Worked solution · Exam P

SOA Exam P Sample Question #714 Solution

This Exam P sample reference tests Continuous Random Variables. Integrating after the substitution u=ln x gives F(x)=(ln x/ln 80)². Inverting F(x)=0.75 yields x=80 to the power (√(0.75))=44.4759, corresponding to choice C.

Independent analysis by ActuaryProof · reviewed July 30, 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 Random Variables question

Setup

Integrate the density from the lower endpoint, using the logarithm as the natural substitution.

u=\ln t,\qquad du=\frac{dt}{t}

Model

The accumulated probability is a squared ratio of logarithms.

F(x)=\int_1^x\frac{2\ln t}{(\ln 80)^2t}\,dt

F(x)=\frac{(\ln x)^2}{(\ln 80)^2},\qquad 1<x<80

Compute

Set the CDF equal to the requested cumulative probability and take the positive logarithmic root.

\left(\frac{\ln x}{\ln 80}\right)^2=0.75

\ln x=\sqrt{0.75}\,\ln 80

x=80^{\sqrt{0.75}}=44.4758518\ldots

Answer

The calculated percentile rounds to the value under choice C.

\boxed{x_{0.75}\approx 44.48\quad\text{(C)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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.

AThis is approximately 80 to the power (√(0.50)), so it computes the median instead of the 75th percentile.
BThis uses ln(x)/ln(80)=0.75 and forgets that the CDF is the square of that ratio.
DThis multiplies the support length 79 by 0.75 and forgets to add the lower endpoint; linear interpolation is not valid for this density.
EThis takes roughly 75% of the upper endpoint, again treating the distribution as though its quantiles were linear in x.

Original practice · fully worked

Original variant: compressed archive size

A compressed research archive has size S, measured in gigabytes, between 1 and 64. Its cumulative distribution function is F(s)=[log_2(s)/6]³ on that interval. Calculate the 12.5th percentile of S.

A
1.414
B
1.682
C
4.351
D
8.000
E
53.401

Answer and calculation

Set the supplied CDF equal to 0.125. Taking the cube root gives log_2(s)/6=0.5, so s=2³=8 and choice D is correct.

Setup

A percentile is found by equating the CDF to its target cumulative probability.

F(s_{0.125})=0.125=\frac18

Model

Insert the given CDF and take the cube root.

\left(\frac{\log_2 s}{6}\right)^3=\frac18

\frac{\log_2 s}{6}=\frac12

Compute

Undo the logarithm after restoring the scale factor.

\log_2 s=3

s=2^3=8

Answer

The archive-size percentile is 8 gigabytes.

\boxed{s_{0.125}=8\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Continuous Random Variables question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: compressed archive size

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables