Worked solution · Exam P

SOA Exam P Sample Question #629 Solution

This Exam P sample reference tests Discrete Random Variables. This is a discrete-moment problem in which the supplied second moment determines an unknown negative support point. That point is -5, the mean is 4, and the variance is 47.8-4²=31.8, selecting choice D.

Independent analysis by ActuaryProof · reviewed July 30, 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 Discrete Random Variables question

Setup

Separate the unknown support point's contribution to the second moment from the three known contributions.

47.8=0.2a^2+0.1(1^2)+0.3(3^2)+0.4(10^2)

Model

Solve the second-moment equation and then use the stated sign restriction to choose the appropriate root.

0.2a^2=47.8-42.8=5

a^2=25,\qquad a=-5

Compute

Calculate the first moment with the negative support value and apply the raw-moment variance identity.

\operatorname{E}[Y]=0.2(-5)+0.1(1)+0.3(3)+0.4(10)=4

\operatorname{Var}(Y)=\operatorname{E}[Y^2]-\operatorname{E}[Y]^2=47.8-4^2=31.8

Answer

The variance is 31.8.

\boxed{31.8\quad\text{(D)}}

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.

AThe value 5.6 is the variance's square root rounded to one decimal place, so it is the standard deviation rather than the requested variance.
BChoosing the positive root a=5 violates the sign restriction and changes the mean to 6; then 47.8-6²=11.8.
CSubtracting a²=25 directly from the second moment gives 22.8, but the variance formula requires the square of the full mean, not a support value.
EAdding the squared mean to the second moment instead of subtracting it gives 47.8+16=63.8.

Original practice · fully worked

Original variant: courier-route time mixture

An autonomous courier uses a direct route with probability 0.70 and an inspection route with probability 0.30. Conditional on the direct route, travel time has mean 8 minutes and variance 4 square minutes. Conditional on the inspection route, it has mean 14 minutes and variance 9 square minutes. Calculate the unconditional variance of travel time.

A
5.29
B
5.50
C
7.56
D
9.80
E
13.06

Answer and calculation

The expected conditional variance is 5.50, while variation between the two route means contributes 7.56. Their sum is 13.06 square minutes, so choice E is correct.

Setup

Let R denote the selected route and T the travel time. First combine the two conditional means.

\operatorname{E}[T]=0.70(8)+0.30(14)=9.8

Model

The law of total variance separates average within-route variability from variability of the route-specific means.

\operatorname{Var}(T)=\operatorname{E}[\operatorname{Var}(T\mid R)]+\operatorname{Var}(\operatorname{E}[T\mid R])

Compute

Evaluate the within-route and between-route components around the overall mean 9.8.

\operatorname{E}[\operatorname{Var}(T\mid R)]=0.70(4)+0.30(9)=5.50

\operatorname{Var}(\operatorname{E}[T\mid R])=0.70(8-9.8)^2+0.30(14-9.8)^2=7.56

\operatorname{Var}(T)=5.50+7.56=13.06

Answer

The unconditional travel-time variance is 13.06 square minutes.

\boxed{13.06\quad\text{(E)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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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: courier-route time mixture

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables