Verified solution · Exam P

SOA Exam P Sample Question #330 Solution

This Exam P sample reference tests Variance of Linear Combinations. Independence removes the covariance term, the coefficient 4 contributes a factor of 16, and constants contribute no variance. The result is 16(2)+3=35, so choice E 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 Variance of Linear Combinations question

Setup

Identify the two random contributions and the deterministic shift.

Z=4X-Y-3,\qquad \operatorname{Var}(X)=2,\qquad \operatorname{Var}(Y)=3

Model

For independent variables, the covariance term vanishes. The negative sign on Y disappears when its coefficient is squared.

\operatorname{Var}(aX+bY+c)=a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)

Compute

Substitute the two variances and coefficients.

\operatorname{Var}(Z)=4^2(2)+(-1)^2(3)=32+3=35

Answer

The variance of the modeled sales quantity is 35.

\boxed{35\quad\text{(E)}}

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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 5 is 2+3. It ignores both coefficients in the linear combination.
BThe value 11 is 4(2)+3. It applies the coefficient 4 only once even though variance scales by its square.
CThe value 29 is 16(2)-3. It carries the minus sign on Y into the variance instead of squaring the coefficient -1.
DThe value 32 is 16(2). It omits the independent Y contribution, whose variance adds 3.

Original practice · fully worked

Original variant: routing pressure across four traffic states

A dispatch controller records a signed congestion index U and a route adjustment V. Four traffic states produce (U,V)=(-1,0), (0,2), (2,-1), and (3,1) with probabilities 0.12, 0.18, 0.27, and 0.43, respectively. The routing pressure score is L=2U+V. Calculate Var(L).

A
1.15
B
8.58
C
9.18
D
9.73
E
10.29

Variant answer in brief

The four traffic states give Var(U)=2.1459, Var(V)=1.1496, and Cov(U,V)=-0.1392. Therefore Var(2U+V)=9.1764, which rounds to 9.18 and choice C.

Setup

Compute the first, second, and mixed moments from the four traffic states.

\operatorname{E}[U]=1.71,\qquad \operatorname{E}[U^2]=5.07

\operatorname{E}[V]=0.52,\qquad \operatorname{E}[V^2]=1.42,\qquad \operatorname{E}[UV]=0.75

Model

Convert the moments to variances and covariance before applying the linear-combination formula.

\operatorname{Var}(U)=5.07-(1.71)^2=2.1459

\operatorname{Var}(V)=1.42-(0.52)^2=1.1496

\operatorname{Cov}(U,V)=0.75-(1.71)(0.52)=-0.1392

Compute

Include the covariance cross term in the variance of the load score.

\operatorname{Var}(L)=4\operatorname{Var}(U)+\operatorname{Var}(V)+4\operatorname{Cov}(U,V)

=4(2.1459)+1.1496+4(-0.1392)=9.1764

Answer

The variance of the routing pressure score rounds to 9.18.

\boxed{9.18\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #330 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Variance of Linear Combinations question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: routing pressure across four traffic states

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution