SOA Exam P Sample Question #648. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #648 — Verified Solution

Independent solution

How to solve this Continuous Random Variables question

Answer in brief

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.

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)}}

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

Variant answer in brief

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)}}

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ActuaryProof. “SOA Exam P Sample Question #648 — Verified Solution.” Verified 2026-07-30.

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