Verified solution · Exam P

SOA Exam P Sample Question #515 Solution

This Exam P sample reference tests Continuous Random Variables. Normalizing the density gives the loss CDF 1-(1-x)⁴. Its median is about 0.159104 million, below the payment limit, so choice A is correct.

Independent analysis by ActuaryProof · reviewed August 4, 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

Normalize the polynomial density on its unit interval.

1=k\int_0^1(1-x)^3\,dx=\frac{k}{4}

k=4

Model

Integrating gives the loss CDF. The probability accumulated below the payment ceiling already exceeds one half, so the ceiling does not change the median.

F_X(x)=1-(1-x)^4,\qquad 0<x<1

F_X(0.225)=1-0.775^4=0.6392496>0.5

Compute

Solve the median equation and convert millions to ordinary currency units.

1-(1-m)^4=\frac12

m=1-2^{-1/4}=0.1591035847

10^6m=159{,}103.58

Answer

The nearest listed median payment is 159,000.

\boxed{159{,}000\quad\text{(A)}}

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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.

BThis uses the exponential approximation (1-x)⁴ approximately exp(−4x); solving exp(−4x)=0.5 gives x=ln(2)/4=0.1733 million, but the exact fourth root is required.
CThis uses the inaccurate numerical value 2⁻¹⁄⁴=0.813 instead of 0.840896, producing 1-0.813=0.187 million and placing about 56.3% of losses below the result.
DThis reports the uncapped mean E[X]=1/(1+4)=0.200 million. A mean is not a median for this right-skewed distribution.
EThis reports the payment ceiling itself. The ceiling carries only probability P(X at least 0.225)=0.775⁴=0.36075, and the CDF has crossed 0.5 before that point.

Original practice · fully worked

Original variant: median excess response time

A server response time T, in seconds, is exponential with mean 4. A monitoring panel records only the excess above one second, so its displayed value is W=max(T-1,0). Calculate the median of W.

A
0.000
B
0.693
C
1.773
D
2.773
E
4.000

Variant answer in brief

Only 22.1% of readings are floored at zero, so the median remains in the positive part. Subtracting one from the exponential median gives 4ln(2)-1=1.773, so choice C is correct.

Setup

Check whether the floor at zero contains at least half of the probability.

\Pr(W=0)=\Pr(T\le 1)=1-e^{-1/4}=0.221199<0.5

Model

Above the floor, W is the increasing transformation T-1. Its median is therefore the raw-time median shifted down by one second.

\Pr(W\le w)=\Pr(T\le w+1)=1-e^{-(w+1)/4},\qquad w>0

Compute

Set the transformed CDF equal to one half.

1-e^{-(w+1)/4}=\frac12

w=4\ln 2-1=1.772588722

Answer

The display has median 1.773 seconds.

\boxed{1.773\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #515 — 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 excess response time

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution