Worked solution · Exam P

SOA Exam P Sample Question #480 Solution

This Exam P sample reference tests Discrete Random Variables. One individual's accident count has mean 0.24, second moment 0.32, and variance 0.2624. Independence makes the group variance 144(0.2624)=37.7856, which rounds to choice C.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Discrete Random Variables question

Setup

Let X be the accident count for one policyholder and let T be the total across the 144 independent policyholders.

T=\sum_{i=1}^{144}X_i

\Pr(X=0)=0.80,\quad \Pr(X=1)=0.16,\quad \Pr(X=2)=0.04

Model

Compute the first two raw moments of the single-policyholder count before forming its variance.

\operatorname{E}[X]=0(0.80)+1(0.16)+2(0.04)=0.24

\operatorname{E}[X^2]=0^2(0.80)+1^2(0.16)+2^2(0.04)=0.32

Compute

Subtract the squared mean, then add identical variances across the independent policies.

\operatorname{Var}(X)=0.32-(0.24)^2=0.2624

\operatorname{Var}(T)=144(0.2624)=37.7856

Answer

The total-count variance is 37.79 to two decimal places.

\boxed{37.79\quad\text{(C)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate 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.

AThis first finds the correct aggregate variance but then divides it by √144 = 12, giving 37.7856/12 = 3.1488. Variances of independent observations add; they do not receive a final standard-error division.
BThis is the expected total, 144 E[X] = 144(0.24) = 34.56, rather than its variance.
DThis uses the raw second moment alone: 144 E[X²] = 144(0.32) = 46.08. It omits the subtraction of (E[X])² for each policyholder.
EThis adds rather than subtracts the squared mean, yielding 144[0.32 + (0.24)²] = 54.3744. The variance identity is E[X²] − (E[X])².

Original practice · fully worked

Original variant: compound shipment credits

A depot receives a Poisson-distributed number N of shipment batches per hour with mean 20. Each batch independently earns a credit W of 0, 1, or 3 points with probabilities 0.50, 0.30, and 0.20, respectively. The batch credits are independent of N. Let S be the total credit earned in one hour. Calculate the variance of S.

A
16.2
B
18.0
C
25.8
D
42.0
E
67.8

Answer and calculation

A compound Poisson total satisfies Var(S)=E[N]E[W²]. Since E[W²]=2.1, the variance is 20(2.1)=42, so choice D is correct.

Setup

Express the hourly credit as a random sum of independent batch credits.

S=\sum_{i=1}^{N}W_i,\qquad N\sim\operatorname{Poisson}(20)

Model

Find both the mean and second raw moment of one credit, which separate the two sources of compound-sum variation.

\operatorname{E}[W]=0(0.50)+1(0.30)+3(0.20)=0.90

\operatorname{E}[W^2]=0+1(0.30)+9(0.20)=2.10

\operatorname{Var}(W)=2.10-(0.90)^2=1.29

Compute

Apply total variance; for a Poisson count, its mean and variance are both 20.

\operatorname{Var}(S)=\operatorname{E}[N]\operatorname{Var}(W)+\operatorname{Var}(N)(\operatorname{E}[W])^2

\operatorname{Var}(S)=20(1.29)+20(0.90)^2=25.8+16.2=42.0

Answer

The one-hour total credit has variance 42.

\boxed{42.0\quad\text{(D)}}

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How to solve this Discrete Random Variables question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: compound shipment credits

Answer and calculation

Setup

Model

Compute

Answer

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Next steps for Exam P Univariate Random Variables