Verified solution · Exam P

SOA Exam P Sample Question #511 Solution

This Exam P sample reference tests Continuous Distributions. Normalizing the power density gives cumulative distribution F(x)=x⁶. The probability that neither of two independent losses exceeds the threshold is therefore d¹². Setting its complement equal to P gives d=(1-P)⁽¹⁄¹²⁾, so choice E is correct.

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 Distributions question

Setup

Normalize the power density and integrate it to obtain the loss distribution function.

1=k\int_0^1x^5\,dx=\frac{k}{6}\quad\Longrightarrow\quad k=6

F(x)=\int_0^x6t^5\,dt=x^6

Model

No payment on either independent observation means both values are at most the threshold.

\Pr(\text{no payment on either})=F(d)^2=(d^6)^2=d^{12}

Compute

Complement the no-payment event and solve for the threshold.

P=1-d^{12}

d^{12}=1-P

d=(1-P)^{1/12}

Answer

The required expression is the twelfth root of one minus P.

\boxed{d=(1-P)^{1/12}\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.

AThis uses the density's exponent 5 as though it were the CDF exponent and also equates the below-threshold probability directly to P.
BThis doubles the incorrect exponent 5 for two observations but still uses P rather than the no-payment probability 1-P.
CThis uses the correct complement 1-P but treats F(d) as d⁵, producing a tenth root instead of a twelfth root.
DThis correctly obtains the exponent 12 but assigns P to the probability that both observations are below the threshold; that event has probability 1-P.

Original practice · fully worked

Original variant: lower quartile of an inspection minimum

A coating score X has density f(x)=2(1-x) for 0<x<1. Four scores are sampled independently, and M is their minimum. Calculate the 25th percentile of M.

A
0.0353
B
0.1340
C
0.1591
D
0.4588
E
0.9647

Variant answer in brief

One score has survival function (1-x)², so the four-score minimum has survival function (1-m)⁸. Setting its CDF to 0.25 gives m=1-0.75⁽¹⁄⁸⁾=0.035321, so choice A is correct.

Setup

Integrate the density and write the survival function of one score.

F_X(x)=1-(1-x)^2

S_X(x)=(1-x)^2

Model

The minimum exceeds m only if all four independent scores exceed m.

\Pr(M>m)=S_X(m)^4=(1-m)^8

F_M(m)=1-(1-m)^8

Compute

Set the minimum's CDF equal to one quarter and solve.

1-(1-m)^8=0.25

m=1-0.75^{1/8}=0.0353213700

Answer

The lower quartile of the minimum rounds to 0.0353.

\boxed{0.0353\quad\text{(A)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #511 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Continuous Distributions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: lower quartile of an inspection minimum

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution