Verified solution · Exam P

SOA Exam P Sample Question #446 Solution

This Exam P sample reference tests Joint Distributions. Marginalizing the afternoon column probabilities gives masses y, 4x, and 3x at lengths 0, 5, and 30. The supplied mean yields x=0.10 and normalization gives y=0.30; therefore E[A²]=280 and Var(A)=280-11²=159, choice A.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Joint Distributions question

Setup

Sum the joint-table entries down each afternoon-length column to obtain its marginal probabilities.

\Pr(A=0)=y,\qquad \Pr(A=5)=4x,\qquad \Pr(A=30)=3x

y+7x=1

Model

Use the stated marginal mean to determine x, then use total probability to recover y.

11=0(y)+5(4x)+30(3x)=110x

x=0.10,\qquad y=1-7(0.10)=0.30

Compute

Calculate the second moment from the recovered marginal distribution and subtract the squared mean.

\operatorname{E}[A^2]=0^2(0.30)+5^2(0.40)+30^2(0.30)=280

\operatorname{Var}(A)=280-11^2=159

Answer

The afternoon-route variance is 159.0 square miles.

\boxed{159.0\quad\text{(A)}}

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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.

BA variance of 168.5 would imply E[A²]=168.5+11²=289.5, but the exact marginal masses give 25(0.40)+900(0.30)=280.
CA variance of 181.5 would require second moment 302.5. This overweights at least one positive-length column relative to the recovered probabilities 0.40 and 0.30.
DThe value 259.0 effectively subtracts only 21 from the correct second moment 280. Variance requires subtracting the full squared mean 11²=121.
EThis computes E[A²]-E[A]=280-11=269. The variance identity uses the square of the mean, not the mean itself.

Original practice · fully worked

Original variant: variance under a latent service mode

A service delay D is generated under one of two modes. Mode A is selected with unknown probability q and produces delays 2 or 6 minutes with equal probability. Mode B produces delays 5 or 15 minutes with equal probability. The overall mean delay is 6.4 minutes. Calculate Var(D).

A
6.40
B
8.64
C
12.40
D
21.04
E
62.00

Variant answer in brief

The conditional means 4 and 10 make 4q+10(1-q)=6.4, so q=0.60. Total variance combines 12.40 of within-mode variance and 8.64 of between-mode variance, giving 21.04 and choice D.

Setup

Compute each mode's conditional mean and variance from its two equally likely delays.

\mu_A=4,\quad \operatorname{Var}(D\mid A)=4

\mu_B=10,\quad \operatorname{Var}(D\mid B)=25

Model

Use the overall mean to recover the mode probability.

6.4=4q+10(1-q)

q=0.60

Compute

Apply the law of total variance, separating within-mode and between-mode variation.

\operatorname{E}[\operatorname{Var}(D\mid M)]=0.60(4)+0.40(25)=12.40

\operatorname{Var}(\operatorname{E}[D\mid M])=0.60(0.40)(10-4)^2=8.64

\operatorname{Var}(D)=12.40+8.64=21.04

Answer

The overall delay variance is 21.04 square minutes.

\boxed{21.04\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #446 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Joint Distributions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: variance under a latent service mode

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution