Verified solution · Exam P

SOA Exam P Sample Question #347 Solution

This Exam P sample reference tests Variance. Marginalizing the joint table gives a positive-test probability of 0.09. A Bernoulli variable with that success probability has standard deviation equal to the square root of 0.09 × 0.91, so its coefficient of variation is approximately 3.18 and choice D.

Independent analysis by ActuaryProof · reviewed August 4, 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 Variance question

Setup

Add the two joint cells in which the test indicator is one.

p=\Pr(Y=1)=0.05+0.04=0.09

Model

The test indicator is Bernoulli, so its mean and variance follow from its positive-test probability.

\operatorname{E}[Y]=p=0.09

\operatorname{Var}(Y)=p(1-p)=0.09(0.91)=0.0819

Compute

Divide the standard deviation by the mean.

\operatorname{CV}(Y)=\frac{\sqrt{0.0819}}{0.09}

=3.179797338\ldots

Answer

The coefficient of variation rounds to 3.18.

\boxed{3.18\quad\text{(D)}}

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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 reverses the coefficient, dividing the mean by the standard deviation, which gives approximately 0.31.
BThis divides the variance by the mean. For a Bernoulli variable that ratio is 0.91, not the coefficient of variation.
CThis is the marginal probability that the underlying condition indicator is zero. It uses the wrong variable and is not a variability measure.
EThis is the coefficient of variation of the underlying condition indicator, whose positive probability is 0.05, rather than of the test indicator.

Original practice · fully worked

Original variant: upload attempts until acceptance

A data terminal repeats independent upload attempts until the first accepted transmission. Each attempt is accepted with probability 0.36. Let N be the total number of attempts, including the accepted attempt. Calculate the coefficient of variation of N.

A
0.36
B
0.64
C
0.80
D
2.22
E
2.78

Variant answer in brief

The attempt count is geometric with mean 25/9 and standard deviation 20/9. Their ratio is 0.80, so choice C.

Setup

Use the geometric distribution that counts the successful attempt.

p=0.36,\qquad q=1-p=0.64

Model

Write the geometric mean and variance.

\operatorname{E}[N]=\frac1p=\frac{25}{9}

\operatorname{Var}(N)=\frac{q}{p^2}=\frac{400}{81}

Compute

Take the square root of the variance and divide by the mean.

\operatorname{SD}(N)=\frac{20}{9}

\operatorname{CV}(N)=\frac{20/9}{25/9}=\frac45=0.80

Answer

The coefficient of variation is 0.80.

\boxed{0.80\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #347 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Variance question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: upload attempts until acceptance

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution