Verified solution · Exam P

SOA Exam P Sample Question #155 Solution

This Exam P sample reference tests Deductible Payments. A deductible between 60 and 200 removes the smallest claim and reduces the other two payments dollar for dollar. Setting the resulting expected payment 40-0.06d equal to 30 gives d=166.67, choice C.

Independent analysis by ActuaryProof · reviewed August 4, 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 Deductible Payments question

Setup

Let d be the ordinary deductible and write the expected payment from the three mutually exclusive loss amounts.

\operatorname{E}[(X-d)_+]=0.10(60-d)_++0.05(200-d)_++0.01(3000-d)_+

Model

A target payment of 30 requires a deductible above 60 but below 200, so only the two larger losses pay.

\operatorname{E}[(X-d)_+]=0.05(200-d)+0.01(3000-d)

\operatorname{E}[(X-d)_+]=40-0.06d

Compute

Equate the expected payment to its target and verify the assumed deductible interval.

40-0.06d=30

d=\frac{10}{0.06}=166.6667

60<166.6667<200

Answer

The deductible lies between 150 and 200.

\boxed{150\le d<200\quad\text{(C)}}

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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 a deductible of 100, the expected payment is 0.05(100)+0.01(2900)=34, not 30.
BFor deductibles from 100 up to but not including 150, the expected payment decreases from 34 toward 31 and never reaches 30.
DFor deductibles from 200 to 250, only the largest loss pays; the expected payment ranges from 28 down to 27.5, already below the target.
EAt deductibles of at least 250, the expected payment is no greater than 27.5 and continues to decrease.

Original practice · fully worked

Original variant: calibrate an exponential deductible

A repair cost follows an exponential distribution with mean 1000. A plan pays the amount of each repair cost above an ordinary deductible d, with no upper limit. Determine d if the expected plan payment per repair is 300.

A
300
B
700
C
1204
D
2303
E
3333

Variant answer in brief

For an exponential cost with mean 1000, the expected excess over d is 1000 exp(-d/1000). Equating this to 300 gives d=1203.97, choice C.

Setup

Use the exponential survival function to express the expected excess payment.

\Pr(X>x)=e^{-x/1000}

\operatorname{E}[(X-d)_+]=\int_d^\infty \Pr(X>x)\,dx

Model

Evaluate the tail integral and set it equal to the target payment.

\operatorname{E}[(X-d)_+]=1000e^{-d/1000}

1000e^{-d/1000}=300

Compute

Take logarithms and solve for the deductible.

e^{-d/1000}=0.3

d=-1000\log(0.3)=1203.9728

Answer

The required deductible rounds to 1204.

\boxed{1204\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #155 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Deductible Payments question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: calibrate an exponential deductible

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution