Worked solution · Exam P

SOA Exam P Sample Question #385 Solution

This Exam P sample reference tests Exponential Distribution. This problem calibrates an exponential survival curve from one failure percentile. Scaling its cumulative hazard to the 80th percentile gives t=14.425135 years, which rounds to 14.4 and choice D.

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 Exponential Distribution question

Setup

Translate the earlier failure probability into an exponential survival equation.

S(2)=0.80=e^{-2/\beta}

\beta=-\frac2{\ln(0.80)}=8.9628402354\ldots

Model

An 80th percentile leaves survival probability 0.20.

S(t)=0.20=e^{-t/\beta}

Compute

Eliminate the scale using the ratio of logarithmic survival levels.

\frac{t}{2}=\frac{\ln(0.20)}{\ln(0.80)}

t=2\frac{\ln(0.20)}{\ln(0.80)}=14.4251348780\ldots

Answer

The requested time rounds to 14.4 years.

\boxed{14.4\ \text{years}\quad\text{(D)}}

Test the method on unseen problems

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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 divides the correct logarithmic survival ratio by the two-year calibration period, producing 3.6063.
BThis reports the logarithmic ratio 7.2126 as years, although it counts two-year exposure blocks.
CThis scales the two-year period directly by the probability ratio four, giving eight years. Exponential probabilities scale through logarithms.
EA rare-event rate approximation of 0.10, combined with the coarse value 1.6 for the natural logarithm of five, gives 16.0 years. The approximation is too crude here.

Original practice · fully worked

Original variant: percentile under a two-stage failure rate

A laboratory component has failure hazard 0.10 per hour during its first five hours of operation. If it survives that long, its hazard becomes 0.20 per hour. Calculate the 80th percentile of its failure time.

A
5.000 hours
B
8.047 hours
C
10.547 hours
D
16.094 hours
E
21.094 hours

Answer and calculation

The first five hours contribute cumulative hazard 0.5, while the 80th percentile requires total hazard equal to the natural logarithm of five. Solving the second stage gives 10.547190 hours and choice C.

Setup

Convert the target CDF level into a cumulative-hazard target.

S(t_{0.80})=0.20

H(t_{0.80})=-\ln(0.20)=\ln5

Model

At the five-hour change point, the cumulative hazard is only one half, so the target lies in the second stage.

H(5)=0.10(5)=0.5<\ln5

H(t)=0.5+0.20(t-5),\qquad t>5

Compute

Set the two-stage cumulative hazard equal to the percentile target and solve.

0.5+0.20(t-5)=\ln5

t=5+\frac{\ln5-0.5}{0.20}=10.5471895622\ldots

Answer

The 80th percentile is approximately 10.547 hours.

\boxed{10.547\ \text{hours}\quad\text{(C)}}

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How to solve this Exponential Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: percentile under a two-stage failure rate

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables