Worked solution · Exam P

SOA Exam P Sample Question #269 Solution

This Exam P sample reference tests Normal Distribution. The sum of the independent normal totals is normal with mean 22 and variance 25. Standardizing 29 gives z=1.4, so the requested lower-tail probability is 0.919243 and choice E.

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.

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Independent solution

How to solve this Normal Distribution question

Setup

Add the two division means.

E[X+Y]=10+12=22

Model

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

\operatorname{Var}(X+Y)=3^2+4^2=25

X+Y\sim N(22,25)

Compute

Standardize the total-claim boundary and evaluate the normal CDF.

z=\frac{29-22}{\sqrt{25}}=\frac75=1.4

\Pr(X+Y\le29)=\Phi(1.4)=0.9192433408\ldots

Answer

The probability rounds to 0.92.

\boxed{0.92\quad\text{(E)}}

Test the method on unseen problems

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30 original multivariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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.

AThe value 0.61 is approximately Φ(7/25)=Φ(0.28)=0.6103. It uses the aggregate variance 25 as though it were the standard deviation.
BThe value 0.69 is Φ(0.5)=0.6915. This results from adding the two standard deviations and then applying an extra factor of two, using 2(3+4)=14 as the scale.
CThe value 0.78 is approximately Φ(7/9)=0.7812. It uses the auto variance 3 squared as the aggregate standard deviation and omits the property uncertainty.
DThe value 0.84 is Φ(7/(3+4))=Φ(1)=0.8413. It adds standard deviations instead of adding independent variances.

Original practice · fully worked

Original variant: calibrating a gain from a normal tail target

Two independent calibration signals X and Y are normally distributed. Signal X has mean 8 and standard deviation 2, while signal Y has mean 3 and standard deviation 1. A controller records S=X+aY for a positive gain a. Determine a if the probability that S is at most 15 equals 0.841344746.

A
1.250
B
1.500
C
1.548
D
1.667
E
3.750

Answer and calculation

The target probability is Φ(1). Standardizing S gives (7-3a)/√(4+a squared)=1. The admissible solution is a=1.5; the other squared-equation root has the wrong sign, so choice B.

Setup

Form the mean and variance of the gain-adjusted signal.

S=X+aY\sim N\!\left(8+3a,\,2^2+a^2(1^2)\right)

Model

Recognize the supplied probability as the standard-normal CDF at one and standardize the threshold.

0.841344746=\Phi(1)

\frac{15-(8+3a)}{\sqrt{4+a^2}}=1

Compute

Square the equation, solve the resulting quadratic, and retain the root satisfying the original positive-sign equation.

(7-3a)^2=4+a^2\quad\Longrightarrow\quad 8a^2-42a+45=0

a=1.5\ \text{or}\ 3.75,\qquad 7-3a\ge0\ \Longrightarrow\ a=1.5

Answer

The controller gain is 1.500.

\boxed{1.500\quad\text{(B)}}

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How to solve this Normal Distribution question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: calibrating a gain from a normal tail target

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables