SOA Exam P Sample Question #732. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #732 — Verified Solution

Independent solution

How to solve this Central Limit Theorem question

Answer in brief

Apply the central limit theorem to the sample average, using an exponential standard deviation equal to its mean. Standardizing the requested cutoff gives z = 1.6 and an upper-tail probability of 0.054799, so the answer is C.

Setup

An exponential observation with the supplied mean has the same numerical standard deviation.

\mu=5,\qquad \sigma=5,\qquad n=64

Model

For a sample this large, approximate the standardized sample average by a standard normal random variable.

\overline X\approx N\!\left(5,\frac{25}{64}\right)

\operatorname{SE}(\overline X)=\frac{5}{\sqrt{64}}=0.625

Compute

Convert the cutoff to a standard-normal score and read the probability to its right.

z=\frac{6-5}{0.625}=1.6

\Pr(\overline X>6)\approx 1-\Phi(1.6)=0.0547992917\ldots

Answer

The matching listed probability is choice C.

\boxed{\Pr(\overline X>6)\approx 0.0548\quad\text{(C)}}

Error diagnosis

Why the other letters tempt

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

AThis is far too small for a cutoff only 1.6 standard errors above the mean; it is consistent with using the raw-observation scale in the denominator incorrectly.
BThis resembles the tail beyond about 2.56 rather than 1.6, which can result from dividing by the variance of the mean instead of its standard deviation.
DThis corresponds to a much smaller standardized distance and is tempting when the square-root adjustment for the sample size is only partly applied.
EThis is close to one-half and ignores how far the requested cutoff lies above the center of the sampling distribution.

Original practice · fully worked

Original variant: plankton detection intervals

A marine laboratory models the time between successive plankton detections by an exponential distribution with mean 8 minutes. During a calibration study, the laboratory records 100 independent intervals. Using the central limit theorem, approximate the probability that their average exceeds 9 minutes.

A
0.0062
B
0.0528
C
0.1056
D
0.3944
E
0.8944

Variant answer in brief

The average of 100 exponential intervals has approximate standard error 0.8 minute. A 9-minute average is 1.25 standard errors above the mean, giving an upper tail of about 0.1056.

Setup

Record the exponential mean, its equal standard deviation, and the study size.

\mu=\sigma=8,\qquad n=100

Model

Use the central limit theorem for the average of the independent intervals.

\overline T\approx N\!\left(8,\frac{64}{100}\right)

\operatorname{SE}(\overline T)=\frac{8}{10}=0.8

Compute

Standardize 9 minutes and evaluate the right-hand normal tail.

z=\frac{9-8}{0.8}=1.25

\Pr(\overline T>9)\approx 1-\Phi(1.25)=0.1056497737\ldots

Answer

Choice C is the nearest listed value.

\boxed{0.1056\quad\text{(C)}}

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 #732 — Verified Solution.” Verified 2026-07-30.

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