SOA Exam P Sample Question #637. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #637

This is a central-limit approximation for a sum of 303 independent daily counts. The standardized cutoff is 0.044191, giving an upper-tail probability of 0.482376 and selecting choice B.

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

Answer in brief

This is a central-limit approximation for a sum of 303 independent daily counts. The standardized cutoff is 0.044191, giving an upper-tail probability of 0.482376 and selecting choice B.

Setup

Let X_i be the count on day i and let T be the aggregate across the full observation window.

T=\sum_{i=1}^{303}X_i

\operatorname{E}[T]=303(3300)=999900

Model

Independence makes the variances additive. With many observations, the central limit theorem gives a normal approximation for T.

\operatorname{Var}(T)=303(130^2)

T\ \dot\sim\ N\!\left(999900,\,303(130^2)\right)

Compute

Standardize the specified aggregate cutoff and evaluate the upper standard-normal tail.

z=\frac{1000000-999900}{130\sqrt{303}}=0.0441911530\ldots

\Pr(T>1000000)\approx 1-\Phi(0.0441911530)=0.4823760170\ldots

Answer

The approximate probability rounds to 0.48.

\boxed{0.48\quad\text{(B)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AUsing 130 as the standard deviation of the entire sum gives z=100/130 and an upper tail near 0.22; the aggregate standard deviation must grow by the square root of 303.
CRounding the aggregate mean of 999900 to the cutoff before standardizing makes the z-score zero and produces 0.50.
DThis is the lower-tail probability Phi(0.044191), whereas the requested aggregate lies above the cutoff.
EThis is approximately the lower-tail complement of choice A, combining the unscaled one-day standard deviation with the wrong tail direction.

Original practice · fully worked

Original variant: culture-vial usable volume

A laboratory prepares 144 culture vials. The usable liquid volume in each vial is independent, with mean 50 milliliters and standard deviation 12 milliliters. Use a normal approximation to find the probability that the combined usable volume exceeds 7300 milliliters.

A
0.000
B
0.244
C
0.477
D
0.756
E
1.000

Variant answer in brief

The combined mean is 7200 milliliters and the combined standard deviation is 144 milliliters. The upper-tail probability at z=100/144 is 0.243702, so choice B is correct.

Setup

Let V_i denote usable volume from vial i and let W be the combined volume.

W=\sum_{i=1}^{144}V_i

\operatorname{E}[W]=144(50)=7200

Model

The independent vial contributions allow variance addition, and the large count supports a central-limit approximation.

\operatorname{SD}(W)=12\sqrt{144}=144

W\ \dot\sim\ N(7200,144^2)

Compute

Convert 7300 milliliters to a standard-normal score and take the upper tail.

z=\frac{7300-7200}{144}=0.6944444444\ldots

\Pr(W>7300)\approx 1-\Phi(0.6944444444)=0.2437017648\ldots

Answer

The requested probability is approximately 0.244.

\boxed{0.244\quad\text{(B)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #637 — Verified Solution.” Verified 2026-07-30.

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