Verified solution · Exam P

SOA Exam P Sample Question #207 Solution

This Exam P sample reference tests Cumulative Distribution Functions. Right continuity requires evaluating the cdf at the threshold with its upper branch, which gives F(V)=0.90. Therefore the strict upper tail is 1-F(V)=0.10, choice B.

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 Cumulative Distribution Functions question

Setup

A strict upper-tail probability uses the cdf value at the boundary, not the left-hand limit.

\Pr(L>V)=1-F(V)

Model

At the boundary, the branch defined for values at least V applies.

F(V)=1-\frac{1}{10}e^{-(V-V)/V}

Compute

Evaluate the exponential factor and take the complement.

F(V)=1-\frac{1}{10}=0.90

\Pr(L>V)=1-0.90=0.10

Answer

The strict upper-tail probability is 0.10.

\boxed{0.10\quad\text{(B)}}

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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 0.00 would incorrectly treat the lower polynomial branch as a bounded distribution and discard the upper-tail branch entirely.
CThe value 0.25 is 1-F(V-)=1-0.75. It equals P(L≥V) and includes both the point mass at V and the strict tail above V.
DThe value 0.75 is F(V-), the probability strictly below V, not the probability strictly above V.
EThe value 0.90 is F(V)=P(L≤V). It is the cdf value whose complement is required.

Original practice · fully worked

Original variant: variance created by a payment floor

A service credit X before adjustment is uniformly distributed between 0 and 200 credits. The issued credit is Y=max(X,100), so every value below the guarantee is raised to 100. Calculate Var(Y).

A
833.33
B
1041.67
C
3333.33
D
15625.00
E
16666.67

Variant answer in brief

The floor creates probability one half at 100 and leaves the original uniform density above 100. Direct first and second moments are 125 and 50000/3, so Var(Y)=50000/3-125²=3125/3=1041.67, choice B.

Setup

The floor collapses the lower half of the uniform distribution into an atom at 100.

Y=\max(X,100)

\Pr(Y=100)=\Pr(X\le100)=\frac12

Model

Combine the atom with the unchanged uniform density on the interval above the floor.

E[Y]=100\left(\frac12\right)+\int_{100}^{200}\frac{x}{200}\,dx

E[Y^2]=100^2\left(\frac12\right)+\int_{100}^{200}\frac{x^2}{200}\,dx

Compute

Evaluate the two moments and center the second moment.

E[Y]=125

E[Y^2]=\frac{50000}{3}

\operatorname{Var}(Y)=\frac{50000}{3}-125^2=\frac{3125}{3}=1041.6667

Answer

The variance of the issued credit is approximately 1041.67 square credits.

\boxed{1041.67\quad\text{(B)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #207 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Cumulative Distribution Functions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: variance created by a payment floor

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution