SOA Exam P Sample Question #649. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #649

Integrating the density gives cumulative distribution 1 minus exp(-4x squared). Setting this equal to one-half yields a median of 0.416277, which rounds to choice C.

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

Answer in brief

Integrating the density gives cumulative distribution 1 minus exp(-4x squared). Setting this equal to one-half yields a median of 0.416277, which rounds to choice C.

Setup

Integrate the density from zero to a positive argument to obtain its cumulative distribution.

F_X(x)=\int_0^x 8t e^{-4t^2}\,dt=1-e^{-4x^2},\qquad x>0

Model

A continuous median m places one-half of the probability at or below m.

F_X(m)=\frac12

1-e^{-4m^2}=\frac12

Compute

Isolate the exponential term, take logarithms, and retain the positive root because the support is positive.

4m^2=\log 2

m=\sqrt{\frac{\log 2}{4}}=0.416277\ldots

Answer

The median rounds to 0.416.

\boxed{m\approx0.416\quad\text{(C)}}

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 0.173 is log(2)/4, which is the computed value of m squared; taking the positive square root is still required.
BThe value 0.268 is approximately the first quartile, obtained by setting the cumulative probability to 0.25 instead of 0.50.
DThe value 0.693 reports log(2) itself, omitting both division by four and the square root.
EThe value 0.833 is sqrt(log(2)); it takes the square root but fails to divide by two afterward.

Original practice · fully worked

Original variant: acoustic detection radius

The distance R, in kilometers, from an ocean sensor to a detected acoustic event has density 18r exp(-9r squared) for positive r and zero otherwise. Determine the 75th percentile of the detection distance.

A
0.278
B
0.333
C
0.392
D
0.462
E
0.680

Variant answer in brief

The cumulative distribution is 1 minus exp(-9r squared). Equating it to 0.75 gives a percentile of 0.392470 kilometer, selecting choice C.

Setup

Integrate the density over its positive support.

F_R(r)=\int_0^r 18t e^{-9t^2}\,dt=1-e^{-9r^2}

Model

Let q be the 75th percentile and set its cumulative probability to 0.75.

1-e^{-9q^2}=0.75

Compute

Solve the exponential equation for the positive distance.

e^{-9q^2}=0.25

q=\sqrt{\frac{\log 4}{9}}=0.392470\ldots

Answer

The 75th-percentile detection radius is approximately 0.392 kilometer.

\boxed{q\approx0.392\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #649 — Verified Solution.” Verified 2026-07-30.

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