Worked solution · Exam P

SOA Exam P Sample Question #221 Solution

This Exam P sample reference tests Mixture Distributions. For a positive median m, the mixture CDF is 0.20+0.80 Φ((m-1000)/400). Setting it to 0.50 requires a conditional normal percentile of 0.375, giving m=872.54 and choice C.

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 m be the unconditional median. It is positive because the atom at zero contains only 20% of the distribution.

F_C(m)=0.20+0.80\Phi\left(\frac{m-1000}{400}\right)

Model

Set the unconditional cumulative probability equal to one half and isolate the percentile inside the positive-loss component.

0.20+0.80\Phi\left(\frac{m-1000}{400}\right)=0.50

\Phi\left(\frac{m-1000}{400}\right)=\frac{0.30}{0.80}=0.375

Compute

Convert the conditional percentile to its standard-normal quantile and rescale.

\Phi^{-1}(0.375)=-0.3186394

m=1000+400(-0.3186394)=872.5443

Answer

The nearest listed median claim amount is 873.

\boxed{873\quad\text{(C)}}

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.

AUsing the no-loss probability 0.20 itself as the percentile inside the normal component gives 1000+400 Phi⁻¹(0.20)=663.35, nearest 663.
BUsing 0.50-0.20=0.30 as the conditional percentile without dividing by the positive-loss probability gives 790.24, nearest 790.
DA value near 994 corresponds to a conditional percentile near 0.494 and makes the mixture CDF about 0.595. It rounds toward the positive-loss median without performing the required probability rescaling.
EThe value 1,000 is the median of the normal component alone. The 20% atom at zero pulls the unconditional median below it.

Original practice · fully worked

Original variant: median of two overlapping score models

A scoring engine is selected at random. Engine A is selected with probability 0.40 and produces a score uniformly distributed from 0 to 1. Engine B is selected otherwise and has conditional distribution function F_B(x)=x² on that same interval. Calculate the median of the unconditional score.

A
0.5000
B
0.6000
C
0.6243
D
0.6385
E
0.7071

Answer and calculation

The mixture CDF is 0.40x+0.60x squared. Setting it to one half gives the positive root x=(-0.40+√(1.36))/1.20=0.6385, choice D.

Setup

Let M be the selected engine and X the resulting score. Average the two conditional CDFs.

F_X(x)=0.40F_{X\mid M=A}(x)+0.60F_{X\mid M=B}(x)

F_X(x)=0.40x+0.60x^2,\qquad 0\le x\le1

Model

Set the unconditional CDF equal to one half and form the quadratic median equation.

0.40m+0.60m^2=0.50

0.60m^2+0.40m-0.50=0

Compute

Use the positive root because the score lies between zero and one.

m=\frac{-0.40+\sqrt{0.40^2+4(0.60)(0.50)}}{2(0.60)}

m=0.6384919825

Answer

The unconditional score median is approximately 0.6385.

\boxed{0.6385\quad\text{(D)}}

Keep practicing

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: median of two overlapping score models

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables