Verified solution · Exam P

SOA Exam P Sample Question #392 Solution

This Exam P sample reference tests Continuous Random Variables. The two linear pieces form a triangle whose total unscaled area is 9, so the normalizing constant is 1/9. Integrating the rising branch over the requested interval gives 4/9=0.444444, choice E.

Independent analysis by ActuaryProof · reviewed August 4, 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

Normalize the two linear density pieces over their support.

1=c\left[\int_5^8(x-5)\,dx+\int_8^{11}(11-x)\,dx\right]

Model

The two integrals are equal triangular areas.

1=c\left(\frac92+\frac92\right)=9c

c=\frac19

Compute

Integrate the increasing branch between the two requested endpoints.

\Pr(6<X<8)=\frac19\int_6^8(x-5)\,dx

=\frac1{18}\left[(8-5)^2-(6-5)^2\right]=\frac49

Answer

The interval probability is approximately 0.444.

\boxed{0.444\quad\text{(E)}}

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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 2f(6)=2/9=0.222 uses a left-endpoint rectangle across the interval. The density rises linearly, so that rectangle misses the area above height f(6).
BThe value 1/2-2/9=5/18=0.278 starts with the correct left-half mass but subtracts the same incorrect rectangle estimate for the lower subinterval instead of its exact triangular area 1/18.
CThe value 1/3 is the peak density f(8), a height rather than the area over the interval.
DAn interval probability of 0.379 would require c=0.379/4=0.09475 because the unscaled requested area is four. That constant gives total mass 9c=0.85275 rather than one.

Original practice · fully worked

Original variant: tail probability in a mixed score model

A diagnostic score X equals zero with an unknown probability p. Conditional on X being positive, its density is f(x)=2x for 0<x<1. The unconditional mean of X is 0.40. Calculate P(X>0.50).

A
0.30
B
0.40
C
0.45
D
0.60
E
0.75

Variant answer in brief

The positive component has mean 2/3, so the overall mean identifies its mixture weight as 0.60. Its conditional tail above 0.50 is 0.75, giving unconditional probability 0.60(0.75)=0.45 and choice C.

Setup

Calculate the mean within the positive continuous component.

E[X\mid X>0]=\int_0^1x(2x)\,dx=\frac23

Model

Let w be the probability of drawing from the positive component and recover it from the unconditional mean.

0.40=w\left(\frac23\right)

w=0.60,\qquad p=0.40

Compute

Evaluate the positive-component tail and then apply its mixture weight.

\Pr(X>0.50\mid X>0)=\int_{0.50}^1 2x\,dx=1-(0.50)^2=0.75

\Pr(X>0.50)=0.60(0.75)=0.45

Answer

The unconditional upper-tail probability is 0.45.

\boxed{0.45\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #392 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

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: tail probability in a mixed score model

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution