2+6^2=100

Model

The requested buffer is the 95th percentile of this difference.

\Pr(D\le b)=0.95

\frac{b-10}{10}=\Phi^{-1}(0.95)

Compute

Insert the standard-normal 95th percentile and return to the original scale.

\Phi^{-1}(0.95)=1.644853627\ldots

b=10+10(1.644853627\ldots)=26.44853627\ldots

Answer

The required buffer is approximately 26.449.

\boxed{b=26.449\quad\text{(E)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #575 — Verified Solution.” Verified 2026-07-30.

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

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: demand-supply buffer

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution