SOA Exam P Sample Question #611. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #611

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+sqrt(5.4)=7.32379, so choice E is correct.

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Independent solution

How to solve this Continuous Random Variables question

Answer in brief

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+sqrt(5.4)=7.32379, so choice E is correct.

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)}}

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

Variant answer in brief

Each fiber has CDF x^2, so the maximum has CDF m^6. 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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ActuaryProof. “SOA Exam P Sample Question #611 — Verified Solution.” Verified 2026-07-30.

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