Worked solution · Exam P

SOA Exam P Sample Question #421 Solution

This Exam P sample reference tests Exponential Distribution. This problem uses the exponential survival curve to scale from a median to a higher percentile. The target survival probability is one eighth, three successive halvings, so its quantile is 3(2.7)=8.10 and choice C.

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

Let beta be the exponential mean and m the supplied median. A median leaves survival probability one half.

e^{-m/\beta}=\frac12

m=2.7

Model

At the requested percentile q, the survival probability is the complement of 0.875.

e^{-q/\beta}=1-0.875=0.125=\frac18

\frac18=\left(\frac12\right)^3

Compute

Because exponential survival powers correspond to proportional times, three median-length intervals produce the required survival.

e^{-q/\beta}=\left(e^{-m/\beta}\right)^3

q=3m=3(2.7)=8.10

Answer

The requested lifetime percentile is 8.10 years.

\boxed{8.10\quad\text{(C)}}

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.

AThe arithmetic 2.7/0.875=3.086 treats elapsed time as inversely proportional to a CDF probability. Direct substitution gives F(3.08)=0.5465, not 0.875.
BAt 4.73 years the calibrated exponential CDF is 1-2⁽⁻⁴·⁷³⁄²·⁷⁾=0.7031, so this value does not reach the requested percentile.
DThe value 10.80 is four median lengths. Its survival is (1/2)⁴=1/16, producing the 93.75th percentile rather than the 87.5th.
EAt 19.68 years the survival probability is 2⁽⁻¹⁹·⁶⁸⁄²·⁷⁾=0.00639, so the CDF is about 0.9936, far above 0.875.

Original practice · fully worked

Original variant: median life of a redundant power unit

A data logger is powered by two independent cells and remains active until both cells have failed. Each cell has an exponentially distributed lifetime with mean 6 hours. Calculate the median of the logger's operating time.

A
2.0794
B
4.1589
C
6.0000
D
7.3677
E
9.0000

Answer and calculation

The logger lifetime is the maximum of the two cell lifetimes, so its CDF is [1-exp(-t/6)]². Setting this equal to one half gives t=-6 ln(1-1/√(2))=7.36768 hours and choice D.

Setup

Let X and Y be the two cell lifetimes. Because either working cell keeps the logger active, the operating time is their maximum.

T=\max(X,Y)

F_X(t)=F_Y(t)=1-e^{-t/6}

Model

The maximum is at most t only when both independent cells have failed by t.

F_T(t)=\Pr(X\le t,Y\le t)

F_T(t)=\left(1-e^{-t/6}\right)^2

Compute

Set the system CDF equal to one half and solve for the positive time.

\left(1-e^{-t/6}\right)^2=\frac12

e^{-t/6}=1-\frac1{\sqrt2}

t=-6\ln\left(1-\frac1{\sqrt2}\right)=7.3676830638

Answer

The logger's median operating time is approximately 7.3677 hours.

\boxed{7.3677\quad\text{(D)}}

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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: median life of a redundant power unit

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables