SOA Exam P Sample Question #652. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #652 — Verified Solution

Independent solution

How to solve this Marginal Distributions question

Answer in brief

Summing the joint mass over the second coordinate gives marginal probabilities 0.1, 0.3, and 0.6 at the positive values of X. The resulting first and second moments are 2.5 and 6.7, so the variance is 0.45 and choice A.

Setup

Marginalize the supplied joint mass function over every allowed value of the second coordinate.

p_X(x)=\sum_{y=0}^{x}\frac{x+2y}{40},\qquad x=0,1,2,3

Model

The finite arithmetic sum reduces to a simple marginal mass function.

\sum_{y=0}^{x}(x+2y)=x(x+1)+x(x+1)=2x(x+1)

p_X(x)=\frac{x(x+1)}{20}

Compute

Calculate both raw moments from the marginal probabilities.

\Pr(X=1)=0.1,\qquad \Pr(X=2)=0.3,\qquad \Pr(X=3)=0.6

\operatorname{E}[X]=1(0.1)+2(0.3)+3(0.6)=2.5

\operatorname{E}[X^2]=1^2(0.1)+2^2(0.3)+3^2(0.6)=6.7

\operatorname{Var}(X)=6.7-(2.5)^2=0.45

Answer

The marginal variance is 0.45.

\boxed{\operatorname{Var}(X)=0.45\quad\text{(A)}}

Error diagnosis

Why the other letters tempt

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

BThis is the raw second moment 6.70 with the decimal point shifted; a variance must subtract the squared mean.
CThis candidate fails the direct mean-centered sum and is too large once the mass concentrated near X = 3 is centered at 2.5.
DThis remains on the scale of the raw second moment because it subtracts only a small correction instead of the full squared mean 6.25.
EThis is E[X²] = 6.70, not Var(X).

Original practice · fully worked

Original variant: inspection tags and manual reviews

A quality team records the number A of priority tags on a production lot and whether a manual review B is requested. Their joint probabilities are P(A=0,B=0)=0.10, P(A=0,B=1)=0.05, P(A=1,B=0)=0.20, P(A=1,B=1)=0.25, P(A=2,B=0)=0.15, and P(A=2,B=1)=0.25. Calculate the variance of A.

A
0.4000
B
0.4500
C
0.4875
D
1.2500
E
2.0500

Variant answer in brief

Marginalizing over the review indicator gives probabilities 0.15, 0.45, and 0.40 for A = 0, 1, and 2. The first two raw moments are 1.25 and 2.05, so the variance is 0.4875 and choice C.

Setup

Add the two joint cells in each row to obtain the distribution of the tag count.

\Pr(A=0)=0.10+0.05=0.15

\Pr(A=1)=0.20+0.25=0.45

\Pr(A=2)=0.15+0.25=0.40

Model

Use the first and second raw moments of this three-point marginal distribution.

\operatorname{Var}(A)=\operatorname{E}[A^2]-\operatorname{E}[A]^2

Compute

Evaluate the moments and subtract the squared mean.

\operatorname{E}[A]=0(0.15)+1(0.45)+2(0.40)=1.25

\operatorname{E}[A^2]=0^2(0.15)+1^2(0.45)+2^2(0.40)=2.05

\operatorname{Var}(A)=2.05-1.25^2=0.4875

Answer

The priority-tag count has variance 0.4875.

\boxed{0.4875\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #652 — Verified Solution.” Verified 2026-07-30.

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