SOA Exam P Sample Question #622. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #622

This problem recovers the endpoints of a uniform distribution from its midpoint and spread. The endpoints are approximately 3.14445 and 29.57555, whose ratio is 9.40563, so choice E is correct.

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Independent solution

How to solve this Uniform Distribution question

Answer in brief

This problem recovers the endpoints of a uniform distribution from its midpoint and spread. The endpoints are approximately 3.14445 and 29.57555, whose ratio is 9.40563, so choice E is correct.

Setup

Write the lower and upper endpoints as a and b. For a continuous uniform variable, the median is the midpoint of the support.

\frac{a+b}{2}=16.36

a+b=32.72

Model

The standard deviation of a uniform distribution equals its support width divided by the square root of twelve.

\frac{b-a}{\sqrt{12}}=7.63

b-a=7.63\sqrt{12}=26.4310953235\ldots

Compute

Solve the sum-and-difference equations for the endpoints, then form the requested quotient.

a=16.36-\frac{26.4310953235}{2}=3.1444523382\ldots

b=16.36+\frac{26.4310953235}{2}=29.5755476618\ldots

\frac{b}{a}=9.4056275880\ldots

Answer

The endpoint ratio rounds to 9.41.

\boxed{9.41\quad\text{(E)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis results from substituting 7.63 itself for the variance in (b-a)^2/12=7.63 instead of first squaring the stated standard deviation. That gives a width of sqrt(12 times 7.63) and a ratio near 1.83.
BThis value understates the support width substantially. It is tempting if the midpoint equation is used correctly but the standard deviation is treated as an endpoint displacement without the uniform-distribution scale factor.
CThis also comes from an interval that is too narrow, typically after mixing a half-width formula with the full-width formula. The standard deviation must be converted to the entire width before the endpoints are recovered.
DUsing (b-a)^2/10 as the variance formula gives a width of 7.63 sqrt(10) and a ratio of about 6.62. The denominator for a uniform distribution is twelve, not ten.

Original practice · fully worked

Original variant: calibration-lag endpoint

A robotics team models calibration lag as uniformly distributed across an unknown interval. Maintenance records give an expected lag of 18 minutes and a variance of 12 square minutes. What is the upper endpoint of the interval?

A
6 minutes
B
12 minutes
C
18 minutes
D
24 minutes
E
30 minutes

Variant answer in brief

A uniform variance of 12 implies a support width of 12 minutes. Centering that interval at the mean of 18 puts its upper endpoint at 24 minutes, so choice D is correct.

Setup

Let l and u denote the unknown endpoints. The mean fixes the center of a uniform interval.

\frac{l+u}{2}=18

Model

Relate the given variance to the full interval width.

\operatorname{Var}(X)=\frac{(u-l)^2}{12}=12

Compute

The width is twelve, so each endpoint lies six minutes from the center.

u-l=\sqrt{144}=12

u=18+\frac{12}{2}=24

Answer

The upper endpoint is 24 minutes.

\boxed{24\text{ minutes}\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #622 — Verified Solution.” Verified 2026-07-30.

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