Worked solution · Exam P

SOA Exam P Sample Question #320 Solution

This Exam P sample reference tests Central Limit Theorem. This is a central-limit approximation for an aggregate of independent losses. The total has mean 100,000 and standard deviation 4,000, so the cutoff is two standard deviations below its mean and has probability 0.02275, selecting choice A.

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

Let S denote the aggregate. Add the individual means and, because the observations are independent, add their variances.

E[S]=100(1000)=100000

\operatorname{Var}(S)=100(400^2)=16000000,\qquad \operatorname{SD}(S)=4000

Model

The central limit theorem supplies a normal approximation for the large independent sum.

S\ \dot\sim\ N(100000,4000^2)

Compute

Standardize the stated aggregate boundary and evaluate the lower standard-normal tail.

z=\frac{92000-100000}{4000}=-2

\Pr(S<92000)\approx\Phi(-2)=0.0227501319

Answer

The requested approximate probability rounds to 0.023.

\boxed{0.023\quad\text{(A)}}

Test the method on unseen problems

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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.

BThe value 0.050 substitutes a stock five-percent lower tail for the calculated standardized boundary. Here the boundary is z=-2, whose tail is about 0.0228.
CThe value 0.421 is approximately Φ(-0.2). That z-score results from dividing the 8,000 shortfall by 100(400)=40,000 instead of the aggregate standard deviation √(100)(400)=4,000.
DThe value 0.579 is approximately Φ(0.2), combining the incorrect 40,000 spread with a reversal of the standardized sign.
EThe value 0.977 is Φ(2), the complement of the required lower-tail probability Φ(-2).

Original practice · fully worked

Original variant: net score from two job groups

An online platform processes 64 standard jobs and 36 adjustment jobs during a billing cycle. The net score is the sum of the standard-job scores minus the sum of the adjustment-job scores. All job scores are independent. A standard-job score has mean 20 and standard deviation 4, while an adjustment-job score has mean 10 and standard deviation 4. Approximate, with a normal model, the chance that the net score is greater than 980.

A
0.0668
B
0.1420
C
0.4332
D
0.8580
E
0.9332

Answer and calculation

The two job groups give net mean 920. Their independent variances add even though one total is subtracted, producing variance 1,600 and standard deviation 40; the boundary is z=1.5 and the upper tail is 0.0668, choice A.

Setup

Write the net score as one group total minus the other and calculate its center.

T=\sum_{i=1}^{64}A_i-\sum_{j=1}^{36}B_j

E[T]=64(20)-36(10)=920

Model

A minus sign changes means but disappears when variances are formed. Independence therefore makes the two group variances additive.

\operatorname{Var}(T)=64(4^2)+36(4^2)=1600

\operatorname{SD}(T)=40

Compute

Apply the normal approximation and evaluate the upper tail at the standardized threshold.

z=\frac{980-920}{40}=1.5

\Pr(T>980)\approx1-\Phi(1.5)=0.0668072013

Answer

The approximate exceedance probability is 0.0668.

\boxed{0.0668\quad\text{(A)}}

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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: net score from two job groups

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables