Worked solution · Exam P

SOA Exam P Sample Question #440 Solution

This Exam P sample reference tests Exponential Distribution. For an exponential variable with mean θ, the median satisfies exp(−m/θ)=1/2. Therefore θ=400/ln2=577.08, which rounds to choice E.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Exponential Distribution question

Setup

Write the survival function of an exponential variable in terms of its mean parameter θ.

\Pr(X>x)=e^{-x/\theta}

Model

At a median, the upper-tail probability is one-half.

e^{-400/\theta}=\frac12

Compute

Take logarithms and isolate the mean parameter.

-\frac{400}{\theta}=\ln\left(\frac12\right)=-\ln2

\theta=\frac{400}{\ln2}=577.0780164\ldots

Answer

The mean rounds to 577.

\boxed{577\quad\text{(E)}}

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.

AThis sets the exponential mean equal to its median. With mean 400, the probability below 400 would be 1 − exp(−1) = 0.6321, not one-half.
BA mean of 446 would give P(X ≤ 400) = 1 − exp(−400 / 446) = 0.5922, so it does not reproduce the defining median probability.
CA mean of 492 would give P(X ≤ 400) = 0.5565. The logarithmic conversion from median to mean is still too small.
DA mean of 533 would give P(X ≤ 400) = 0.5279. Only division by ln 2 makes the cumulative probability exactly one-half.

Original practice · fully worked

Original variant: first failure among parallel components

A component lifetime is exponentially distributed, and the probability that it survives beyond 80 hours is 0.20. Three independent components begin operation together, and a system stops when the first component fails. Calculate the expected system lifetime.

A
9.94
B
16.57
C
26.67
D
49.71
E
80.00

Answer and calculation

The individual rate is ln5/80. The minimum of three independent exponential lifetimes has three times that rate, so its mean is 80/(3ln5)=16.57, choice B.

Setup

Let λ be the individual failure rate and calibrate it from the survival probability.

e^{-80\lambda}=0.20

\lambda=\frac{\ln5}{80}

Model

The first failure time is the minimum of the three independent lifetimes. Minimum exponential rates add.

M=\min(T_1,T_2,T_3)

M\sim\operatorname{Exponential}(3\lambda)

Compute

Invert the combined rate to obtain the expected system lifetime.

\operatorname{E}[M]=\frac1{3\lambda}=\frac{80}{3\ln5}

\operatorname{E}[M]=16.56893159\ldots

Answer

The expected time to the first component failure is 16.57 hours.

\boxed{16.57\quad\text{(B)}}

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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: first failure among parallel components

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables