Worked solution · Exam P

SOA Exam P Sample Question #137 Solution

This Exam P sample reference tests Poisson Mode and Tail. Two adjacent Poisson modes at 2 and 3 imply mean 3. The tails are P(N>3)=0.3528 and P(N>4)=0.1847, so the smallest qualifying K is 4, choice D.

Independent analysis by ActuaryProof · reviewed October 6, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

Independent solution

How to solve this Poisson Mode and Tail question

Setup

A Poisson distribution has two adjacent modes only when its mean is the larger of those two modes. The stated modes therefore determine mean three.

\text{modes }2,3\Longrightarrow \lambda=3

Model

The requested cutoff is the first integer whose upper tail is below 0.25. The tail beyond three is still approximately 0.35277.

\Pr(N>3)=1-13e^{-3}=0.35277

Compute

The tail beyond four is approximately 0.18474, below 0.25. Comparing these adjacent cutoffs proves that four is minimal.

\Pr(N>4)=1-\left(13+\frac{81}{24}\right)e^{-3}=0.18474<0.25

Answer

The smallest cutoff with tail below 25% is 4.

\boxed{4\quad\text{(D)}}

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.

AA cutoff of 1 is below the mode and leaves an upper tail far greater than 0.25.
BA cutoff of 2 is also below the larger mode and cannot be the first point at which the upper tail falls below 0.25.
CThe value 3 is the Poisson mean and larger mode, but its upper tail is approximately 0.35277, still above 0.25.
EA cutoff of 5 also has an upper tail below 0.25, but cutoff 4 already satisfies the condition, so 5 is not minimal.

Original practice · fully worked

Original variant: tail cutoff from two Poisson modes

A Poisson count has two modes at 4 and 5. Find the smallest integer K for which the probability that the count exceeds K is less than 0.20.

Answer and calculation

The adjacent modes imply mean 5. The tail above 6 is 0.2378, while the tail above 7 is 0.1334, so K=7, choice C.

Setup

Adjacent Poisson modes at four and five determine a mean of five.

\text{modes }4,5\Longrightarrow \lambda=5

Model

At cutoff six, the upper tail is approximately 0.237817, which is still above 0.20.

\Pr(N>6)=0.237817>0.20

Compute

At cutoff seven, the upper tail falls to approximately 0.133372. Thus seven is the first qualifying integer.

\Pr(N>7)=0.133372<0.20

Answer

The smallest qualifying cutoff is 7.

\boxed{7\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Poisson Mode and Tail question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: tail cutoff from two Poisson modes

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables