Worked solution · Exam P

SOA Exam P Sample Question #482 Solution

This Exam P sample reference tests Uniform Distribution. For a uniform interval, the mean is its midpoint and the standard deviation is its width divided by √(12). The width is 2√(3), so the lower endpoint is 3-√(3)≈1.27, which is 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 a and b be the lower and upper endpoints of the continuous uniform support.

\operatorname{E}[X]=\frac{a+b}{2}=3

\operatorname{SD}(X)=\frac{b-a}{\sqrt{12}}=1

Model

Translate the two moment conditions into the endpoint sum and interval width.

a+b=6

b-a=\sqrt{12}=2\sqrt{3}

Compute

Subtract the width from the endpoint sum and divide by two to isolate the lower endpoint.

a=\frac{(a+b)-(b-a)}{2}

a=\frac{6-2\sqrt{3}}{2}=3-\sqrt{3}=1.267949\ldots

Answer

The shortest possible value rounds to 1.27.

\boxed{1.27\quad\text{(B)}}

Test the method on unseen problems

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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 reports 1/√3 = 0.577…, the reciprocal of the correct half-width factor. The half-width is √3 standard deviations, and it must be subtracted from the mean.
CThis is the half-width √3 = 1.732… by itself. The requested endpoint is the midpoint minus that half-width.
DThis value is 4/√3 = 2.309…, obtained by dividing an unsupported four-hour span by the uniform half-width factor. The supplied equations instead determine both the sum 6 and width 2√3.
EThis uses a reciprocal half-width and computes 3 − 1/√3 = 2.423…. From Var(X) = (b − a)²/12, the half-width is √3, not 1/√3.

Original practice · fully worked

Original variant: recover spread from a tail fraction

A laboratory reading X follows a continuous uniform law over an unknown interval. The reading averages 10, while only 20% of readings are above 14. Determine Var(X).

A
3.85
B
10.00
C
13.33
D
14.81
E
177.78

Answer and calculation

The threshold 14 is four units above the midpoint and is the 80th percentile. That four-unit shift equals 0.30 of the support width, giving width 40/3 and variance 400/27≈14.81, choice D.

Setup

Write the support in terms of its known midpoint 10 and unknown width w.

X\sim\operatorname{Uniform}\left(10-\frac{w}{2},10+\frac{w}{2}\right)

Model

A 0.20 upper tail makes 14 the 80th percentile, located 0.30w above the midpoint.

14=10-\frac{w}{2}+0.80w=10+0.30w

Compute

Solve for the support width and insert it into the continuous-uniform variance formula.

w=\frac{4}{0.30}=\frac{40}{3}

\operatorname{Var}(X)=\frac{w^2}{12}=\frac{(40/3)^2}{12}=\frac{400}{27}=14.814814\ldots

Answer

The variance rounds to 14.81.

\boxed{14.81\quad\text{(D)}}

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Next steps for Exam P Univariate Random Variables

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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: recover spread from a tail fraction

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables