Worked solution · Exam P

SOA Exam P Sample Question #220 Solution

This Exam P sample reference tests Mixture Distributions. The median lies in the lower mixture component, so 0.75(672/b)=0.50 and b=1,008. Weighting the two conditional uniform means gives 0.75(b/2)+0.25(2b)=882, choice D.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Mixture Distributions question

Setup

Let X denote the loss and use the fact that the lower component alone carries probability 0.75.

\Pr(X\le x)=0.75\frac{x}{b},\qquad 0\le x\le b

Model

Because the lower component has more than half of the total mass, the stated median is inside its interval.

0.50=0.75\frac{672}{b}

b=672\left(\frac{0.75}{0.50}\right)=1008

Compute

Average the midpoint of each uniform component using its mixture probability.

\operatorname{E}[X\mid\text{minor}]=\frac b2

\operatorname{E}[X\mid\text{major}]=\frac{b+3b}{2}=2b

\operatorname{E}[X]=0.75\left(\frac b2\right)+0.25(2b)=0.875b=882

Answer

The unconditional mean loss is 882.

\boxed{882\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.

ATreating the entire loss as uniform from 0 to 3b makes its median 1.5b, so b=448. Applying the actual mixture-mean factor 0.875 to that wrong scale gives 392.
BThe lower-component mean is b/2=504, whose nearest listed value is 512. Reporting it ignores both the component probability and the larger losses in the upper component.
CThe value 672 repeats the supplied median. A skewed two-component mixture need not have equal mean and median.
EThe value 1,008 is the recovered scale b. It is an interval endpoint, not the unconditional mean of the loss.

Original practice · fully worked

Original variant: variance of a calibrated exponential mixture

A processing time T follows a fast regime with an unknown probability p and a slow regime otherwise. Conditional on the fast regime, T is exponential with mean 1 minute; conditional on the slow regime, it is exponential with mean 2 minutes. The unconditional probability that T is at most ln(2) minutes is 0.4171573. Calculate Var(T).

A
0.6000
B
1.4000
C
2.2000
D
2.4400
E
4.4000

Answer and calculation

At ln(2), the fast and slow conditional CDFs are 0.5 and 1-1/√(2). The supplied mixture probability gives p=0.60. The first two unconditional moments are 1.4 and 4.4, producing variance 4.4-1.4 squared=2.44 and choice D.

Setup

Evaluate each regime's conditional CDF at the supplied time.

F_{\mathrm{fast}}(\ln2)=1-e^{-\ln2}=\frac12

F_{\mathrm{slow}}(\ln2)=1-e^{-\ln2/2}=1-\frac1{\sqrt2}=0.2928932

Model

Recover the fast-regime probability from the unconditional CDF value.

0.5p+0.2928932(1-p)=0.4171573

p=0.6000

Compute

Average the exponential first and second moments across regimes, then center the second moment.

E[T]=0.60(1)+0.40(2)=1.40

E[T^2]=0.60\{2(1^2)\}+0.40\{2(2^2)\}=4.40

\operatorname{Var}(T)=4.40-1.40^2=2.44

Answer

The unconditional processing-time variance is 2.44 square minutes.

\boxed{2.4400\quad\text{(D)}}

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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: variance of a calibrated exponential mixture

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables