Worked solution · Exam P

SOA Exam P Sample Question #310 Solution

This Exam P sample reference tests Expected Value. The limited policy has expected payment 284/75, while the ordinary-deductible policy has expected payment 72/25. Their absolute difference is 68/75=0.906667, which rounds to 0.91 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 Expected Value question

Setup

Write each claim payment as a piecewise function of the loss.

Y_1=\min(X,4),\qquad Y_2=(X-4)_+

Model

Integrate each payment against the loss density on the regions where its formula changes.

\operatorname{E}[Y_1]=\int_0^4 x\frac{x}{50}\,dx+\int_4^{10}4\frac{x}{50}\,dx

\operatorname{E}[Y_2]=\int_4^{10}(x-4)\frac{x}{50}\,dx

Compute

Evaluate the two expected payments and subtract.

\operatorname{E}[Y_1]=\frac{284}{75}=3.7866667\ldots

\operatorname{E}[Y_2]=\frac{72}{25}=2.88

\left|\operatorname{E}[Y_1]-\operatorname{E}[Y_2]\right|=\frac{68}{75}=0.9066667\ldots

Answer

The expected-payment difference rounds to 0.91.

\boxed{0.91\quad\text{(D)}}

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.

AThe value 0.32 is twice F_X(4)=0.16. It manipulates the probability of a sublimit loss instead of integrating the two payment amounts.
BThe value 0.64 is 4F_X(4). It assigns the limit to losses below four and omits the actual payments above four, so it is not either expected payment or their difference.
CThe value 0.79 results from rounding the deductible-plan expectation 2.88 to 3 before subtracting it from 3.7867. The expected payments must retain precision until the final difference.
EThe value 1.12 is 4-2.88. It assumes the limited policy always pays its maximum 4, even though losses below four receive their actual smaller amounts.

Original practice · fully worked

Original variant: expected reimbursement from a three-part support rule

A repair cost X, measured in thousands, equals 2, 7, or 12 with probabilities 0.22, 0.51, and 0.27. A support contract reimburses 60% of the first 6, plus 30% of any amount above 6. It also adds 1.5 when the cost exceeds 10. Calculate the expected reimbursement.

A
0.405
B
0.639
C
3.072
D
3.975
E
4.116

Answer and calculation

The three possible reimbursements are 1.2, 3.9, and 6.9. Weighting them by probabilities 0.22, 0.51, and 0.27 gives an expected reimbursement of 4.116, which is choice E.

Setup

Express the three-part reimbursement as one payment function.

R(x)=0.6\min(x,6)+0.3(x-6)_++1.5I(x>10)

Model

Evaluate the payment function at each possible repair cost.

R(2)=1.2,\qquad R(7)=3.6+0.3=3.9

R(12)=3.6+1.8+1.5=6.9

Compute

Weight the three reimbursements by the cost probabilities.

\operatorname{E}[R]=0.22(1.2)+0.51(3.9)+0.27(6.9)

=0.264+1.989+1.863=4.116

Answer

The expected reimbursement is 4.116 thousand.

\boxed{4.116\quad\text{(E)}}

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How to solve this Expected Value question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: expected reimbursement from a three-part support rule

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables