Worked 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 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 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)}}

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 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

Answer and calculation

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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Next steps for Exam P Univariate Random Variables

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How to solve this Variance question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: upload attempts until acceptance

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables