Verified solution · Exam P

SOA Exam P Sample Question #479 Solution

This Exam P sample reference tests Uniform Distribution. For a continuous uniform interval beginning at zero, the standard deviation is the upper endpoint divided by √(12). The first standard deviation fixes c=24; the second interval therefore has width 54 and standard deviation 9√(3)≈15.6, so choice D is correct.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Uniform Distribution question

Setup

Use the variance formula for a continuous uniform variable on an interval of width w.

\operatorname{Var}(U)=\frac{w^2}{12}

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

Model

The first interval has width c. Equate its stated standard deviation to the uniform formula and solve for the positive endpoint.

\frac{c}{\sqrt{12}}=4\sqrt{3}

c=(4\sqrt{3})(2\sqrt{3})=24

Compute

Insert the recovered constant into the width of the second interval and apply the same standard-deviation formula.

w_2=3c-18=3(24)-18=54

\operatorname{SD}_2=\frac{54}{\sqrt{12}}=\frac{54}{2\sqrt{3}}=9\sqrt{3}

9\sqrt{3}=15.588457\ldots

Answer

The second standard deviation rounds to 15.6.

\boxed{15.6\quad\text{(D)}}

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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 comes from the invalid operation |3(4√3) − 18| = 2.784…, which applies an endpoint shift directly to a standard deviation. Constants inside an interval endpoint change its width before the uniform formula is used.
BThis treats the recovered width 54 as though it were a variance and reports √54 = 7.348…. A uniform interval of width 54 has variance 54²/12, not 54.
CThis uses the incorrect rule SD = w/(3√3), producing 54/(3√3) = 6√3 ≈ 10.4. The continuous-uniform denominator is √12 = 2√3.
EThis ignores the 18-unit reduction in the second interval width. Using width 3c = 72 gives 72/√12 = 12√3 ≈ 20.8.

Original practice · fully worked

Original variant: calibrate a transformed sensor reading

A sensor reading X is uniformly distributed on an unknown interval. Its mean is 30, and its 90th percentile is 42. A processed reading is defined by Y=2X-5. Calculate the variance of Y.

A
60
B
75
C
150
D
300
E
900

Variant answer in brief

The 90th percentile is 0.4 interval widths above the midpoint, so the width is 30. Thus Var(X)=30²⁄¹²=75, and the scale factor 2 multiplies variance by 4, giving 300 and choice D.

Setup

Write the unknown support as an interval with midpoint 30 and width w.

X\sim\operatorname{Uniform}\left(30-\frac{w}{2},30+\frac{w}{2}\right)

Model

A 90th percentile lies 90% of the width above the lower endpoint, hence 40% of the width above the midpoint.

42=30-\frac{w}{2}+0.90w=30+0.40w

Compute

Recover the width, compute the uniform variance, and apply the squared scale factor in the affine transformation.

w=\frac{12}{0.40}=30

\operatorname{Var}(X)=\frac{30^2}{12}=75

\operatorname{Var}(Y)=\operatorname{Var}(2X-5)=2^2(75)=300

Answer

The processed reading has variance 300.

\boxed{300\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #479 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Uniform Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: calibrate a transformed sensor reading

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution