Worked solution · Exam P

SOA Exam P Sample Question #326 Solution

This Exam P sample reference tests Exponential Distribution. This is an exponential interval probability centered at the mean. The distribution has mean and standard deviation 20, so the interval is from 10 to 30; subtracting the two exponential tails gives e to the minus one-half minus e to the minus three-halves, or 0.3834 and choice B.

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

Identify the exponential scale from the density and use its equal mean and standard deviation.

X\sim\operatorname{Exponential}(\text{mean }20)

E[X]=20,\qquad \operatorname{SD}(X)=20

Model

Half a standard deviation is ten, so translate the centered-distance condition into endpoints on the positive support.

\lvert X-20\rvert<10\quad\Longleftrightarrow\quad 10<X<30

Compute

Use the exponential survival function at the two endpoints.

\Pr(10<X<30)=\Pr(X>10)-\Pr(X>30)

=e^{-10/20}-e^{-30/20}=e^{-1/2}-e^{-3/2}=0.3834004996

Answer

The interval probability rounds to 0.38.

\boxed{0.38\quad\text{(B)}}

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.

AThe value 0.34 is approximately Φ(1)-Φ(0). It substitutes a normal model and uses a one-sided interval one full standard deviation wide, rather than the stated exponential half-standard-deviation band.
CThe value 0.50 assumes that a mean-centered interval automatically contains half the probability. An exponential distribution is right-skewed, so its mass is not symmetric about its mean.
DThe value 0.68 is the familiar normal probability within one standard deviation of a mean. Both the distribution and the stated half-width differ here.
EThe value 0.95 invokes the normal two-standard-deviation rule. It does not evaluate the exponential probability between 10 and 30.

Original practice · fully worked

Original variant: mean absolute deviation of a self-check time

A device's self-check time X, measured in minutes, follows an exponential distribution with mean 12. Calculate the expected absolute difference between the self-check time and its mean, E[absolute value of X-12].

A
4.4146 minutes
B
8.8291 minutes
C
12.0000 minutes
D
17.6582 minutes
E
24.0000 minutes

Answer and calculation

For an exponential time with mean θ, the expected excess above θ is the integrated survival tail θ/e. Centering implies the expected shortfall below the mean is equal, so the mean absolute deviation is 2theta/e=24/e=8.8291 minutes and choice B.

Setup

Center the exponential time at its mean and split absolute deviation into positive and negative parts.

Y=X-12

\lvert Y\rvert=Y_++(-Y)_+

Model

Because E[Y]=0, the positive and negative deviation parts have equal expectations.

E[Y_+]=E[(-Y)_+]

E[\lvert Y\rvert]=2E[Y_+]

Compute

Evaluate the expected excess above the mean with the exponential survival function.

E[(X-12)_+]=\int_{12}^{\infty}\Pr(X>t)\,dt

=\int_{12}^{\infty}e^{-t/12}\,dt=12e^{-1}

E[\lvert X-12\rvert]=\frac{24}{e}=8.829106588

Answer

The mean absolute deviation is approximately 8.8291 minutes.

\boxed{8.8291\text{ minutes}\quad\text{(B)}}

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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: mean absolute deviation of a self-check time

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables