Worked solution · Exam P

SOA Exam P Sample Question #489 Solution

This Exam P sample reference tests Uniform Distribution. For a limit m between 10 and 60, the limited payment has mean m-(m-10)²⁄¹⁰⁰. Setting this equal to 31 gives feasible root m=40, so choice C is correct.

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 X denote the loss and m the benefit limit. Express the limited payment by subtracting the amount by which the limit exceeds a smaller loss.

Y=\min(X,m)=m-(m-X)_+

X\sim\operatorname{Uniform}(10,60)

Model

For a limit inside the support, the expected shortfall below m is the area of a triangle under the uniform density.

\operatorname{E}[(m-X)_+]=\int_{10}^{m}(m-x)\frac{1}{50}\,dx

\operatorname{E}[(m-X)_+]=\frac{(m-10)^2}{100}

Compute

Set the expected limited payment equal to the supplied value and solve the resulting quadratic.

m-\frac{(m-10)^2}{100}=31

m^2-120m+3200=0

m\in\{40,80\}

Answer

Only 40 lies within the loss support and can be the stated maximum benefit.

\boxed{40\quad\text{(C)}}

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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 is the uncapped mean, (10+60)/2=35. It does not solve for the benefit limit that lowers the expected payment to 31.
BNormalizing the triangular shortfall by the upper endpoint 60 instead of the support width 50 gives a spurious root near 37.1 and can lead to choice 38.
DThe expected reduction is 35-31=4. Setting only the conditional mean excess (60-m)/2 equal to 4 gives m=52 but omits the probability that the limit is reached.
ESetting the maximum possible reduction 60-m equal to the expected reduction 4 gives m=56. A maximum shortfall is not an expected shortfall.

Original practice · fully worked

Original variant: calibrate a sensor saturation point

A sensor's uncapped load reading L is exponential with mean 8 units. The display records D=min(L,c), where c is an unknown saturation point. Engineers measure E[D]=6. Determine c.

A
2.301
B
5.545
C
8.000
D
11.090
E
16.636

Answer and calculation

The expected limited exponential reading is 8(1-exp(−c/8)). Equating this to 6 gives exp(−c/8)=1/4 and c=8 ln 4=11.090, so choice D is correct.

Setup

Use the tail-integral formula for a nonnegative reading capped at c.

\operatorname{E}[\min(L,c)]=\int_0^c\Pr(L>t)\,dt

\Pr(L>t)=e^{-t/8}

Model

Integrate the exponential survival function up to the saturation point.

\operatorname{E}[D]=\int_0^c e^{-t/8}\,dt

\operatorname{E}[D]=8\left(1-e^{-c/8}\right)

Compute

Set the limited mean equal to six and invert the exponential.

8\left(1-e^{-c/8}\right)=6

e^{-c/8}=\frac14

c=8\ln 4=11.0903549\ldots

Answer

The display saturates at approximately 11.090 units.

\boxed{11.090\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: calibrate a sensor saturation point

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables