Verified solution · Exam P

SOA Exam P Sample Question #450 Solution

This Exam P sample reference tests Poisson Distribution. This problem uses independence to compute the mean and variance of a weighted Poisson sum. Solving the coefficient-of-variation equation gives λ=0.4420, which selects 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.

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Independent solution

How to solve this Poisson Distribution question

Setup

Each count has both mean and variance λ. Apply linearity to the weighted total.

E[U]=(100+150+200)\lambda=450\lambda

Model

Independence makes the weighted variances add; each coefficient is squared.

\operatorname{Var}(U)=(100^2+150^2+200^2)\lambda

\operatorname{Var}(U)=72{,}500\lambda

Compute

Set standard deviation divided by mean equal to the stated coefficient of variation and solve.

0.90=\frac{\sqrt{72{,}500\lambda}}{450\lambda}

\lambda=\frac{72{,}500}{0.90^2(450)^2}=0.4420057918\ldots

Answer

The common Poisson parameter rounds to 0.44.

\boxed{0.44\quad\text{(A)}}

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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.

BAt λ=0.82, the stated model gives CV=√(72,500λ)/(450λ)=0.6608, not 0.90. The dependence on λ is 1/√(λ), which is lost if the square root is handled inconsistently.
CTreating the weighted total itself as Poisson gives λ approximately 1/0.90²=1.23. Unequal event weights prevent that shortcut.
DSubstituting λ=1.50 gives CV=0.4886, so this candidate fails the defining equation; the squared weights must produce variance 72,500λ.
ESubstituting λ=2.25 gives CV=0.3989, so this candidate also fails the defining equation.

Original practice · fully worked

Original variant: infer a two-speed mixture from its variability

A document-processing task takes 2 minutes in normal mode and 8 minutes in long mode. Let p be the probability of long mode. The processing time has coefficient of variation 0.50, and its mean exceeds 5 minutes. Calculate p.

A
0.0352
B
0.2500
C
0.5000
D
0.6315
E
0.7500

Variant answer in brief

The two-point processing time has mean 2+6p and variance 36p(1-p). The coefficient-of-variation equation has two roots; the condition that the mean exceeds 5 selects p=(5+2sqrt(5))/15=0.6315 and choice D.

Setup

Let T be the processing time. Compute its first two moments from the two possible durations.

E[T]=2(1-p)+8p=2+6p

E[T^2]=4(1-p)+64p=4+60p

Model

Center the second moment to obtain the variance.

\operatorname{Var}(T)=4+60p-(2+6p)^2

\operatorname{Var}(T)=36p(1-p)

Compute

Square the coefficient-of-variation equation and solve the resulting quadratic.

\frac{36p(1-p)}{(2+6p)^2}=0.50^2

45p^2-30p+1=0

p=\frac{5\pm2\sqrt5}{15}

Answer

The smaller root gives a mean below 5. The larger root is therefore required.

p=\frac{5+2\sqrt5}{15}=0.6314757303\ldots

\boxed{0.6315\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #450 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

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: infer a two-speed mixture from its variability

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution