Worked solution · Exam P

SOA Exam P Sample Question #174 Solution

This Exam P sample reference tests Mixture Distributions. Equal territory weights produce marginal claim probabilities 0.80, 0.13, and 0.07. Their first two moments are 215 and 110500, so the standard deviation is √(110500-215²)=253.525, choice A.

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.

Read the solution Open official source

Independent solution

How to solve this Mixture Distributions question

Setup

Average each amount probability across the three equally represented territories.

\Pr(X=100)=0.80,\quad \Pr(X=500)=0.13,\quad \Pr(X=1000)=0.07

Model

Compute the first and second raw moments of the resulting mixture.

E[X]=100(0.80)+500(0.13)+1000(0.07)=215

E[X^2]=100^2(0.80)+500^2(0.13)+1000^2(0.07)=110500

Compute

Subtract the squared mean and take the square root.

\operatorname{Var}(X)=110500-215^2=64275

\operatorname{SD}(X)=\sqrt{64275}=253.5251

Answer

The listed standard deviation is 254.

\boxed{254\quad\text{(A)}}

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.

BThe value 291 is close to the standard deviation 290 within territory 3 alone. A randomly selected claim mixes all three equally represented territories.
CThe value 332 is √(E[X²])=√(110500)=332.4, the root-mean-square amount with the squared mean left unsubtracted.
DA standard deviation of 368 would imply second moment 368²+215²=181649, far above the verified mixture second moment 110500.
EA standard deviation of 396 would imply second moment 396²+215²=203041, again incompatible with the marginal probabilities.

Original practice · fully worked

Original variant: separate within-mode and between-mode variance

A production reading comes from mode A with probability 0.30 and mode B with probability 0.70. Conditional on the mode, the reading is normal with mean 10 and standard deviation 2 for A, or mean 20 and standard deviation 3 for B. Calculate the unconditional variance of the reading.

A
7.29
B
7.50
C
21.00
D
28.50
E
317.50

Answer and calculation

Average conditional variance contributes 7.5, while variance of the two conditional means contributes 21. Their sum is 28.5, choice D.

Setup

Separate variability within modes from variability between their means.

\operatorname{Var}(X)=E[\operatorname{Var}(X\mid M)]+\operatorname{Var}(E[X\mid M])

Model

Compute both terms of the total-variance identity.

E[\operatorname{Var}(X\mid M)]=0.3(2^2)+0.7(3^2)=7.5

\operatorname{Var}(E[X\mid M])=0.3(0.7)(20-10)^2=21

Compute

Add the within-mode and between-mode components.

\operatorname{Var}(X)=7.5+21=28.5

Answer

The unconditional variance is 28.50.

\boxed{28.50\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

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

How to solve this Mixture Distributions question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: separate within-mode and between-mode variance

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables