Worked solution · Exam P

SOA Exam P Sample Question #628 Solution

This Exam P sample reference tests Continuous Random Variables. This is a parameter-identification problem for a power density on the unit interval. The stated mean fixes the exponent at two, after which the upper-tail probability above one-half is 1-(1/2)³=0.875, so choice E is correct.

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

Write the mean as an integral under the stated one-parameter density.

\operatorname{E}[R]=\int_0^1 r(\alpha+1)r^\alpha\,dr=\frac{\alpha+1}{\alpha+2}

Model

Equate the theoretical mean to the supplied value and solve for the positive parameter.

\frac{\alpha+1}{\alpha+2}=\frac34

4(\alpha+1)=3(\alpha+2)

Compute

The parameter is two, so the cumulative distribution function on the unit interval is the third power. Take the complement at one-half.

\alpha=2,\qquad F_R(r)=r^3

\Pr(R>1/2)=1-F_R(1/2)=1-\left(\frac12\right)^3=\frac78=0.875

Answer

The probability rounds to 0.88.

\boxed{0.88\quad\text{(E)}}

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.

See the cover and sample pages

Error diagnosis

Why the other letters tempt

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

AThis treats the variable as uniform and copies the one-half cutoff into the answer, ignoring the increasing power density and the supplied mean.
BThis is approximately the probability above 0.75, since 1-0.75³=0.578125; it uses the mean as the cutoff instead of the required one-half.
CSubstituting the numerical mean 0.75 directly for the parameter gives 1-0.5¹·⁷⁵, approximately 0.70; the parameter must first be solved from the mean equation.
DThis simply reports the given expected value. A mean is not the same as the probability that the variable exceeds one-half.

Original practice · fully worked

Original variant: transformed optical-clarity score

A scanner assigns raw optical clarity U a density f(u)=2u for 0<u<1. Its dashboard reports the transformed score C=√(U), which expands the lower end of the scale. Calculate the probability that the displayed score exceeds 0.80.

A
0.2000
B
0.3600
C
0.4096
D
0.5904
E
0.6400

Answer and calculation

The displayed-score event C>0.80 is equivalent to U>0.64. Integrating the raw density above 0.64 gives 1-0.64²=0.5904, which selects choice D.

Setup

The raw clarity distribution is already normalized on the unit interval, and the dashboard transformation is increasing.

f_U(u)=2u,\qquad 0<u<1

C=\sqrt{U}

Model

Square the displayed-score threshold to express the event on the raw scale.

\{C>0.80\}=\{\sqrt{U}>0.80\}=\{U>0.64\}

Compute

Integrate the raw density over the converted upper-tail region.

\Pr(C>0.80)=\int_{0.64}^{1}2u\,du

\Pr(C>0.80)=1-0.64^2=1-0.4096=0.5904

Answer

The probability that the displayed score exceeds 0.80 is 0.5904.

\boxed{0.5904\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Continuous Random Variables question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: transformed optical-clarity score

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables