Worked solution · Exam P

SOA Exam P Sample Question #708 Solution

This Exam P sample reference tests Continuous Random Variables. The first density segment contains probability 0.40, so the median lies in the next segment. Accumulating another 0.10 at density 0.30 requires one-third of a year beyond 0.5, giving a median of 5/6 = 0.833333 year and choice B.

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 Continuous Random Variables question

Setup

A continuous median m satisfies F(m) = 0.5. First locate the density segment in which the cumulative probability reaches one-half.

\Pr(X\le 0.5)=0.80(0.5)=0.40

Model

Because 0.40 is below one-half, continue through the following segment, where the density is constant at 0.30.

F(m)=0.40+0.30(m-0.5),\qquad 0.5<m\le 1.5

Compute

Set the segment-specific cumulative distribution equal to one-half and solve.

0.40+0.30(m-0.5)=0.50

m-0.5=\frac{0.10}{0.30}=\frac{1}{3}

m=\frac{5}{6}=0.833333\ldots

Answer

The median lifetime rounds to 0.83 year.

\boxed{m\approx 0.83\text{ year}\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.

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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.

AAt 0.39 year the cumulative probability is only 0.80(0.39) = 0.312, so this value has not accumulated the required probability 0.50.
CAt 0.94 year the cumulative probability is 0.40 + 0.30(0.44) = 0.532; continuing too far into the second segment overshoots the median.
DThe value 1.38 is the rounded mean: direct integration gives E[X] = 1.375. A skewed distribution need not have its mean equal to its median.
EThe point 1.50 is a density breakpoint, but F(1.50) = 0.40 + 0.30(1.00) = 0.70, not 0.50.

Original practice · fully worked

Original variant: greenhouse sensor drift

The time R, in hundreds of operating hours, until a greenhouse sensor requires recalibration has density 0.30 for 0 ≤ R ≤ 1, density 0.20 for 1 < R ≤ 3, density 0.10 for 3 < R ≤ 6, and zero elsewhere. Calculate the median recalibration time.

A
1.2
B
1.6
C
2.0
D
2.4
E
2.8

Answer and calculation

The first interval contributes probability 0.30. Accumulating the remaining 0.20 at density 0.20 takes one more unit, so the median is 2.0 hundreds of hours and choice C.

Setup

Confirm that the stated pieces form a probability density and identify the cumulative mass at each breakpoint.

0.30(1)+0.20(2)+0.10(3)=0.30+0.40+0.30=1

Model

The first interval contains 0.30 probability, so the 50th percentile occurs in the second interval.

F(r)=0.30+0.20(r-1),\qquad 1<r\le 3

Compute

Solve the median equation inside that interval.

0.30+0.20(r-1)=0.50

r-1=1,\qquad r=2

Answer

The median is 2.0 hundreds of operating hours, or 200 hours.

\boxed{r=2.0\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Continuous Random Variables question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: greenhouse sensor drift

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables