Worked solution · Exam P

SOA Exam P Sample Question #407 Solution

This Exam P sample reference tests Exponential Distribution. The median determines an exponential mean of 4.3281. Memorylessness leaves the residual lifetime exponential with that same mean, so its variance is 18.7323 and 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 Exponential Distribution question

Setup

Relate the exponential survival function to the supplied median.

\Pr(X>3)=e^{-3/\beta}=0.5

Model

Solve for the exponential mean parameter.

\beta=\frac{3}{\ln2}=4.328085123\ldots

Compute

By memorylessness, the future lifetime after any survived age has the original exponential distribution and variance.

X-a\mid X>a\sim\operatorname{Exp}(\text{mean }\beta)

\operatorname{Var}(X-a\mid X>a)=\beta^2

\beta^2=18.73232083\ldots

Answer

The remaining-lifetime variance rounds to 18.7.

\boxed{18.7\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.

See the cover and sample pages

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis subtracts the elapsed half-year from the unconditional mean. That is not valid for an exponential residual lifetime, and the result is not a variance.
BThe value 4.3 is the exponential mean, not its variance.
CThe value 9.0 squares the median 3; exponential variance is the square of the mean, not the median.
DThis first subtracts the elapsed half-year from the mean and then squares the result. Memorylessness does not reduce the exponential mean by the elapsed age.

Original practice · fully worked

Original variant: variability after a fixed initialization

A laboratory cycle consists of a fixed four-minute initialization followed by an independent exponentially distributed processing time with mean six minutes. Calculate the coefficient of variation of the total cycle time.

A
0.36
B
0.40
C
0.60
D
1.00
E
1.50

Answer and calculation

The total time has mean 4+6=10 and standard deviation 6 because the fixed initialization adds no variance. Its coefficient of variation is 6/10=0.60, choice C.

Setup

Let Z be the exponential processing time and T the full cycle time.

T=4+Z,\qquad E[Z]=6,\qquad \operatorname{Var}(Z)=36

Model

A fixed shift changes the mean but not the variance.

E[T]=4+6=10

\operatorname{SD}(T)=\sqrt{36}=6

Compute

Form standard deviation divided by mean.

\operatorname{CV}(T)=\frac{6}{10}=0.60

Answer

The coefficient of variation is 0.60.

\boxed{0.60\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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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: variability after a fixed initialization

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables