SOA Exam P Sample Question #610. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #610 — Verified Solution

Independent solution

How to solve this Uniform Distribution question

Answer in brief

This is an inverse limited-expected-value problem for a uniform loss. Equating m-m^2/4000 to 910 gives candidate roots 1400 and 2600, and only 1400 lies within the loss support, so choice C is correct.

Setup

Let X be the uniform loss and let Y be the payment limited to the unknown maximum.

X\sim\operatorname{Uniform}(0,2000)

Y=\min(X,m)

Model

Split the expectation between losses below the maximum and losses that reach the maximum.

\operatorname{E}[Y]=\int_0^m\frac{x}{2000}\,dx+m\Pr(X>m)

\operatorname{E}[Y]=\frac{m^2}{4000}+m\left(1-\frac{m}{2000}\right)=m-\frac{m^2}{4000}

Compute

Set the limited expectation equal to the supplied value and solve the quadratic.

m-\frac{m^2}{4000}=910

m^2-4000m+3640000=0

m=1400\ \text{or}\ 2600

Answer

A policy maximum above 2000 would be outside the stated loss range, leaving 1400 as the admissible solution.

\boxed{m=1400\quad\text{(C)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis copies the expected payment as the policy maximum. Because some losses exceed the maximum, the payment mean is not generally equal to the maximum.
BSubstitution gives 1150-1150^2/4000=819.375, so this proposed maximum does not reproduce the stated expectation.
DSubstitution gives 1600-1600^2/4000=960, which is larger than the stated expected payment.
EThis sets m/2=910 as though every payment were uniform on [0,m]. Losses above m instead create a probability mass at the maximum.

Original practice · fully worked

Original variant: coinsured repair layer

Equipment repair loss X is uniform from 0 to 1000. A contract pays 80% of the portion of a loss between 200 and 700, so its payment is Y=0.8 min((X-200)+,500). Calculate the variance of Y.

A
162.07
B
26,266.67
C
41,041.67
D
48,400.00
E
74,666.67

Variant answer in brief

Before coinsurance, the layer payment has first moment 275 and second moment 350000/3. Its variance is 123125/3, and multiplying by 0.8^2 gives Var(Y)=78800/3=26266.67, so choice B is correct.

Setup

Separate the three-part contract design into an underlying layer amount Z and the 80% coinsurance factor.

Z=\min\!\left((X-200)_+,500\right)

Y=0.8Z

Model

For losses between 200 and 700, Z increases continuously from zero to 500. Losses at or above 700 produce a point mass at the layer maximum.

\Pr(X\ge700)=0.30

Compute

Calculate the first two layer moments, form its variance, and then square the coinsurance factor.

\operatorname{E}[Z]=\int_0^{500}\frac{z}{1000}\,dz+500(0.30)=275

\operatorname{E}[Z^2]=\int_0^{500}\frac{z^2}{1000}\,dz+500^2(0.30)=\frac{350000}{3}

\operatorname{Var}(Y)=0.8^2\left(\frac{350000}{3}-275^2\right)=\frac{78800}{3}=26266.6667\ldots

Answer

The contract-payment variance is approximately 26,266.67 square currency units.

\boxed{26266.67\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #610 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.