Worked solution · Exam P

SOA Exam P Sample Question #526 Solution

This Exam P sample reference tests Central Limit Theorem. The total has mean 5,000,000 and standard deviation 707,106.78. A normal approximation places the threshold 0.7071 standard deviations above the mean, giving an upper-tail probability of 0.23975 and choice C.

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

For one exponential component, the standard deviation equals its mean. Add the means and variances of the 50 independent components.

E[S]=50(100{,}000)=5{,}000{,}000

\operatorname{Var}(S)=50(100{,}000)^2,\qquad \operatorname{SD}(S)=707{,}106.7812

Model

Apply the central limit theorem to approximate the standardized total by a standard normal variable.

Z=\frac{S-E[S]}{\operatorname{SD}(S)}\approx N(0,1)

Compute

Standardize the target level and evaluate the normal upper tail.

z=\frac{5{,}500{,}000-5{,}000{,}000}{707{,}106.7812}=0.7071068

\Pr(S>5{,}500{,}000)\approx 1-\Phi(0.7071068)=0.239750

Answer

The approximate probability is 0.24.

\boxed{0.24\quad\text{(C)}}

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.

AThis reports the relative standard deviation of the total, 1/√50=0.1414, as though it were the requested tail probability.
BThis badly underestimates √50 as about 5.5, producing z≈0.91 and an upper tail near 0.18.
DThis rounds √50 up to 10, inflating the total standard deviation to 1,000,000; the resulting tail 1-Φ(0.5)=0.3085 is pushed toward 0.33.
EThis adds component standard deviations as if all losses moved together. Using a 5,000,000 standard deviation gives 1-Φ(0.1)=0.4602.

Original practice · fully worked

Original variant: average component tolerance

A production line records 64 independent component tolerances from a stable process. Each reading has mean 12 units and standard deviation 4 units, but its distribution is otherwise unspecified. Approximate the probability that the sample average lies between 11.5 and 12.8 units.

A
0.0548
B
0.1587
C
0.7865
D
0.8413
E
0.9452

Answer and calculation

The average has approximate standard error 4/√(64)=0.5. Its two bounds standardize to -1.0 and 1.6, so the probability is Φ(1.6)-Φ(-1)=0.7865, giving choice C.

Setup

Compute the mean and standard error of the 64-reading average.

E[\overline X]=12,\qquad \operatorname{SD}(\overline X)=\frac{4}{\sqrt{64}}=0.5

Model

The central limit theorem gives an approximately normal sampling distribution for the average.

\frac{\overline X-12}{0.5}\approx N(0,1)

Compute

Standardize both endpoints and subtract the two cumulative probabilities.

z_L=\frac{11.5-12}{0.5}=-1,\qquad z_U=\frac{12.8-12}{0.5}=1.6

\Pr(11.5<\overline X<12.8)\approx\Phi(1.6)-\Phi(-1)=0.786545

Answer

The requested interval probability is approximately 0.7865.

\boxed{0.7865\quad\text{(C)}}

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How to solve this Central Limit Theorem question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: average component tolerance

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables