SOA Exam P Sample Question #587. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #587 — Verified Solution

Independent solution

How to solve this Joint Distributions question

Answer in brief

This is a marginal-distribution variance calculation. The stated mean fixes p=0.12, normalization gives q=0.16, and the afternoon-distance second moment is 48, so the variance is 48-6^2=12 and choice B is correct.

Setup

Collapse the joint table over the morning categories. The afternoon distance A has masses q, 4p, and 3p at 0, 5, and 10, respectively.

\Pr(A=0)=q,\qquad \Pr(A=5)=4p,\qquad \Pr(A=10)=3p

Model

Use the supplied mean to determine p, then use total probability to determine q.

E[A]=5(4p)+10(3p)=50p=6

7p+q=1

Compute

After solving for the two table parameters, calculate the second moment and subtract the square of the mean.

p=0.12,\qquad q=0.16

E[A^2]=25(4p)+100(3p)=400p=48

\operatorname{Var}(A)=48-6^2=12

Answer

The variance of the afternoon distance is 12 square miles.

\boxed{12.0\quad\text{(B)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis results if the second moment is understated by one unit before subtracting the squared mean; the marginal probabilities should be recomputed from all nonzero table cells.
CThis can result from failing to combine both joint-table cells that contribute to the five-mile afternoon outcome.
DThis treats the symbols in the table as final probabilities without first imposing both the mean and normalization conditions.
EThis is E[A^2]-E[A]=48-6=42, which subtracts the mean rather than the square of the mean.

Original practice · fully worked

Original variant: correlated sensor adjustment

Two calibrated sensor errors X and Y have standard deviations 2 and 3, with correlation 0.5. A correction score is defined as Z=2X-Y. Calculate the variance of Z.

A
1
B
9
C
13
D
25
E
37

Variant answer in brief

The covariance is 0.5(2)(3)=3. Applying the variance formula for 2X-Y gives 4(4)+9-4(3)=13, so choice C is correct.

Setup

Convert the standard deviations and correlation into variances and covariance.

\operatorname{Var}(X)=4,\qquad \operatorname{Var}(Y)=9

\operatorname{Cov}(X,Y)=0.5(2)(3)=3

Model

For a linear combination, include both squared coefficients and the covariance cross term.

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

Compute

Substitute a=2 and b=-1 into the linear-combination formula.

\operatorname{Var}(Z)=4(4)+9-4(3)=13

Answer

The correction score has variance 13.

\boxed{13\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #587 — Verified Solution.” Verified 2026-07-30.

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