Verified solution · Exam P

SOA Exam P Sample Question #459 Solution

This Exam P sample reference tests Central Limit Theorem. The sample average has mean 16 and standard deviation √(16/64)=0.5. A normal approximation transforms the two endpoints to -2 and 4, giving Φ(4)-Φ(-2)=0.9772, so choice D is correct.

Independent analysis by ActuaryProof · reviewed August 4, 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

Let the selected group counts be independent copies of X and write their average as the statistic of interest.

\operatorname{E}[X]=16,\qquad \operatorname{Var}(X)=16

\overline X=\frac1{64}\sum_{i=1}^{64}X_i

Model

Compute the center and spread of the average, then use the central limit theorem.

\operatorname{E}[\overline X]=16

\operatorname{Var}(\overline X)=\frac{16}{64}=0.25

\operatorname{SD}(\overline X)=0.5

Compute

Standardize both interval endpoints using the sampling standard deviation.

\Pr(15<\overline X<18)\approx\Pr\left(\frac{15-16}{0.5}<Z<\frac{18-16}{0.5}\right)

\Pr(-2<Z<4)=\Phi(4)-\Phi(-2)

\Phi(4)-\Phi(-2)=0.9772181968

Answer

The approximated interval probability rounds to 0.98.

\boxed{0.98\quad\text{(D)}}

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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 uses the single-group standard deviation 4 instead of the sampling standard deviation 0.5: Φ(0.5) − Φ(−0.25) = 0.2902, near 0.29.
BThis applies a one-group Poisson calculation to the integer range 15 through 18, obtaining approximately 0.3748. The random quantity here is the average of 64 groups, not one group count.
CThis replaces the asymmetric standardized interval (−2, 4) by (−2, 2), giving Φ(2) − Φ(−2) = 0.9545, near 0.95.
EThis rounds both the upper tail beyond 4 and the lower tail below -2 to zero. The lower tail is about 0.0228 and cannot be discarded when rounding the final probability.

Original practice · fully worked

Original variant: completion time for a service queue

A technician completes 100 independent calibration tasks. Each task duration is exponentially distributed with mean 2 minutes. Using a normal approximation, calculate the probability that the total duration is at most 220 minutes.

A
0.159
B
0.500
C
0.841
D
0.921
E
1.000

Variant answer in brief

The total has mean 200 and standard deviation 20. The threshold is one standard deviation above its mean, so the approximation is Φ(1)=0.8413, selecting choice C.

Setup

Let T be the sum of the 100 independent task durations. An exponential duration with mean 2 also has standard deviation 2.

T=\sum_{i=1}^{100}Y_i

\operatorname{E}[Y_i]=2,\qquad \operatorname{Var}(Y_i)=2^2=4

Model

Add the means and variances, then approximate the standardized total by a normal variable.

\operatorname{E}[T]=100(2)=200

\operatorname{Var}(T)=100(4)=400

\operatorname{SD}(T)=20

Compute

Convert the time threshold to a standard normal score.

\Pr(T\le220)\approx\Phi\left(\frac{220-200}{20}\right)

\Pr(T\le220)\approx\Phi(1)=0.8413447461

Answer

The normal approximation gives probability 0.841.

\boxed{0.841\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #459 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Central Limit Theorem question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: completion time for a service queue

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution