SOA Exam P Sample Question #692. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #692 — Verified Solution

Independent solution

How to solve this Central Limit Theorem question

Answer in brief

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.

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)}}

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, \(\Pr(S\le 5750)\approx\Phi(-2.5)=0.0062\), 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

Variant answer in brief

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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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #692 — Verified Solution.” Verified 2026-07-30.

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