Worked solution · Exam P

SOA Exam P Sample Question #344 Solution

This Exam P sample reference tests Order Statistics. The population cumulative distribution is the square of the observation. The median of three therefore has density 12 times the difference between its third and fifth powers. Its first two moments are 24/35 and 1/2, giving variance approximately 0.030 and choice B.

Independent analysis by ActuaryProof · reviewed July 31, 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 Order Statistics question

Setup

Find the population cumulative distribution on the unit interval.

F(x)=\int_0^x2t\,dt=x^2

Model

Use the density of the second order statistic from a sample of three.

g(y)=6F(y)[1-F(y)]f(y)

g(y)=12y^3(1-y^2),\qquad 0<y<1

Compute

Calculate the first two moments and subtract the squared mean.

\operatorname{E}[Y]=12\int_0^1(y^4-y^6)\,dy=\frac{24}{35}

\operatorname{E}[Y^2]=12\int_0^1(y^5-y^7)\,dy=\frac12

\operatorname{Var}(Y)=\frac12-\left(\frac{24}{35}\right)^2=\frac{73}{2450}=0.0297959\ldots

Answer

The variance of the sample median is approximately 0.030.

\boxed{0.030\quad\text{(B)}}

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.

AThe value 0.019 is approximately the population variance divided by three, the variance of a sample mean rather than a sample median.
CThe value 0.056 is the variance of one observation from the original distribution, not the variance of the median order statistic.
DThe value 0.500 is the second moment of the sample median. The squared mean must still be subtracted.
EThe value 0.714 cannot be the variance of a variable confined to the unit interval; any such variance is at most one quarter.

Original practice · fully worked

Original variant: expected span between inspection marks

Four inspection marks are placed independently at random along a cable ten metres long. The covered span is the distance between the leftmost and rightmost marks. Calculate the expected length of this span.

Answer and calculation

For four uniform marks along the cable, the expected leftmost position is 2 metres and the expected rightmost position is 8 metres. Linearity of expectation makes the expected span 6 metres, choice D.

Setup

Write the distribution functions of the rightmost and leftmost mark positions.

\Pr(M\le x)=\left(\frac{x}{10}\right)^4

\Pr(m>x)=\left(1-\frac{x}{10}\right)^4

Model

Use tail integrals for the two nonnegative order statistics.

\operatorname{E}[M]=\int_0^{10}\left[1-\left(\frac{x}{10}\right)^4\right]dx=8

\operatorname{E}[m]=\int_0^{10}\left(1-\frac{x}{10}\right)^4dx=2

Compute

Apply linearity of expectation to the range.

\operatorname{E}[M-m]=\operatorname{E}[M]-\operatorname{E}[m]=8-2=6

Answer

The expected covered span is 6 metres.

\boxed{6\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Order Statistics question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: expected span between inspection marks

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables