SOA Exam P Sample Question #628. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #628 — Verified Solution

Independent solution

How to solve this Continuous Random Variables question

Answer in brief

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.

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)}}

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 1-0.75^3=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 1-0.5^1.75, 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=sqrt(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

Variant answer in brief

The displayed-score event C>0.80 is equivalent to U>0.64. Integrating the raw density above 0.64 gives 1-0.64^2=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)}}

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 #628 — Verified Solution.” Verified 2026-07-30.

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