Worked solution · Exam P

SOA Exam P Sample Question #557 Solution

This Exam P sample reference tests Exponential Distribution. An exponential lifetime has standard deviation equal to its mean. Solving the supplied five-year failure probability gives a mean of 9.788 years, so choice D is correct.

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

Let beta denote the exponential mean. The corresponding survival probability at five years is the complement of the supplied failure probability.

\Pr(X>5)=1-0.40=0.60

Model

Use the exponential survival function to connect this probability to the mean parameter.

\Pr(X>5)=e^{-5/\beta}=0.60

Compute

Take logarithms and isolate the positive scale parameter.

-\frac{5}{\beta}=\ln(0.60)

\beta=-\frac{5}{\ln(0.60)}=9.788075945

Answer

For an exponential variable, the standard deviation equals beta.

\boxed{\operatorname{SD}(X)=9.79\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 computes the rate, -ln(0.60)/5=0.1022, but reports that rate as the standard deviation.
BThis takes an extra square root, √(9.788)=3.128, as though the exponential mean were a variance.
CThis treats 0.40 as the survival probability instead of the failure probability, giving -5/ln(0.40)=5.46.
EThis divides 5 by 0.40 to get 12.50, incorrectly replacing the exponential distribution function by a linear proportion.

Original practice · fully worked

Original variant: calibration survival after inspection

The time until a monitoring unit loses calibration is exponentially distributed. Its 75th percentile is 10 days. A unit has already remained calibrated for 6 days. Calculate the probability that it remains calibrated for at least 5 additional days.

A
0.2176
B
0.2500
C
0.3750
D
0.5000
E
0.7500

Answer and calculation

The percentile statement gives S(10)=0.25. Memorylessness reduces the requested conditional probability to S(5)=√(0.25)=0.50, so choice D is correct.

Setup

Translate the 75th percentile into a survival probability.

F(10)=0.75,\qquad S(10)=0.25

Model

Exponential memorylessness makes the additional five-day survival probability independent of the six days already observed.

\Pr(T>11\mid T>6)=\Pr(T>5)=S(5)

Compute

The exponential survival over half as much time is the square root of the ten-day survival.

S(5)=S(10)^{5/10}=0.25^{1/2}=0.50

Answer

The unit has a one-half chance of lasting at least five more days.

\boxed{0.5000\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Exponential Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: calibration survival after inspection

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables