Verified solution · Exam P
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 solution
Setup
Let S denote the aggregate. Add the individual means and, because the observations are independent, add their variances.
Model
The central limit theorem supplies a normal approximation for the large independent sum.
Compute
Standardize the stated aggregate boundary and evaluate the lower standard-normal tail.
Answer
The requested approximate probability rounds to 0.023.
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Original practice · fully worked
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.
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.
Model
A minus sign changes means but disappears when variances are formed. Independence therefore makes the two group variances additive.
Compute
Apply the normal approximation and evaluate the upper tail at the standardized threshold.
Answer
The approximate exceedance probability is 0.0668.
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ActuaryProof. “SOA Exam P Sample Question #320 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
