Worked solution · Exam P

SOA Exam P Sample Question #143 Solution

This Exam P sample reference tests Continuous Distributions. The unknown normalizing constant does not affect the maximizer. Differentiating the density kernel gives a sign change at x=cube root of 2, or 1.25992, which is choice C.

Independent analysis by ActuaryProof · reviewed October 6, 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

Maximize the density kernel because multiplying it by a positive normalizing constant cannot move its mode.

h(x)=\frac{x^2}{1+x^3},\qquad 0<x<5

Model

Differentiate the kernel and simplify the numerator.

h'(x)=\frac{2x(1+x^3)-3x^4}{(1+x^3)^2}

h'(x)=\frac{x(2-x^3)}{(1+x^3)^2}

Compute

The derivative is positive before the interior critical point and negative afterward.

2-x^3=0\iff x=\sqrt[3]{2}

\sqrt[3]{2}=1.2599210

Answer

The density reaches its unique maximum at approximately 1.26.

\boxed{1.26\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.

AAt the lower endpoint the density kernel tends to zero, so 0.00 cannot maximize it.
BThe value 0.79 is approximately the cube root of 1/2. It comes from inverting the stationary equation x³=2.
DAt 4.42 the derivative is negative and the kernel is about 0.22365, far below its value 0.52913 at the critical point.
EChoosing the upper endpoint amounts to maximizing the numerator alone. The denominator grows cubically, and the kernel at 5 is only 25/126.

Original practice · fully worked

Original variant: calibrate a density exponent from its mode

For x>0, a workload index has density c xᵃ exp(-x/2), where c is a normalizing constant and a is positive. Calibration records show that the density has its unique mode at x=6. Determine the exponent a.

Answer and calculation

The log-density derivative is a/x-1/2, so the mode is located at x=2a. Requiring that location to be 6 gives a=3, choice C.

Setup

Work with the logarithm of the density kernel; the normalizing constant does not depend on x.

\ell(x)=a\ln x-\frac{x}{2}+\text{constant}

Model

Differentiate and locate the stationary point in terms of the unknown exponent.

\ell'(x)=\frac{a}{x}-\frac12

\ell'(x)=0\iff x=2a

Compute

Match the modeled stationary point to the observed modal location.

2a=6

a=3

Answer

The calibrated density exponent is 3.

\boxed{a=3\quad\text{(C)}}

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Next steps for Exam P Univariate Random Variables

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How to solve this Continuous Distributions question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: calibrate a density exponent from its mode

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables