Verified solution · Exam P

SOA Exam P Sample Question #615 Solution

This Exam P sample reference tests Poisson Distribution. The weekly mean gives a daily Poisson mean of 9 and standard deviation 3. A count at least two standard deviations below the mean is at most 3, whose Poisson cumulative probability is 0.02122649, so choice B is correct.

Independent analysis by ActuaryProof · reviewed August 4, 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 Poisson Distribution question

Setup

Convert the weekly expected count to a daily Poisson mean.

\lambda=\frac{63}{7}=9

D\sim\operatorname{Poisson}(9)

Model

A Poisson count has variance equal to its mean, so determine the requested lower cutoff.

\operatorname{SD}(D)=\sqrt9=3

9-2(3)=3

Compute

Sum the Poisson masses through the integer cutoff.

\Pr(D\le3)=e^{-9}\sum_{k=0}^{3}\frac{9^k}{k!}

\Pr(D\le3)=0.0212264863\ldots

Answer

The probability rounds to 0.021.

\boxed{0.021\quad\text{(B)}}

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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 uses a strict cutoff below 3 and computes P(D≤2)=0.006232 instead of including count 3.
CThis uses the higher cutoff 5, giving P(D≤5)=0.115691 rather than moving two full standard deviations below the mean.
DThis results from an inappropriate approximation or a noninteger cutoff rather than evaluating the exact Poisson lower tail at 3.
EThis subtracts only one standard deviation, using cutoff 6 and obtaining P(D≤6)=0.206781.

Original practice · fully worked

Original variant: transformed incident score

A cycle's incident count N is Poisson. An operational score is defined by Y=2N+5, and Var(Y)=36. Calculate the probability that Y is at least 29.

A
0.073
B
0.197
C
0.294
D
0.413
E
0.803

Variant answer in brief

Because Var(2N+5)=4Var(N), the Poisson mean is 9. The event Y≥29 is equivalent to N≥12, whose Poisson upper-tail probability is 0.19699162, so choice B is correct.

Setup

Use the variance of the transformed score to recover the Poisson parameter.

\operatorname{Var}(Y)=2^2\operatorname{Var}(N)=4\lambda=36

\lambda=9

Model

Translate the score threshold into an integer count threshold.

2N+5\ge29

N\ge12

Compute

Evaluate the Poisson upper tail by complementing the cumulative probability through 11.

\Pr(Y\ge29)=\Pr(N\ge12)

\Pr(N\ge12)=1-e^{-9}\sum_{k=0}^{11}\frac{9^k}{k!}

\Pr(N\ge12)=0.1969916175\ldots

Answer

The upper-tail probability rounds to 0.197.

\boxed{0.197\quad\text{(B)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #615 — Verified Solution.” Verified 2026-07-30.

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

How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: transformed incident score

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution