Worked solution · Exam P

SOA Exam P Sample Question #435 Solution

This Exam P sample reference tests Exponential Distribution. For an exponential variable, the mean and standard deviation are the same scale parameter. The scale is √(7225)=85 and the median is 85 ln(2), so median minus mean is -26.0825 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 Exponential Distribution question

Setup

Let beta denote the exponential mean. Its variance is beta squared.

\beta^2=7225

\beta=85

Model

The median m leaves survival probability one half.

e^{-m/\beta}=\frac12

m=\beta\ln2=85\ln2

Compute

Subtract the mean from the median in the requested order.

m-\operatorname{E}[X]=85\ln2-85

85(\ln2-1)=-26.0824896524

Answer

The median is approximately 26 units below the mean.

\boxed{-26\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.

AThe value 7225(ln2-1)=-2217.0 uses the variance itself as the exponential scale instead of taking its square root.
CThe value zero treats the exponential distribution as symmetric with equal mean and median. Its right skew makes the median smaller.
DThe positive value 26 reverses the requested subtraction and computes mean minus median.
EThe value 2217 both uses the variance as the scale and reverses the subtraction order.

Original practice · fully worked

Original variant: recover a competing alarm's mean

Two independent alarms can end a laboratory run. The first alarm has an exponential waiting time with mean 4 minutes. The second alarm also has an exponential waiting time, with unknown mean m. The median time until the first alarm of either kind is 1 minute. Calculate m.

A
1.060
B
1.443
C
2.000
D
2.257
E
4.000

Answer and calculation

The minimum has rate 1/4+1/m and median ln(2)/(1/4+1/m). Setting that median to one gives m=1/(ln2-1/4)=2.25659, so choice D.

Setup

Convert the two exponential means to rates and add them for the minimum waiting time M.

\lambda_M=\frac14+\frac1m

M\sim\operatorname{Exponential}(\lambda_M)

Model

A median of one minute means the minimum survives one minute with probability one half.

\Pr(M>1)=e^{-\lambda_M}=\frac12

\lambda_M=\ln2

Compute

Remove the known alarm rate and invert the remaining rate.

\frac1m=\ln2-\frac14

m=\frac1{\ln2-1/4}=2.2565866237

Answer

The second alarm's mean waiting time is approximately 2.257 minutes.

\boxed{2.257\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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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: recover a competing alarm's mean

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables