Worked solution · Exam P

SOA Exam P Sample Question #163 Solution

This Exam P sample reference tests Uniform Distribution. For a uniform loss with endpoint b above the deductible, the limited expected value at 180 is 180-16,200/b. Equating this to 144 gives b=450, choice D.

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

The unreimbursed portion is the loss capped at the deductible. If b were at most 180, its mean would be b/2=144, forcing b=288 and contradicting that assumption; hence b exceeds 180.

U=\min(X,180),\qquad X\sim\operatorname{Uniform}(0,b)

b\le180\ \Longrightarrow\ E[U]=\frac b2=144\ \Longrightarrow\ b=288\quad\text{(contradiction)}

Model

Use the survival-integral identity for a nonnegative limited variable.

E[U]=\int_0^{180}\Pr(X>x)\,dx

\Pr(X>x)=1-\frac{x}{b}\quad(0\le x\le180)

Compute

Integrate the linear survival function and solve for the endpoint.

144=\int_0^{180}\left(1-\frac{x}{b}\right)dx=180-\frac{180^2}{2b}

b=\frac{16200}{36}=450

Answer

The upper endpoint of the loss distribution is 450.

\boxed{450\quad\text{(D)}}

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.

AAt b=236 the limited mean would be 180-16,200/236=111.36, not 144. This endpoint fails the supplied moment by a wide margin.
BThe value 288 sets b/2=144, treating the limited loss as the uncapped loss. The cap at 180 creates an additional point mass at the cap.
CAt b=388 the limited mean is 138.25. This near miss comes from understating the survival mass above the deductible.
EThe value 468 equals 180+2(144), as if the unreimbursed amount were a uniform residual above the deductible. Substitution gives 145.38 rather than 144.

Original practice · fully worked

Original variant: capped excess reimbursement for a uniform repair cost

A service shop observes repair bills spread evenly across the interval from zero through 600 dollars. A protection plan pays the portion above 150 dollars, subject to a maximum payment of 300 dollars. Determine the plan's expected payment.

A
$75.00
B
$112.50
C
$150.00
D
$168.75
E
$225.00

Answer and calculation

The payment is min((X-150)+,300). Integrating its survival probability from 0 to 300 gives an expected payment of $150, choice C.

Setup

Write the plan payment as a capped excess variable.

Y=\min\bigl((X-150)_+,300\bigr)

Model

For payment level t between zero and the cap, Y exceeds t exactly when the cost exceeds 150+t.

\Pr(Y>t)=\Pr(X>150+t)=\frac{450-t}{600},\qquad 0\le t<300

Compute

Integrate the payment survival curve through the cap.

E[Y]=\int_0^{300}\frac{450-t}{600}\,dt

E[Y]=\frac{450(300)-300^2/2}{600}=150

Answer

The protection plan pays $150 on average.

\boxed{\$150.00\quad\text{(C)}}

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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: capped excess reimbursement for a uniform repair cost

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables