Worked solution · Exam P

SOA Exam P Sample Question #595 Solution

This Exam P sample reference tests Uniform Distribution. The smallest threshold exceeded by no more than 5% of observations is the 95th percentile. For a uniform interval from 1.1 to 8.6, this is 1.1+0.95(7.5)=8.225, so choice D is correct.

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 Uniform Distribution question

Setup

Translate the upper-tail requirement into a cumulative probability.

\Pr(X>y)\le0.05

F(y)\ge0.95

Model

At the minimum admissible threshold, equality holds in the continuous uniform CDF.

F(y)=\frac{y-1.1}{8.6-1.1}=0.95

Compute

Solve the percentile equation.

y=1.1+0.95(7.5)=8.225

Answer

The minimum threshold is 8.225 minutes.

\boxed{8.225\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.

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.

AThis calculates the 5th percentile, 1.1+0.05(7.5), instead of the 95th percentile required by an upper 5% tail.
BThis multiplies the interval width by 0.95 but forgets to add the lower endpoint 1.1.
CThis computes 0.95(8.6), treating the interval as though it began at zero.
EThis subtracts 5% of the lower endpoint from 8.6 instead of subtracting 5% of the full interval width.

Original practice · fully worked

Original variant: spread from uniform quartiles

A sensor latency is uniformly distributed on an unknown interval. Its 25th percentile is 14 milliseconds and its 75th percentile is 26 milliseconds. Calculate the standard deviation of the latency.

A
6.000
B
6.928
C
12.000
D
13.856
E
24.000

Answer and calculation

For a uniform distribution, the interquartile range equals one-half of the support width. The 12-millisecond interquartile range therefore gives width 24 and standard deviation 24/√(12)=6.9282, so choice B is correct.

Setup

Relate uniform percentiles to the unknown support width w.

q_{0.75}-q_{0.25}=(0.75-0.25)w

Model

Use the observed interquartile range to recover the full width.

26-14=0.5w

w=24

Compute

Apply the continuous uniform standard-deviation formula.

\operatorname{SD}(X)=\frac{w}{\sqrt{12}}

\operatorname{SD}(X)=\frac{24}{\sqrt{12}}=6.9282032303\ldots

Answer

The latency standard deviation rounds to 6.928 milliseconds.

\boxed{6.928\quad\text{(B)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Uniform Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: spread from uniform quartiles

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables