Worked solution · Exam P

SOA Exam P Sample Question #168 Solution

This Exam P sample reference tests Central Limit Theorem. The affine update changes the individual mean to 109 and standard deviation to 26. A sample mean of 25 observations has standard error 5.2, so the standardized interval probability is 0.5345, choice B.

Independent analysis by ActuaryProof · reviewed July 31, 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

Transform the first two moments of an individual observation.

E[Y]=1.04(100)+5=109

\operatorname{SD}(Y)=1.04(25)=26

Model

Because the individual transformed claims are normal and independent, their average is exactly normal.

\overline Y\sim N\left(109,\frac{26^2}{25}\right)

\operatorname{SD}(\overline Y)=\frac{26}{5}=5.2

Compute

Standardize both endpoints and subtract the two normal CDF values.

z_L=\frac{100-109}{5.2}=-1.7307692,\qquad z_U=\frac{110-109}{5.2}=0.1923077

\Pr(100<\overline Y<110)\approx\Phi(0.1923077)-\Phi(-1.7307692)=0.5345029

Answer

The approximate interval probability rounds to 0.53.

\boxed{0.53\quad\text{(B)}}

Test the method on unseen problems

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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.

AThe value 0.48 is Φ(2)-Φ(0)=0.4773, obtained by applying the 100-to-110 interval directly to the untransformed sample mean.
CThe value 0.54 results from transforming the mean to 109 but leaving the standard error at 25/√(25)=5 instead of scaling it by 1.04.
DThe value 0.67 results from using mean 105, as though only the added 5 affected the mean, while still using the transformed standard error 5.2.
EThe value 0.68 is the symmetric probability Φ(1)-Φ(-1), obtained from mean 105 and standard error 5 while ignoring the multiplicative transformation.

Original practice · fully worked

Original variant: choose a sample size after affine calibration

A laboratory's unreported instrument output follows a Gaussian law centered at 50 with standard deviation 12. The displayed value equals three quarters of the raw output plus 8. Find the least number of independent displayed values whose average is within 3 units of its center with probability at least 0.95.

A
25
B
34
C
35
D
36
E
62

Answer and calculation

The reported standard deviation is 9, so the 95% requirement is 3 √(n)/9 at least z_0.975. This gives n at least 34.573, making 35 the smallest integer, choice C.

Setup

Transform the individual mean and standard deviation.

E[Y]=0.75(50)+8=45.5

\operatorname{SD}(Y)=0.75(12)=9

Model

Because the readings are normal, the sample mean is exactly normal for every positive n.

\overline Y\sim N\left(45.5,\frac{9^2}{n}\right)

\Pr(|\overline Y-45.5|\le3)=2\Phi\left(\frac{3\sqrt n}{9}\right)-1

Compute

Match the central probability to the 97.5th-percentile z-score and round the sample-size requirement upward.

\frac{3\sqrt n}{9}\ge z_{0.975}=1.959964

n\ge\left(\frac{1.959964(9)}{3}\right)^2=34.5731294

Answer

The smallest qualifying integer sample size is 35.

\boxed{35\quad\text{(C)}}

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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: choose a sample size after affine calibration

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables