SOA Exam P Sample Question #599. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #599

An exponential variable with mean mu has percentile q_p=-mu ln(1-p). The 75th and 25th percentiles are mu ln 4 and mu ln(4/3), whose difference simplifies to mu ln 3, so choice D is correct.

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

Independent solution

How to solve this Exponential Distribution question

Answer in brief

An exponential variable with mean mu has percentile q_p=-mu ln(1-p). The 75th and 25th percentiles are mu ln 4 and mu ln(4/3), whose difference simplifies to mu ln 3, so choice D is correct.

Setup

Write the exponential distribution function using its mean as the scale parameter.

F(x)=1-e^{-x/\mu},\qquad x\ge0

Model

Solve the percentile equation once in terms of a general cumulative probability.

p=1-e^{-q_p/\mu}

q_p=-\mu\ln(1-p)

Compute

Evaluate the two quartiles and simplify their difference.

q_{0.75}=\mu\ln4,\qquad q_{0.25}=\mu\ln\!\left(\frac43\right)

q_{0.75}-q_{0.25}=\mu\ln\!\left(\frac{4}{4/3}\right)=\mu\ln3

Answer

The interquartile range is mu times the natural logarithm of 3.

\boxed{\mu\ln3\quad\text{(D)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis places the mean in the denominator as though mu were a rate; it also has units inconsistent with a percentile difference.
BThis assumes the interquartile range equals the exponential mean and drops the logarithmic factor.
CThe quantity mu ln 2 is the exponential median and also the gap from the median to the 75th percentile, not the full interquartile range.
EThis confuses a squared scale such as variance with a difference of percentiles, which must have the same units as the random variable.

Original practice · fully worked

Original variant: earliest beacon signal

Three independent emergency beacons have exponential times to their first signal. Each beacon's individual median signal time is 12 minutes. Calculate the 75th percentile of the time until the first of the three beacons signals.

A
4 minutes
B
8 minutes
C
12 minutes
D
24 minutes
E
36 minutes

Variant answer in brief

An individual median of 12 implies rate ln(2)/12. The minimum of three independent exponential times has three times that rate. Its 75th percentile solves survival 0.25 and equals 8 minutes, so choice B is correct.

Setup

Convert the individual median into an exponential rate.

e^{-12\lambda}=0.5

\lambda=\frac{\ln2}{12}

Model

The earliest of three independent exponential times is exponential with the sum of their rates.

T_{\min}\sim\operatorname{Exponential}(3\lambda)

\Pr(T_{\min}>t)=e^{-3\lambda t}

Compute

At a 75th percentile, the survival probability is one-quarter.

e^{-3(\ln2/12)t}=0.25

t=\frac{\ln4}{3\ln2/12}=8

Answer

The 75th percentile of the earliest signal time is 8 minutes.

\boxed{8\text{ minutes}\quad\text{(B)}}

Continue by topic

Build a complete proof notebook

The Probability Proof Manual reorganizes verified solutions by syllabus skill and adds formula proofs, error patterns, and original practice.

Explore study products Get the free formula sheet

Cite this solution

ActuaryProof. “SOA Exam P Sample Question #599 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.