Worked 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 July 30, 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)}}

Test the method on unseen problems

Probability Multivariate & Risk Models Practice Workbook

30 original multivariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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

Answer and calculation

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)}}

Keep practicing

Next steps for Exam P Multivariate Random Variables

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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: capsule shipment weight

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables