Verified solution · Exam P

SOA Exam P Sample Question #709 Solution

This Exam P sample reference tests Continuous Random Variables. The unknown proportionality constant cancels when the upper-tail integral is divided by the integral over the full support. The ratio is ln(5)/ln(10) = 0.698970, which rounds to 0.70 and matches choice D.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Continuous Random Variables question

Setup

Represent the density by a positive constant times its stated kernel over the finite support.

f(x)=c\frac{x}{1+x^2},\qquad 0<x<3

Model

For a probability ratio, the same normalizing constant appears in the numerator and denominator and therefore cancels.

\Pr(X>1)=\frac{\int_1^3 x/(1+x^2)\,dx}{\int_0^3 x/(1+x^2)\,dx}

\int \frac{x}{1+x^2}\,dx=\frac{1}{2}\ln(1+x^2)

Compute

Evaluate the logarithmic antiderivative at the two sets of limits and simplify the ratio.

\Pr(X>1)=\frac{\tfrac12[\ln(10)-\ln(2)]}{\tfrac12[\ln(10)-\ln(1)]}

\Pr(X>1)=\frac{\ln(5)}{\ln(10)}=0.6989700043\ldots

Answer

The requested upper-tail probability rounds to 0.70.

\boxed{\Pr(X>1)\approx 0.70\quad\text{(D)}}

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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 0.15 is close to one-half of log base 10 of 2. It both keeps an antiderivative factor that should cancel and reports a lower-tail quantity instead of the requested upper tail.
BDropping the numerator x changes the kernel to 1/(1+x²); its normalized upper tail is about 0.3712, which points to B but solves a different density.
CReplacing the correct logarithmic ratio 10/2 = 5 by (3²-1)/(1+1²) = 4 gives log base 10 of 4 = 0.6021. The plus one in the antiderivative cannot be changed to a subtraction.
EKeeping an extra one-half in the lower-tail probability gives 0.5 log base 10 of 2 = 0.1505; taking its complement then gives 0.8495, near E, but the one-half cancels from the normalized ratio.

Original practice · fully worked

Original variant: acoustic signal intensity

A marine recorder assigns a dimensionless intensity score S between 0 and 4 to each detected acoustic event. The score has a continuous density proportional to s on that interval. Calculate the probability that a detected event has an intensity score greater than 3.

A
0.0625
B
0.2500
C
0.4375
D
0.5625
E
0.7500

Variant answer in brief

Normalizing a density proportional to s gives f(s) = s/8. Its probability above 3 is the ratio (4 squared − 3 squared)/4 squared = 7/16 = 0.4375, so choice C is correct.

Setup

Introduce the normalizing constant for the linearly increasing density.

f(s)=cs,\qquad 0<s<4

Model

Choose c so that the total area under the density is one.

1=c\int_0^4 s\,ds=c\frac{4^2}{2}=8c

c=\frac{1}{8}

Compute

Integrate the normalized density over scores above 3.

\Pr(S>3)=\int_3^4\frac{s}{8}\,ds

\Pr(S>3)=\frac{4^2-3^2}{16}=\frac{7}{16}=0.4375

Answer

The upper-tail probability is 0.4375.

\boxed{\Pr(S>3)=0.4375\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #709 — Verified Solution.” Verified 2026-07-30.

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

How to solve this Continuous Random Variables question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: acoustic signal intensity

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution