Verified solution · Exam P

SOA Exam P Sample Question #65 Solution

This Exam P sample reference tests Central Limit Theorem. The total has mean 6,328,125 and standard deviation 11,250. A full-precision normal 90th percentile is about 6,342,542, which rounds to choice C.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Central Limit Theorem question

Setup

Let S be the sum of 2025 independent contributions, each with mean 3125 and standard deviation 250.

S=\sum_{i=1}^{2025}X_i

E[X_i]=3125,\quad\operatorname{SD}(X_i)=250

Model

Independence makes the aggregate mean scale by 2025 and its standard deviation scale by the square root of 2025.

E[S]=2025(3125)=6328125

\operatorname{SD}(S)=250\sqrt{2025}=11250

Compute

Add 1.2815516 aggregate standard deviations to the mean to obtain the normal 90th percentile, 6,342,542.5.

q_{0.90}\approx6328125+1.2815516(11250)=6342542.5

Answer

The listed value nearest the approximate percentile is 6,343,000, choice C.

\boxed{6{,}343{,}000\quad\text{(C)}}

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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 6,328,000 is the rounded aggregate mean. It omits the positive 90th-percentile loading.
EThe value 6,977,000 is produced approximately by scaling the individual standard deviation linearly with 2025 rather than by its square root.

Original practice · fully worked

Original variant: upper quantile of combined solar output

A solar farm operates 400 independent panels. Daily panel output has mean 12.0 kilowatt-hours and standard deviation 3.0 kilowatt-hours. Using a normal approximation, estimate the 95th percentile of total daily output.

D
4800.0 kWh
B
4860.0 kWh
C
4880.0 kWh
A
4898.7 kWh
E
4986.9 kWh

Variant answer in brief

The aggregate mean is 4800 and standard deviation 60. Adding 1.64485 standard deviations gives 4898.69 kWh.

Setup

Let T be the total output of 400 independent panels with individual mean 12.0 kWh and standard deviation 3.0 kWh.

T=\sum_{i=1}^{400}X_i

Model

The sum has mean 400 × 12 and standard deviation 3 times the square root of 400.

E[T]=400(12)=4800

\operatorname{SD}(T)=3\sqrt{400}=60

Compute

Adding 1.6448536 aggregate standard deviations to 4800 gives the approximate 95th percentile 4898.6912 kWh.

q_{0.95}\approx4800+1.6448536(60)=4898.6912

Answer

The 95th percentile of total output is approximately 4898.7 kWh, corresponding to choice A.

\boxed{4898.7\ \text{kWh}\quad\text{(A)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #65 — 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: upper quantile of combined solar output

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution