Worked solution · Exam P

SOA Exam P Sample Question #447 Solution

This Exam P sample reference tests Uniform Distribution. The expected amount above the threshold is 180: the threshold is exceeded with probability 0.60 and the conditional mean excess is 300. Therefore the payment mean is 180p=90, giving p=0.50 and 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

Let p be the payment fraction written as a decimal. The payment is p times the positive part of the loss above the threshold.

Y=p(X-400)_+

Model

Integrate the excess against the constant density on the loss support.

\operatorname{E}[Y]=p\int_{400}^{1000}(x-400)\frac{1}{1000}\,dx

Compute

Evaluate the expected excess and solve the supplied mean equation for p.

\operatorname{E}[(X-400)_+]=\frac{(1000-400)^2}{2(1000)}=180

90=180p

p=0.50

Answer

The payment fraction is 50 percent.

\boxed{50\%\quad\text{(D)}}

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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 divides 90 by the maximum possible excess 1000-400=600 and obtains 0.15. A maximum is not an expected excess.
BThis divides 90 by the unconditional mean loss 500 and obtains 0.18, ignoring both the threshold and the positive-part payment.
CThis divides by the conditional mean excess 300 and obtains 0.30, but omits the 0.60 probability that the threshold is exceeded.
EUsing the nonexceedance probability 0.40 instead of the exceedance probability 0.60 gives an erroneous expected excess 0.40(300)=120 and 90/120=0.75.

Original practice · fully worked

Original variant: calibrate an overflow threshold

A solar generator's hourly output X is exponentially distributed with mean 10 units. A controller diverts one half of any output above a threshold c, so the diverted amount is D=0.5(X-c)+. The mean diverted amount is 2 units. Calculate c.

A
2.000
B
5.000
C
6.931
D
9.163
E
10.000

Answer and calculation

For an exponential output with mean 10, E[(X-c)+]=10exp(-c/10). Thus 5exp(-c/10)=2 and c=10ln(2.5)=9.163, so choice D is correct.

Setup

Use the tail-integral formula for the expected positive-part excess.

\operatorname{E}[(X-c)_+]=\int_c^\infty \Pr(X>t)\,dt

\Pr(X>t)=e^{-t/10}

Model

Integrate the exponential tail and apply the one-half diversion fraction.

\operatorname{E}[(X-c)_+]=10e^{-c/10}

\operatorname{E}[D]=5e^{-c/10}

Compute

Set the diverted mean equal to two and invert the exponential.

5e^{-c/10}=2

e^{-c/10}=0.40

c=-10\ln(0.40)=10\ln(2.5)=9.162907319\ldots

Answer

The overflow threshold is approximately 9.163 units.

\boxed{9.163\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 an overflow threshold

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables