Verified solution · Exam P

SOA Exam P Sample Question #636 Solution

This Exam P sample reference tests Central Limit Theorem. The aggregate has mean 3,000,000 and standard deviation 50,000 times the square root of 60. A central-limit approximation gives an upper-tail probability of 0.09835, so the nearest listed value is choice A.

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 S be the aggregate of the 60 independent exponential loss variables.

S=\sum_{i=1}^{60}X_i

E[S]=60(50000)=3000000

Model

An exponential variable has standard deviation equal to its mean. Independence makes the component variances additive, and the central limit theorem supplies a normal approximation.

\operatorname{SD}(S)=50000\sqrt{60}=387298.3346\ldots

S\ \dot\sim\ N\!\left(3000000,\,60(50000)^2\right)

Compute

Standardize the aggregate threshold and evaluate the upper standard-normal tail.

z=\frac{3500000-3000000}{50000\sqrt{60}}=\frac{10}{\sqrt{60}}=1.2909944487\ldots

\Pr(S>3500000)\approx1-\Phi(1.2909944487)=0.0983528012\ldots

Answer

The approximation is closest to 0.10.

\boxed{0.10\quad\text{(A)}}

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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.

BThis can result from rounding the standardized threshold too aggressively before taking the normal tail.
CThis understates the standardized distance by using an inflated aggregate standard deviation.
DThis divides the threshold excess by the aggregate mean rather than by its standard deviation, producing a small z-score and an upper tail near 0.43.
EThis treats the threshold as though it were equal to the aggregate mean.

Original practice · fully worked

Original variant: capsule shipment weight

A shipment contains 50 independent sensor capsules and 30 independent shielding clips. Capsule weights have mean 12 grams and standard deviation 2 grams; clip weights have mean 20 grams and standard deviation 3 grams. All item weights are mutually independent. Using a normal approximation, estimate the probability that the shipment weighs between 1170 and 1230 grams.

A
0.083
B
0.417
C
0.834
D
0.917
E
0.970

Variant answer in brief

The two item groups give total mean 1200 grams and variance 470 square grams. The interval endpoints are 1.383797 standard deviations below and above the mean, giving central normal probability 0.8335793 and selecting choice C.

Setup

Add the expected weights from the two item groups.

T=\sum_{i=1}^{50}X_i+\sum_{j=1}^{30}Y_j

E[T]=50(12)+30(20)=1200

Model

Mutual independence makes every item variance additive. Approximate the heterogeneous sum by a normal distribution.

\operatorname{Var}(T)=50(2^2)+30(3^2)=470

T\ \dot\sim\ N(1200,470)

Compute

Standardize both interval endpoints and evaluate the central normal probability.

z=\frac{30}{\sqrt{470}}=1.3837968120\ldots

\Pr(1170<T<1230)\approx\Phi(z)-\Phi(-z)

=0.8335793205\ldots

Answer

The approximate interval probability rounds to 0.834.

\boxed{0.834\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #636 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.

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: capsule shipment weight

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution