Worked solution · Exam P

SOA Exam P Sample Question #215 Solution

This Exam P sample reference tests Exponential Distribution. The payment median is positive because the zero-payment probability is below one half. The corresponding loss threshold is the loss median 4 ln(2), so subtracting the deductible gives 1.7726, choice A.

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 Exponential Distribution question

Setup

Write the ordinary-deductible payment as a positive-part transformation of the loss.

Y=(X-1)_+

\Pr(Y=0)=\Pr(X\le1)=1-e^{-1/4}=0.2212<0.5

Model

For a positive candidate median y, the event Y at most y is the same as X at most y+1.

\Pr(Y\le y)=\Pr(X\le y+1)=1-e^{-(y+1)/4}

Compute

Set the payment distribution function equal to one half and solve.

1-e^{-(m+1)/4}=0.5

m=4\ln2-1=1.7725887

Answer

The median payment rounds to 1.77.

\boxed{1.77\quad\text{(A)}}

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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.

BThe value 2.08 is approximately (4-1)ln(2). It incorrectly reduces the exponential mean by the deductible before calculating a median.
CThe value 2.12 is approximately 4exp(-1/4)-1. It first computes the mean payment and then subtracts the deductible a second time.
DThe value 2.77 is the median loss 4ln(2). It does not subtract the deductible from the positive loss threshold.
EThe value 3.12 is the mean payment E[(X-1)+]=4exp(-1/4), not its median.

Original practice · fully worked

Original variant: expected recording time after logger activation

A battery lifetime X is exponentially distributed with mean 4 hours. A diagnostic logger activates if the battery remains operating after hour 2 and then records for at most 3 additional hours. Given that the logger activates, calculate its expected recording time.

A
0.6065
B
1.2801
C
2.1105
D
2.4261
E
3.0000

Answer and calculation

Conditional on reaching hour 2, exponential memorylessness makes the remaining lifetime exponential with mean 4. Recording at most 3 additional hours gives expectation 4(1-exp(−3/4))=2.1105, choice C.

Setup

Write the capped recording duration and condition on logger activation.

R=\min\left((X-2)_+,3\right)

\{\text{logger activates}\}=\{X>2\}

Model

Memorylessness makes the remaining lifetime a fresh exponential variable Z with mean 4.

Z=(X-2)\mid X>2\sim\operatorname{Exponential}(\text{mean }4)

R\mid X>2=\min(Z,3)

Compute

Use the survival-integral formula for the limited mean.

E[R\mid X>2]=\int_0^3\Pr(Z>t)\,dt

=\int_0^3e^{-t/4}\,dt=4\left(1-e^{-3/4}\right)=2.1105338

Answer

Given activation, the expected recording time is approximately 2.1105 hours.

\boxed{2.1105\quad\text{(C)}}

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How to solve this Exponential Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: expected recording time after logger activation

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables