Verified solution · Exam P

SOA Exam P Sample Question #161 Solution

This Exam P sample reference tests Continuous Distributions. Normalizing the power density gives a CDF proportional to x raised to a+1. The supplied CDF value calibrates that exponent near 2.5, and the upper tail beyond 4 is 0.4276, 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 Continuous Distributions question

Setup

Normalize the density over its finite support before using the probability statement.

1=\int_0^5 cx^a\,dx=\frac{c5^{a+1}}{a+1}

Model

After normalization, the CDF depends only on the ratio of the point to the upper endpoint.

F(x)=\left(\frac{x}{5}\right)^{a+1},\qquad 0<x<5

0.4871=(0.75)^{a+1}

Compute

Recover the power from logarithms and evaluate the complementary CDF at 4.

a+1=\frac{\log(0.4871)}{\log(0.75)}=2.5002804

\Pr(X>4)=1-(0.8)^{2.5002804}=0.4276024

Answer

The requested upper-tail probability rounds to 0.428.

\boxed{0.428\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 0.404 corresponds to using an exponent near 2.31 rather than the exponent fixed by the supplied CDF point. The log calibration gives 2.5003.
CThe value 0.500 treats 4 as a median without deriving that claim from the density. The calibrated CDF at 4 is about 0.5724.
DThe value 0.572 is F(4), the probability below 4. The question requires its complement.
EThe value 0.596 is the complement of distractor A and inherits the same miscalibrated power before also selecting the wrong side of the cutoff.

Original practice · fully worked

Original variant: infer a bounded power law from a conditional fraction

A quality score X lies between 0 and 12 and has cumulative distribution F(x)=(x/12)ᵏ on that interval, where k is positive. Among scores no greater than 9, the fraction no greater than 6 is 4/9. Determine the median score.

A
6.000
B
8.000
C
8.485
D
9.000
E
10.392

Variant answer in brief

The conditional fraction gives (2/3)ᵏ=4/9, so k=2. Solving F(m)=1/2 then gives m=12/√(2)=8.485, choice C.

Setup

Express the supplied conditional fraction as a ratio of CDF values.

\Pr(X\le6\mid X\le9)=\frac{F(6)}{F(9)}=\frac49

Model

Substitute the power-form CDF; the common endpoint cancels.

\frac{(6/12)^k}{(9/12)^k}=\left(\frac23\right)^k=\frac49

Compute

The calibration gives k=2, after which the median equation is immediate.

k=2

\left(\frac{m}{12}\right)^2=\frac12

m=\frac{12}{\sqrt2}=8.4852814

Answer

The median quality score is approximately 8.485.

\boxed{8.485\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #161 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Continuous Distributions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: infer a bounded power law from a conditional fraction

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution