Worked solution · Exam P

SOA Exam P Sample Question #219 Solution

This Exam P sample reference tests Uniform Distribution. The unreimbursed amount is min(X,d). For a uniform loss on [0,450], its mean is d-d squared/900. Setting this to 56 gives roots 60 and 840; only 60 lies in the loss support, so choice A.

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 policyholder retains the smaller of the loss and the deductible.

R=\min(X,d),\qquad X\sim\operatorname{Uniform}(0,450)

Model

Split the expectation between losses below the deductible and losses capped at the deductible.

\operatorname{E}[R]=\int_0^d x\frac{dx}{450}+d\Pr(X>d)

\operatorname{E}[R]=\frac{d^2}{900}+d\left(1-\frac d{450}\right)=d-\frac{d^2}{900}

Compute

Set the retained mean to the supplied amount and solve the quadratic.

d-\frac{d^2}{900}=56

d^2-900d+50400=0

d=60\quad\text{or}\quad d=840

Answer

A deductible above the maximum possible loss is incompatible with a retained mean of 56, so the admissible root is 60.

\boxed{60\quad\text{(A)}}

Test the method on unseen problems

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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.

BSubstituting d=87 into the correct limited-mean formula gives 87-87 squared/900=78.59, not 56. This value does not satisfy the required loss partition.
CThe value 112 follows from the shortcut E[min(X,d)] approximately d/2. That treats every retained amount as uniform from zero to d and ignores losses above d that contribute exactly d.
DThe value 169 is approximately three times 56, corresponding to an unsupported d/3 mean approximation. The actual retained distribution has an atom at d.
EKeeping only the below-deductible integral gives d squared/900=56 and d=224.5, near 224. It omits all losses above d, each of which contributes the full deductible to the unreimbursed amount.

Original practice · fully worked

Original variant: retained amount under an increasing loss density

A repair severity X has density f(x)=x/5000 for x between 0 and 100, and zero elsewhere. A service plan applies a deductible of 60 credits, so the customer retains min(X,60). Calculate the expected amount retained by the customer.

A
13.867 credits
B
21.600 credits
C
52.800 credits
D
60.000 credits
E
66.667 credits

Answer and calculation

The below-deductible contribution is the integral of x squared/5000 from 0 to 60, equal to 14.4. The probability above 60 is 0.64, adding 38.4, so the retained mean is 52.8, choice C.

Setup

Separate the continuous retained amount below 60 from the capped amount paid by every larger loss.

R=\min(X,60)

Model

Compute the probability that the loss crosses the deductible.

F_X(x)=\int_0^x\frac{u}{5000}\,du=\frac{x^2}{10000}

\Pr(X>60)=1-\frac{60^2}{10000}=0.64

Compute

Add the sub-deductible first moment and the capped contribution.

\operatorname{E}[R]=\int_0^{60}x\frac{x}{5000}\,dx+60(0.64)

\operatorname{E}[R]=14.4+38.4=52.8

Answer

The customer retains an average of 52.8 credits.

\boxed{52.800\quad\text{credits (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: retained amount under an increasing loss density

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables