SOA Exam P Sample Question #595. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #595

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.

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

Answer in brief

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.

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)}}

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

Variant answer in brief

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/sqrt(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)}}

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ActuaryProof. “SOA Exam P Sample Question #595 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.