Worked solution · Exam P

SOA Exam P Sample Question #453 Solution

This Exam P sample reference tests Uniform Distribution. This problem first finds the variance of the payment above the deductible before coinsurance. That variance is 6000, so variance scaling gives p=√(2000/6000)=0.57735 and choice C.

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 Z be the amount above the deductible before applying the selected percentage, and let X=pZ be the paid amount.

Z=(L-240)_+,\qquad X=pZ

Model

Half of the uniform losses produce zero. On the positive range, integrate with the original uniform density 1/480.

E[Z]=\int_{240}^{480}(l-240)\frac{1}{480}\,dl=60

E[Z^2]=\int_{240}^{480}(l-240)^2\frac{1}{480}\,dl=9600

Compute

Compute the base variance, then use the square-law for a scaled random variable.

\operatorname{Var}(Z)=9600-60^2=6000

2000=\operatorname{Var}(pZ)=p^2(6000)

p=\sqrt{\frac{2000}{6000}}=0.5773502692\ldots

Answer

The selected percentage is approximately 57.7%.

\boxed{57.7\%\quad\text{(C)}}

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 p=0.111, the payment variance would be p²(6000)=73.9, not 2000; the candidate neither uses the correct square-root scaling nor satisfies the target.
BThis is 2000/6000=1/3. Variance scales with the square of the percentage, so the square root is required.
DAt p=0.645, the payment variance would be p²(6000)=2496.2, which is above the 2000 target.
EAt p=0.913, the payment variance would be p²(6000)=5001.4, which is far above the 2000 target.

Original practice · fully worked

Original variant: coefficient of variation of a service charge

The number W of follow-up tasks on a service case equals 0, 2, or 5 with probabilities 0.5, 0.3, and 0.2. The total charge is C=10+3W dollars. Calculate the coefficient of variation of C.

A
0.1289
B
0.1496
C
0.3867
D
1.0722
E
1.1924

Answer and calculation

The task count has mean 1.6 and variance 3.64. Thus the charge has mean 14.8 and variance 32.76, giving CV=√(32.76)/14.8=0.3867 and choice C.

Setup

Compute the first two moments of the discrete task count.

E[W]=0(0.5)+2(0.3)+5(0.2)=1.6

E[W^2]=0^2(0.5)+2^2(0.3)+5^2(0.2)=6.2

Model

Apply the mean and variance rules for an affine transformation.

\operatorname{Var}(W)=6.2-1.6^2=3.64

E[C]=10+3(1.6)=14.8

\operatorname{Var}(C)=3^2(3.64)=32.76

Compute

Divide the charge standard deviation by its mean.

\operatorname{CV}(C)=\frac{\sqrt{\operatorname{Var}(C)}}{E[C]}

\operatorname{CV}(C)=\frac{\sqrt{32.76}}{14.8}=0.3867321087\ldots

Answer

The coefficient of variation is about 0.3867.

\boxed{0.3867\quad\text{(C)}}

Keep practicing

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: coefficient of variation of a service charge

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables