Verified solution · Exam P

SOA Exam P Sample Question #539 Solution

This Exam P sample reference tests Normal Linear Combination. The average of the three independent normal variables is normal with mean 100 and variance (13.5+27+40.5)/9=9. The cutoff is two standard deviations above the mean, giving the upper tail 1-Φ(2) and choice D.

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 Linear Combination question

Setup

Let A be the arithmetic average of the three independent measurements.

A=\frac{X_1+X_2+X_3}{3}

Model

A linear combination of independent normal variables is normal. Add the component variances, then square the averaging coefficient.

E[A]=100

\operatorname{Var}(A)=\frac{13.5+27+40.5}{3^2}=9

A\sim N(100,9)

Compute

Standardize the requested upper-tail event.

\Pr(A>106)=\Pr\!\left(Z>\frac{106-100}{3}\right)

\Pr(A>106)=1-\Phi(2)

Answer

The required expression is the standard-normal upper tail beyond 2.

\boxed{1-\Phi(2)\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 finds the correct standardized cutoff 2 but uses the lower tail Φ(2) instead of the upper tail.
BThis incorrectly assigns the sum variance 81 to the average, obtains z=2/3, and then uses the lower tail.
CThis uses the upper tail but still assigns the sum variance 81 to the average, producing the wrong z-value 2/3.
EThis divides the sum variance by 3 instead of 3², giving variance 27 for the average and the incorrect cutoff 2/√(3).

Original practice · fully worked

Original variant: weighted calibration score

Three independent calibration readings A, B, and C are normal. Their means are 2, -1, and 0, and their variances are 4, 9, and 16, respectively. A diagnostic score is W=A-2B+0.5C. Calculate the 90th percentile of W.

A
-4.501
B
4.000
C
10.901
D
12.501
E
16.816

Variant answer in brief

The score is normal with mean 4 and variance 4+4(9)+0.25(16)=44. Its 90th percentile is 4+1.2815516sqrt(44)=12.5009, so choice D is correct.

Setup

Apply the score coefficients to the three component means.

E[W]=2-2(-1)+0.5(0)=4

Model

Independence eliminates covariance terms, while each variance is multiplied by the square of its coefficient.

\operatorname{Var}(W)=4+(-2)^2(9)+(0.5)^2(16)=44

W\sim N(4,44)

Compute

Use the standard-normal 90th-percentile value.

z_{0.90}=1.2815515655

q_{0.90}=4+z_{0.90}\sqrt{44}=12.50085138

Answer

The score's 90th percentile rounds to 12.501.

\boxed{12.501\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #539 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Normal Linear Combination question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: weighted calibration score

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution