Worked solution · Exam P

SOA Exam P Sample Question #648 Solution

This Exam P sample reference tests Continuous Random Variables. Differentiate the cumulative distribution to obtain density 2x/25 on the interval from zero to five. Integrating x squared against this density gives a second raw moment of 12.50, so choice D 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

Differentiate the nonconstant part of the cumulative distribution to recover the density on its support.

f_X(x)=F_X'(x)=\frac{2x}{25},\qquad 0<x<5

Model

The requested second moment is the expectation of the square, not the variance or the square of the mean.

\operatorname{E}[X^2]=\int_0^5 x^2 f_X(x)\,dx

Compute

Substitute the density and evaluate the power integral.

\operatorname{E}[X^2]=\frac{2}{25}\int_0^5 x^3\,dx

\operatorname{E}[X^2]=\frac{2}{25}\left[\frac{x^4}{4}\right]_0^5=12.50

Answer

The second raw moment is 12.50.

\boxed{\operatorname{E}[X^2]=12.50\quad\text{(D)}}

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.

AThe value 1.39 is the variance: 12.50-(10/3)²=25/18. A second moment does not subtract the squared mean.
BThe value 3.33 is the mean E[X]=10/3, so it uses one power of X instead of the requested two.
CThe value 6.25 is the square of the support midpoint. The distribution is not concentrated at that midpoint.
EThe value 25.00 is the square of the upper endpoint, an upper bound on X squared rather than its average.

Original practice · fully worked

Original variant: coating depth profile

A laboratory measures the penetration depth X, in micrometers, of a protective coating. Its cumulative distribution is zero below 0, equals (x/8)³ from 0 through 8, and equals one above 8. Calculate the second moment of the penetration depth.

A
2.400
B
21.333
C
36.000
D
38.400
E
64.000

Answer and calculation

Differentiation gives density 3x squared divided by 512 on the support. The second-moment integral is 38.4 square micrometers, which selects choice D.

Setup

Differentiate the interior cumulative-distribution formula.

f_X(x)=\frac{3x^2}{8^3}=\frac{3x^2}{512},\qquad 0<x<8

Model

Weight the square of the depth by this density over the full support.

\operatorname{E}[X^2]=\int_0^8 x^2\frac{3x^2}{512}\,dx

Compute

Evaluate the fifth-power antiderivative.

\operatorname{E}[X^2]=\frac{3}{512}\left[\frac{x^5}{5}\right]_0^8

\operatorname{E}[X^2]=\frac{3(8^5)}{5(8^3)}=\frac{192}{5}=38.4

Answer

The second moment is 38.400 square micrometers.

\boxed{\operatorname{E}[X^2]=38.400\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: coating depth profile

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables