SOA Exam P Sample Question #730. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #730

Exponential quantiles convert the stated percentile gap into the distribution mean through a logarithmic ratio. The resulting probability is approximately 0.776868, corresponding to choice A.

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

Answer in brief

Exponential quantiles convert the stated percentile gap into the distribution mean through a logarithmic ratio. The resulting probability is approximately 0.776868, corresponding to choice A.

Setup

Write the exponential quantile function in terms of the mean beta.

q_p=-\beta\log(1-p)

Model

Subtract the median from the 75th percentile and simplify the logarithms.

q_{0.75}-q_{0.50}=\beta\bigl(\log4-\log2\bigr)=\beta\log2

\beta\log2=1.3863

Compute

Recover the mean and evaluate the exponential distribution function at three.

\beta=\frac{1.3863}{\log2}=2.000008135\ldots

\Pr(X<3)=1-e^{-3/\beta}=0.7768684784\ldots

Answer

Rounding to three decimals selects choice A.

\boxed{\Pr(X<3)\approx0.777\quad\text{(A)}}

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 0.793 implies an exponential mean of about 1.9047 through beta=-3/log(1-p); that mean does not reproduce the given percentile gap.
CThe value 0.810 implies a mean near 1.8064, again contradicting beta log(2)=1.3863.
DThe value 0.864 implies a mean near 1.5037 and results from an incorrect quantile-spacing conversion.
ETreating 1.3863 itself as the mean gives 1-e^{-3/1.3863}=0.885, but the supplied number is a percentile difference, not beta.

Original practice · fully worked

Original variant: protective coating lifetime

The time T until a protective coating develops its first microfracture is exponential. Laboratory records show that the 90th percentile exceeds the median by 12.8755 hours. Calculate the probability that a coating develops a microfracture within six hours.

A
0.372
B
0.472
C
0.528
D
0.658
E
0.777

Variant answer in brief

The 90th-percentile-to-median gap equals theta log(5), so the exponential mean is approximately eight hours. Evaluating the six-hour CDF gives approximately 0.527634, which selects choice C.

Setup

Express the two relevant quantiles using the exponential mean theta.

q_{0.90}=-\theta\log(0.10),\qquad q_{0.50}=-\theta\log(0.50)

Model

Take their difference and reduce the logarithmic ratio.

q_{0.90}-q_{0.50}=\theta\log\left(\frac{0.50}{0.10}\right)=\theta\log5

\theta\log5=12.8755

Compute

Solve for the mean and evaluate the probability of failure by six hours.

\theta=\frac{12.8755}{\log5}=7.999997950\ldots

\Pr(T<6)=1-e^{-6/\theta}=0.5276335380\ldots

Answer

The probability rounds to the value under choice C.

\boxed{\Pr(T<6)\approx0.528\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #730 — Verified Solution.” Verified 2026-07-30.

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