SOA Exam P Sample Question #609. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #609 — Verified Solution

Independent solution

How to solve this Distribution Functions question

Answer in brief

This is a distribution-function normalization followed by an upper-tail calculation. The endpoint condition gives c=1/30, the loss distribution has F(3.2)=0.448, and payment occurs in the complementary event with probability 0.552, so choice D is correct.

Setup

A cumulative distribution must reach one at the upper endpoint of its finite support. Use that requirement to determine the constant.

F(5)=c(5^2+5)=30c=1

c=\frac1{30}

Model

A positive payment under an ordinary deductible occurs exactly when the loss exceeds the deductible.

\Pr(\text{positive payment})=\Pr(X>3.2)=1-F(3.2)

Compute

Evaluate the distribution function at the deductible and take its complement.

F(3.2)=\frac{3.2^2+3.2}{30}=\frac{13.44}{30}=0.448

\Pr(X>3.2)=1-0.448=0.552

Answer

The insurer makes a positive payment with probability 0.552.

\boxed{0.552\quad\text{(D)}}

Error diagnosis

Why the other letters tempt

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

ATreating the supplied polynomial CDF as though it were a density and normalizing its integral gives an erroneous lower-tail probability near 0.296.
BThis treats the loss as uniform on the support and uses the remaining length (5-3.2)/5=0.36, ignoring the stated distribution function.
CThis is F(3.2), the probability that the loss does not exceed the deductible. It is the complement of the positive-payment event.
EThis is the complement of choice A, obtained by treating the CDF expression as a density before forming the upper tail.

Original practice · fully worked

Original variant: transformed polymer response

A normalized polymer response X has density f(x)=k x(2-x) for 0<x<2 and zero elsewhere. A monitoring system reports Y=(X-1)^2 rather than X itself. Calculate the probability that the reported value is at most 1/4.

A
0.2500
B
0.3125
C
0.5000
D
0.6875
E
0.7500

Variant answer in brief

Normalizing the response density gives k=3/4. The event Y at most 1/4 is the central interval 1/2 at most X at most 3/2, whose integrated probability is 11/16=0.6875, so choice D is correct.

Setup

Determine the density constant from its total integral over the support.

1=k\int_0^2 x(2-x)\,dx=k\left(\frac43\right)

k=\frac34

Model

Translate the nonmonotone transformation event back into an interval for X.

(X-1)^2\le\frac14

\frac12\le X\le\frac32

Compute

Integrate the normalized density across the central interval.

\Pr\!\left(Y\le\frac14\right)=\frac34\int_{1/2}^{3/2}x(2-x)\,dx

\Pr\!\left(Y\le\frac14\right)=\frac{11}{16}=0.6875

Answer

The transformed response is at most one-quarter with probability 0.6875.

\boxed{0.6875\quad\text{(D)}}

Continue by topic

Build a complete proof notebook

The Probability Proof Manual reorganizes verified solutions by syllabus skill and adds formula proofs, error patterns, and original practice.

Explore study products Get the free formula sheet

Cite this solution

ActuaryProof. “SOA Exam P Sample Question #609 — Verified Solution.” Verified 2026-07-30.

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