Worked solution · Exam P

SOA Exam P Sample Question #672 Solution

This Exam P sample reference tests Exponential Distribution. For an exponential lifetime with mean θ, the median is θ times log 2. The stated gap determines θ as 12.3838, so the variance is 153.358 and choice D.

Independent analysis by ActuaryProof · reviewed July 30, 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

Parameterize the exponential lifetime by its mean θ. Its variance is θ squared, so first recover θ from the mean-median gap.

\operatorname{E}[T]=\theta,\qquad \operatorname{Var}(T)=\theta^2

Model

The median solves a one-half survival probability and therefore equals θ times the natural logarithm of two.

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

m=\theta\ln 2

Compute

Use the stated difference to solve for the mean and then square that mean.

\theta-\theta\ln2=3.80

\theta=\frac{3.80}{1-\ln2}=12.38378714\ldots

\operatorname{Var}(T)=\theta^2=153.35818399\ldots

Answer

The variance rounds to 153 at the precision of the choices.

\boxed{\operatorname{Var}(T)\approx153\quad\text{(D)}}

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.

AThis is the recovered mean 12.38 rounded to a whole number; the question asks for the variance, which is the square of the mean.
BThis is approximately the square of the stated gap, 3.80² = 14.44, rather than the square of the exponential mean.
CEquating the gap to the median gives θ = 3.80 / ln(2) and a variance near 30; the gap is actually mean minus median.
EApproximating 1 − ln(2) by 0.25 gives θ ≈ 15.2 and variance near 231, but that approximation is far too coarse.

Original practice · fully worked

Original variant: maintenance ticket duration

The resolution time for an automated maintenance ticket is exponential. Its 75th percentile is 3.86294 hours greater than its mean. Calculate the variance, in squared hours, of the resolution time.

A
38.6
B
62.5
C
81.0
D
100.0
E
192.2

Answer and calculation

The exponential 75th percentile is θ log 4. The stated excess equals θ times (log 4 minus 1), giving θ approximately 10 and variance 100, so choice D.

Setup

Let θ be the mean resolution time and write the exponential quantile formula.

q_p=-\theta\ln(1-p)

q_{0.75}=\theta\ln4

Model

Subtract the mean θ from the 75th percentile and equate the result to the observed gap.

q_{0.75}-\operatorname{E}[T]=\theta(\ln4-1)=3.86294

Compute

Solve for θ and use the exponential variance identity.

\theta=\frac{3.86294}{\ln4-1}=9.99999\ldots

\operatorname{Var}(T)=\theta^2=99.9998\ldots

Answer

The variance is approximately 100 squared hours.

\boxed{100.0\quad\text{(D)}}

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: maintenance ticket duration

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables