Verified solution · Exam P

SOA Exam P Sample Question #379 Solution

This Exam P sample reference tests Continuous Random Variables. This is a median calculation for a decreasing density on a bounded interval. Integrating the upper tail gives survival (4-m)⁴⁄²⁵⁶; setting it to one half yields m=4 minus the fourth root of 128, approximately 0.636, so choice B.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Continuous Random Variables question

Setup

For a continuous variable, a median m leaves probability one half above and one half below. The upper-tail integral is especially short here.

S(x)=\int_x^4\frac{(4-u)^3}{64}\,du

Model

Evaluate the survival function and impose the median condition.

S(x)=\frac{(4-x)^4}{256}

S(m)=\frac12

Compute

Solve the resulting fourth-power equation using the root that lies in the support.

\frac{(4-m)^4}{256}=\frac12

(4-m)^4=128

m=4-\sqrt[4]{128}=0.6364143390

Answer

The exact median is four minus the fourth root of 128.

\boxed{4-\sqrt[4]{128}\quad\text{(B)}}

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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.

AZero is the lower support boundary, where no cumulative probability has yet accrued. A median must divide the total probability into equal halves.
CThe value 0.8 is the mean of this scaled Beta(1,4) distribution. The decreasing density is right-skewed, so its mean and median differ.
DSetting the density height f(m) equal to 0.5 gives (4-m)³=32 and hence 4-cuberoot(32). A median is defined by cumulative probability, not density height.
EThe support midpoint is 2, which would be the median for a symmetric uniform model. The supplied density places much more mass near zero.

Original practice · fully worked

Original variant: median absolute calibration error

A calibration error X is uniformly distributed from -3 to 5. Let Y=|X| be the absolute calibration error. Calculate the median of Y.

A
1.0
B
2.0
C
2.5
D
3.0
E
4.0

Variant answer in brief

For 0≤y≤3, the event |X|≤y covers the interval from -y to y, of length 2y within the eight-unit support. Setting 2y/8=1/2 gives median 2, choice B.

Setup

Translate an upper bound on the absolute error back to an interval for X.

\{Y\le y\}=\{-y\le X\le y\}

Model

For y between zero and three, both interval endpoints lie inside the support of X.

F_Y(y)=\frac{y-(-y)}{5-(-3)}=\frac y4,\qquad 0\le y\le3

Compute

Set the absolute-error CDF equal to one half.

F_Y(m)=\frac m4=\frac12

m=2

Answer

The median absolute calibration error is 2.

\boxed{2.0\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #379 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Continuous Random Variables question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: median absolute calibration error

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution