Worked 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 July 30, 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)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

See the cover and sample pages

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

Answer and calculation

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)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: transformed incident score

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables