Worked solution · Exam P

SOA Exam P Sample Question #692 Solution

This Exam P sample reference tests Central Limit Theorem. The aggregate has approximate mean 6250 and standard deviation 200. The threshold is 2.5 standard deviations below the mean, so the upper-tail probability is 0.993790 and choice E.

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 sum of the 625 independent payments. Means add, while variances add for independent observations.

\operatorname{E}[S]=625(10)=6250

\operatorname{Var}(S)=625(8^2)=40000

Model

Apply the normal approximation to the aggregate, using the square root of the aggregate variance as its standard deviation.

S\ \dot\sim\ N(6250,200^2)

Z=\frac{S-6250}{200}

Compute

Standardize the payment threshold and convert the requested upper tail by symmetry of the standard normal distribution.

z=\frac{5750-6250}{200}=-2.5

\Pr(S>5750)\approx\Pr(Z>-2.5)=\Phi(2.5)=0.9937903347\ldots

Answer

Because the threshold lies far below the aggregate mean, a probability close to one is also the necessary sanity check.

\boxed{\Pr(S>5750)\approx0.9938\quad\text{(E)}}

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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 is the opposite tail, P(S ≤ 5750) ≈ Φ(−2.5) = 0.0062, rather than the probability of exceeding the threshold.
BA probability below one-half contradicts the fact that 5750 is below the aggregate mean of 6250; it reflects a tail-direction or centering error.
CThis value is only slightly above one-half and corresponds to using a standardized distance near 0.10 instead of the correct distance 2.5.
DThis value corresponds to an upper tail at approximately z = −0.40, understating the concentration created by summing 625 independent payments.

Original practice · fully worked

Original variant: bike-share battery reimbursements

A city bike-share program will process 400 independent battery-replacement reimbursements this quarter. Each reimbursement has mean 15 dollars and standard deviation 6 dollars. Using a normal approximation, calculate the probability that the total reimbursement amount exceeds 6240 dollars.

A
0.0228
B
0.1587
C
0.5000
D
0.9772
E
0.9987

Answer and calculation

The total has approximate mean 6000 and standard deviation 120. The threshold has standardized value 2, giving an upper tail of 0.022750 and choice A.

Setup

Write T for the total of the 400 independent reimbursements and aggregate the first two moments.

\operatorname{E}[T]=400(15)=6000

\operatorname{Var}(T)=400(6^2)=14400

Model

The normal approximation uses standard deviation 120 for the quarterly total.

T\ \dot\sim\ N(6000,120^2)

Compute

Standardize 6240 and evaluate the standard normal upper tail.

z=\frac{6240-6000}{120}=2

\Pr(T>6240)\approx1-\Phi(2)=0.0227501319\ldots

Answer

The probability that the quarter exceeds the stated reimbursement budget is about 2.28 percent.

\boxed{0.0228\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: bike-share battery reimbursements

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables