Worked solution · Exam P

SOA Exam P Sample Question #611 Solution

This Exam P sample reference tests Continuous Random Variables. This is a percentile calculation under a symmetric triangular density. Normalization gives c=1/9, and solving (x-5)²⁄¹⁸=0.30 on the rising half gives x=5+√(5.4)=7.32379, so choice E is correct.

Independent analysis by ActuaryProof · reviewed July 30, 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 pieces. Geometrically they form two congruent triangles, each with base three and height 3c.

1=2\left(\frac12\right)(3)(3c)=9c

c=\frac19

Model

Half of the total density lies below the center at eight, so the 30th percentile belongs to the rising left branch.

F(8)=\frac12

F(x)=\int_5^x\frac{t-5}{9}\,dt=\frac{(x-5)^2}{18},\qquad 5\le x\le8

Compute

Set the left-branch cumulative probability equal to 0.30 and use the positive distance from the lower endpoint.

\frac{(x-5)^2}{18}=0.30

(x-5)^2=5.4

x=5+\sqrt{5.4}=7.3237900077\ldots

Answer

The 30th percentile rounds to 7.32.

\boxed{7.32\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.

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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.

AThis places the percentile almost at the lower endpoint, where the triangular density is near zero and too little probability has accumulated.
BThis also understates the quantile by treating the rising density too much like a constant height instead of integrating its triangular area.
CThis is the 30%-of-range location 5+0.30(11-5)=6.80, which would apply to a uniform density rather than the stated triangular density.
DThis results from underestimating the area needed on the rising branch; the correct CDF is quadratic in the distance x-5, not linear.

Original practice · fully worked

Original variant: maximum strength-ratio median

Three independently manufactured fibers have strength ratios X1, X2, and X3, each with density f(x)=2x for 0<x<1. A batch is summarized by its strongest fiber, M=max(X1,X2,X3). Calculate the median of M.

A
0.5000
B
0.7071
C
0.7937
D
0.8909
E
0.9439

Answer and calculation

Each fiber has CDF x², so the maximum has CDF m⁶. Setting that CDF equal to one-half gives median 2⁻¹⁄⁶=0.890899, which selects choice D.

Setup

Integrate the common fiber density to obtain its distribution function.

F_X(x)=\int_0^x2t\,dt=x^2,\qquad 0<x<1

Model

The maximum is at most m exactly when all three independent fibers are at most m.

F_M(m)=\Pr(X_1\le m,X_2\le m,X_3\le m)

F_M(m)=F_X(m)^3=m^6

Compute

Set the maximum's CDF equal to one-half and solve for its median.

m^6=\frac12

m=2^{-1/6}=0.8908987181\ldots

Answer

The median strongest-fiber ratio is approximately 0.8909.

\boxed{0.8909\quad\text{(D)}}

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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: maximum strength-ratio median

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables