Worked 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 July 31, 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)}}

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.

See the cover and sample pages

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

Answer and calculation

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)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Continuous Random Variables question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: tail probability in a mixed score model

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables