Worked solution · Exam P

SOA Exam P Sample Question #396 Solution

This Exam P sample reference tests Continuous Random Variables. This problem forms the loss distribution conditional on exceeding the deductible and then inverts its 90th percentile. The verified conditional quantile is 93.007768, which rounds to 93 and selects 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

Integrating the power density and applying normalization gives the loss CDF on its support.

F(x)=\left(\frac{x}{100}\right)^{5/4},\qquad 0\le x\le100

F(20)=\left(\frac{20}{100}\right)^{5/4}=0.1337480610\ldots

Model

Let p be the 90th percentile after restricting the distribution to losses above 20.

0.90=\Pr(X\le p\mid X>20)

0.90=\frac{F(p)-F(20)}{1-F(20)}

Compute

Solve first for the unconditional CDF level corresponding to the conditional percentile, then invert the power CDF.

F(p)=0.90+0.10F(20)=0.9133748061\ldots

p=100\left(0.9133748061\ldots\right)^{4/5}

p=93.0077680477\ldots

Answer

The conditional 90th percentile rounds to 93.

\boxed{93\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.

AIncreasing the density exponent 1/4 only to 1/2 during integration, instead of to 5/4, makes the same conditional calculation give 89.2498 and leads to 89.
BThis takes 90% of the upper endpoint, 0.90(100)=90, ignoring both the nonuniform density and the conditioning above the deductible.
CIgnoring the conditioning gives the unconditional 90th percentile 91.9166; truncating that value downward rather than rounding it gives 91.
DThis is the unconditional 90th percentile 100(0.90)⁽⁴⁄⁵⁾=91.9166 rounded to 92. Conditioning on X>20 shifts the percentile upward.

Original practice · fully worked

Original variant: percentile of a zero-truncated flag count

Four independent modules are each flagged with probability 0.50. Let N be the number flagged. Given that at least one module is flagged, calculate the 80th percentile of N.

A
0 flagged modules
B
1 flagged module
C
2 flagged modules
D
3 flagged modules
E
4 flagged modules

Answer and calculation

Conditioning removes the all-clear outcome and leaves 15 equally likely nonempty flag patterns. The conditional CDF is 10/15 at N=2 and 14/15 at N=3, so the 80th percentile is 3 and choice D.

Setup

With four fair independent flag indicators, all sixteen flag patterns are equally likely. The condition removes only the empty pattern.

N\sim\operatorname{Binomial}(4,0.50)

\Pr(N\ge1)=1-\frac1{16}=\frac{15}{16}

Model

Compute the conditional cumulative probabilities on either side of the 80% threshold.

\Pr(N\le2\mid N\ge1)=\frac{\binom41+\binom42}{15}=\frac{10}{15}

\Pr(N\le3\mid N\ge1)=\frac{\binom41+\binom42+\binom43}{15}=\frac{14}{15}

Compute

The conditional CDF remains below 0.80 at two and first exceeds 0.80 at three.

\frac{10}{15}<0.80<\frac{14}{15}

q_{0.80}=3

Answer

The conditional 80th percentile is three flagged modules.

\boxed{3\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: percentile of a zero-truncated flag count

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables