Worked solution · Exam P

SOA Exam P Sample Question #388 Solution

This Exam P sample reference tests Mixture Distributions. The payment is a zero-inflated exponential variable. Its mean is 2 and its raw second moment is 32, so the variance is 32-2 squared=28 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 Mixture Distributions question

Setup

Let I indicate that a positive payment occurs, and let Z be the positive exponential amount.

\Pr(I=1)=0.25,\qquad E[Z]=8,\qquad E[Z^2]=2(8^2)=128

Y=IZ

Model

Average the first two conditional moments across the zero and positive components.

E[Y]=0.25(8)=2

E[Y^2]=0.25(128)=32

Compute

Convert the raw second moment to variance.

\operatorname{Var}(Y)=E[Y^2]-E[Y]^2

\operatorname{Var}(Y)=32-2^2=28

Answer

The payment variance is 28.

\boxed{28\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.

AThe value 2 is the mean payment, not its variance.
BSubtracting the mean 2 from the within-component contribution 16 gives 14. Variance instead uses the full raw second moment and subtracts the squared mean.
CThe value 16 retains only the average conditional variance within the positive-payment state and omits variation between zero and positive states.
EThe value 32 is the raw second moment. The squared overall mean, 4, still must be subtracted.

Original practice · fully worked

Original variant: a randomly reversed calibration reading

A raw calibration adjustment X has mean 3 and variance 4. Independently, a recorder reports X with probability 0.75 and its negative with probability 0.25. Calculate the variance of the reported value Y.

A
1.50
B
4.00
C
6.75
D
10.75
E
13.00

Answer and calculation

A random sign leaves the second moment at 13. The sign has mean 0.5, so the reported value has mean 1.5 and variance 13 minus 1.5 squared, or 10.75, choice D.

Setup

Represent the reported value as the raw adjustment multiplied by an independent random sign.

\Pr(S=1)=0.75,\qquad \Pr(S=-1)=0.25

E[S]=0.50,\qquad S^2=1

Model

Recover the raw second moment of X and use independence for the first moment of Y.

E[X^2]=\operatorname{Var}(X)+E[X]^2=4+9=13

E[Y]=E[S]E[X]=0.50(3)=1.50

Compute

Because the sign disappears when squared, Y and X share the same second moment.

E[Y^2]=E[S^2X^2]=E[X^2]=13

\operatorname{Var}(Y)=13-(1.50)^2=10.75

Answer

The reported-value variance is 10.75.

\boxed{10.75\quad\text{(D)}}

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Next steps for Exam P Univariate Random Variables

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How to solve this Mixture Distributions question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: a randomly reversed calibration reading

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables