Verified solution · Exam P
This Exam P sample reference tests Poisson Distribution. For a Poisson count, a zero-count probability q implies mean -ln(q). Applying this to the three groups and weighting by their population sizes gives 36.134873 thousand expected claims, which rounds to 36 and choice C.
Independent solution
Setup
Invert each Poisson zero-count probability to recover the corresponding mean claim count.
Model
Multiply each per-policy mean by its group population, measured in thousands.
Compute
Evaluate and add the three expected group totals.
Answer
The expected total rounds to 36 thousand claims.
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Original practice · fully worked
An array contains identical independent sensors. The daily alert count from each sensor is Poisson, and one sensor has no alerts on a day with probability 0.82. The expected total number of daily alerts from the array is 29.76764. Calculate the number of sensors in the array.
The zero-alert probability gives a per-sensor mean of -ln(0.82)=0.198451. Dividing the array's expected total by that mean gives 149.999996, so the array contains 150 sensors and choice C.
Setup
Use the Poisson zero-count identity to determine one sensor's mean alert rate.
Model
For n identical sensors, linearity of expectation makes the array mean n times the individual mean.
Compute
Divide by the recovered individual rate.
Answer
The integer array size is 150 sensors.
Keep practicing
These references share syllabus concepts or sit nearby in the official sequence. All 388 Exam P univariate random variables practice problems.
Drill the method: Exam P Poisson Distribution problems · Exam P Expected Value problems
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ActuaryProof. “SOA Exam P Sample Question #299 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
