Worked solution · Exam P

SOA Exam P Sample Question #442 Solution

This Exam P sample reference tests Uniform Distribution. The two thresholds are three units apart while their upper-tail probabilities differ by 0.40. Uniform probability is proportional to length, so the support width is 3/0.40=7.5, and the variance is 7.5²⁄¹²=4.6875, choice B.

Independent analysis by ActuaryProof · reviewed July 31, 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 Uniform Distribution question

Setup

Let w=b-a be the uniform support width. An upper-tail probability inside the support is its remaining length divided by w.

\Pr(X>t)=\frac{b-t}{w}

Model

Subtract the two given tail equations so the unknown upper endpoint cancels.

\frac{b-8}{w}=0.60,\qquad \frac{b-11}{w}=0.20

\frac{3}{w}=0.40

Compute

Recover the width and substitute it into the continuous-uniform variance formula.

w=\frac3{0.40}=7.5

\operatorname{Var}(X)=\frac{w^2}{12}=\frac{7.5^2}{12}=4.6875

Answer

The variance rounds to 4.69.

\boxed{4.69\quad\text{(B)}}

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.

AA variance of 3.70 would imply width √(12 × 3.70) = 6.66. Across a three-unit threshold change, that width would change the tail by 0.45 rather than the stated 0.40.
CThis uses the incorrect formula w² / 9, producing 7.5² / 9 = 6.25. A continuous uniform variance has denominator 12.
DA variance of 7.24 implies width √(12 × 7.24) = 9.32, which would make the three-unit tail change only 0.322 rather than 0.40.
EThis is the mean of the recovered support: the tail equations give a = 5 and b = 12.5, so (a + b) / 2 = 8.75. The question asks for variance.

Original practice · fully worked

Original variant: conditional tail of a bounded response time

A response time T is uniformly distributed from 0 to an unknown upper endpoint L, where L is greater than 10. Given that T exceeds 6, the conditional probability that T exceeds 10 is 1/3. Calculate the 90th percentile of T.

A
4.0
B
6.0
C
10.0
D
10.8
E
12.0

Answer and calculation

The conditional tail ratio is (L-10)/(L-6)=1/3, which gives L=12. A uniform 90th percentile is 0.9L=10.8, so choice D is correct.

Setup

Convert the conditional probability into a ratio of interval lengths within the uniform support.

\Pr(T>10\mid T>6)=\frac{\Pr(T>10)}{\Pr(T>6)}=\frac{L-10}{L-6}

Model

Set the length ratio equal to the supplied conditional probability and solve for the endpoint.

\frac{L-10}{L-6}=\frac13

3L-30=L-6

Compute

Recover the endpoint and move 90% of the way across the support.

L=12

Q(0.90)=0+0.90(12)=10.8

Answer

The 90th percentile is 10.8.

\boxed{10.8\quad\text{(D)}}

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How to solve this Uniform Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: conditional tail of a bounded response time

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables