Worked solution · Exam P

SOA Exam P Sample Question #568 Solution

This Exam P sample reference tests Normal Distribution. Dividing the overall exceedance probability by the 0.8 claim probability gives a conditional normal tail of 0.229625. Its z-score is 0.7400824, so c=5000/0.7400824=6756, selecting D.

Independent analysis by ActuaryProof · reviewed July 30, 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

Separate claim occurrence from the conditional severity distribution.

0.1837=0.80\Pr(Y>55000\mid\text{claim})

Model

Convert the conditional exceedance into a standard normal tail.

\Pr\left(Z>\frac{55000-50000}{c}\right)=\frac{0.1837}{0.80}=0.229625

Compute

Invert the normal distribution and solve for the unknown standard deviation.

\frac{5000}{c}=\Phi^{-1}(0.770375)=0.7400823955

c=\frac{5000}{0.7400823955}=6756.0045

Answer

The standard deviation rounds to 6756.

\boxed{6756\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.

AThis ignores the 0.80 claim-occurrence factor, uses 0.1837 as the conditional normal tail, and obtains z about 0.90 and c about 5550.
BThis mistakes the overall nonexceedance probability 1-0.1837=0.8163 for a z-score, giving 5000/0.8163=6125.
CThis correctly forms the conditional CDF 0.770375 but then uses that probability itself as z, giving 5000/0.770375=6490.
EThis reads z as about 0.70, whose upper tail is about 0.242 rather than 0.229625; 5000/0.70 is about 7143, represented by the last choice.

Original practice · fully worked

Original variant: infer inactive time

A monitoring unit is either inactive or active on a given day. On an active day, its peak load is normally distributed with mean 80 and standard deviation 6. Across all days, the probability that the unit is active and its peak load exceeds 89 is 0.0534458. Calculate the probability that the unit is inactive.

A
0.0534
B
0.0668
C
0.2000
D
0.8000
E
0.9466

Answer and calculation

On an active day the threshold has z=(89-80)/6=1.5 and upper tail 0.0668072. Thus the active probability is 0.0534458/0.0668072=0.8, leaving inactive probability 0.2 and choice C.

Setup

First compute the threshold's standardized distance on an active day.

z=\frac{89-80}{6}=1.5

Model

Let a denote the active-day probability and factor the joint exceedance.

\Pr(Z>1.5)=0.0668072013

0.0534457610=a(0.0668072013)

Compute

Recover the active fraction and take its complement.

a=0.8000000000

\Pr(\text{inactive})=1-a=0.2000000000

Answer

The unit is inactive on 20% of days.

\boxed{0.2000\quad\text{(C)}}

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Next steps for Exam P Univariate Random Variables

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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: infer inactive time

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables