Worked solution · Exam P

SOA Exam P Sample Question #222 Solution

This Exam P sample reference tests Normal Distribution. The deductible has standardized value -1.1111 and cumulative probability 0.13326. A conditional 95th percentile therefore has unconditional CDF 0.13326+0.95(0.86674)=0.956663, giving a loss of 27,709 and choice B.

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 Normal Distribution question

Setup

Let d be the deductible and q the requested percentile among losses above d.

d=15000,\qquad z_d=\frac{15000-20000}{4500}=-1.1111111

F_X(d)=\Phi(z_d)=0.1332603

Model

Translate the conditional percentile equation into an unconditional cumulative probability.

\Pr(X\le q\mid X>d)=\frac{F_X(q)-F_X(d)}{1-F_X(d)}=0.95

F_X(q)=F_X(d)+0.95\left(1-F_X(d)\right)

Compute

Evaluate the adjusted normal percentile and transform back to the loss scale.

F_X(q)=0.9566630

z_q=\Phi^{-1}(0.9566630)=1.7132097

q=20000+4500(1.7132097)=27709.44

Answer

Rounded to the nearest hundred, the conditional percentile is 27,700.

\boxed{27700\quad\text{(B)}}

Test the method on unseen problems

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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 27,400 is 20,000+1.645(4,500), the unconditional 95th percentile. It ignores that the distribution is first truncated below the deductible.
CAt 28,100 the conditional CDF is approximately 0.95855. This choice comes from using the rough standardized value 1.80 instead of solving the adjusted-tail equation.
DAt 28,400 the remaining unconditional upper tail is about 0.03097, whereas the required tail is 0.05 P(X>15000)=0.04334. It removes too much upper-tail probability.
EThe value 28,800 is approximately 20,000+1.96(4,500). It uses the two-sided 95% critical value, which is the 97.5th unconditional percentile, not the requested conditional percentile.

Original practice · fully worked

Original variant: median after two-sided screening

A diagnostic score X has density f(x)=2x for values between 0 and 1, and zero elsewhere. Records are retained only when their scores lie strictly between 0.2 and 0.8. Calculate the median score among the retained records.

A
0.2000
B
0.5000
C
0.5477
D
0.5831
E
0.7071

Answer and calculation

The CDF is F(x)=x squared. The retained interval has probability 0.64-0.04=0.60, so its median has original CDF 0.04+0.30=0.34. Thus the conditional median is √(0.34)=0.5831, choice D.

Setup

Integrate the density and measure the probability retained by the two-sided screen.

F_X(x)=\int_0^x2u\,du=x^2

\Pr(0.2<X<0.8)=0.8^2-0.2^2=0.60

Model

Place half of the retained probability between the lower screen and the conditional median m.

\frac{F_X(m)-F_X(0.2)}{F_X(0.8)-F_X(0.2)}=0.50

\frac{m^2-0.04}{0.60}=0.50

Compute

Solve the conditional-median equation within the retained interval.

m^2=0.04+0.30=0.34

m=\sqrt{0.34}=0.5830952

Answer

The median retained score is approximately 0.5831.

\boxed{0.5831\quad\text{(D)}}

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How to solve this Normal Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: median after two-sided screening

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables