Worked solution · Exam P

SOA Exam P Sample Question #257 Solution

This Exam P sample reference tests Uniform Distribution. This is a uniform-moment inversion followed by a percentile calculation. The stated second moment forces the first lower endpoint to be 60, making the second interval [75,100] and its 80th percentile 95, so choice E.

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

Use the raw second moment formula for a uniform variable whose endpoints are a and 100.

E[X^2]=\frac{a^2+100a+100^2}{3}=\frac{19600}{3}

Model

Solve the resulting quadratic and retain the root that forms a valid interval below 100.

a^2+100a-9600=0

(a-60)(a+160)=0

Compute

The valid lower endpoint is 60, so the second model begins at 1.25 × 60. Move 80% of the way across that interval.

a=60,\qquad 1.25a=75

q_{0.80}(Y)=75+0.80(100-75)=95

Answer

The requested percentile is 95.

\boxed{95\quad\text{(E)}}

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.

AThe value 80 is 75+0.20(25), the 20th percentile of Y. It uses the upper-tail complement 1-0.80 as the lower-tail percentile.
BThe value 85 is 75+0.40(25), only 40% of the way across Y's support. It results from halving the requested percentile position.
CThe value 90 is 75+0.60(25), the 60th percentile of Y. It stops before the required 80% position.
DThe value 92 is 60+0.80(40), the 80th percentile of X. It does not replace X's lower endpoint by the specified endpoint for Y.

Original practice · fully worked

Original variant: expected correction above a threshold

A calibration drift U is uniformly distributed on [a,16], where a is less than 16, and its second raw moment is 112. A correction amount is C=max(U-10,0). Calculate E[C].

A
0.50 units
B
1.50 units
C
3.00 units
D
6.00 units
E
10.00 units

Answer and calculation

The uniform second-moment equation gives the valid lower endpoint a=4. Averaging the positive excess U-10 over the resulting interval [4,16] gives E[C]=1.5 units, so choice B.

Setup

Use the raw second moment of a uniform variable to recover the unknown lower endpoint.

E[U^2]=\frac{a^2+16a+16^2}{3}=112

a^2+16a-80=0

Model

Retain the endpoint that produces a valid ordered support.

(a-4)(a+20)=0

a=4,\qquad U\sim\operatorname{Uniform}(4,16)

Compute

Integrate the positive correction over the portion of the support above the threshold.

E[C]=\frac1{16-4}\int_{10}^{16}(u-10)\,du

=\frac1{12}\left[\frac{(u-10)^2}{2}\right]_{10}^{16}=\frac{18}{12}=1.5

Answer

The expected correction amount is 1.50 units.

\boxed{1.50\text{ units}\quad\text{(B)}}

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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: expected correction above a threshold

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables