Worked solution · Exam P

SOA Exam P Sample Question #732 Solution

This Exam P sample reference tests Central Limit Theorem. 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.

Independent analysis by ActuaryProof · reviewed July 30, 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 Central Limit Theorem question

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)}}

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 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

Answer and calculation

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)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Central Limit Theorem question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: plankton detection intervals

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables