Verified solution · Exam P

SOA Exam P Sample Question #573 Solution

This Exam P sample reference tests Poisson Distribution. The independent individual counts combine to a Poisson total with mean and variance 40,000(0.16)=6,400. Its standard deviation is 80, so the coefficient of variation is 80/6,400=0.0125, which rounds to choice A.

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.

Read the solution Open official source

Independent solution

How to solve this Poisson Distribution question

Setup

Let S be the aggregate count across all insured drivers. Independent Poisson variables add by summing their parameters.

S\sim\operatorname{Poisson}(40000\cdot0.16)

Model

A Poisson variable has equal mean and variance.

E[S]=\operatorname{Var}(S)=6400

Compute

Take the square root of the variance, then divide by the mean.

\operatorname{SD}(S)=\sqrt{6400}=80

\operatorname{CV}(S)=\frac{80}{6400}=0.0125

Answer

At the precision of the listed choices, the coefficient of variation is 0.01.

\boxed{0.01\quad\text{(A)}}

Continue without hunting through PDFs

All 718 Exam P sample solutions in syllabus order

The searchable 3108-page manual includes this complete solution, its error analysis, and one original worked variant for every active reference.

See inside the Exam P study manual PDF

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

BThis reports the single-driver Poisson mean 0.16 rather than the aggregate coefficient of variation.
CThis is the single-driver standard deviation √(0.16)=0.40 and neither aggregates the drivers nor divides by the aggregate mean.
DThis uses the Poisson variance-to-mean ratio, which equals 1, in place of standard deviation divided by mean.
EThis is the single-driver coefficient of variation 1/√(0.16)=2.50 and ignores the diversification from 40,000 independent counts.

Original practice · fully worked

Original variant: planning a monitoring window

Events arrive according to a Poisson process at a rate of 18 per hour. Determine the observation time, in hours, required for the event count to have coefficient of variation 0.10.

A
0.056
B
0.556
C
5.556
D
10.000
E
100.000

Variant answer in brief

Over t hours the count has mean 18t and coefficient of variation 1/√(18t). Setting this equal to 0.10 gives 18t=100 and t=50/9=5.556 hours, so choice C is correct.

Setup

Write the distribution of the count accumulated over an unknown observation length t.

N(t)\sim\operatorname{Poisson}(18t)

Model

For a Poisson count with mean m, the coefficient of variation is one divided by the square root of m.

\operatorname{CV}(N(t))=\frac{\sqrt{18t}}{18t}=\frac1{\sqrt{18t}}

Compute

Set the coefficient of variation to its target and solve for the exposure time.

\frac1{\sqrt{18t}}=0.10

18t=100

t=\frac{50}{9}=5.555555\ldots

Answer

The required observation window is approximately 5.556 hours.

\boxed{5.556\text{ hours}\quad\text{(C)}}

Keep practicing

Turn this solution into a complete study sequence

The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.

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

Cite this solution

ActuaryProof. “SOA Exam P Sample Question #573 — 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: planning a monitoring window

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution