SOA Exam P Sample Question #613. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #613 — Verified Solution

Independent solution

How to solve this Poisson Distribution question

Answer in brief

The first Poisson mean is the square of its standard deviation, or 2.25. The ratio of zero-count probabilities gives the second mean as 2.25-ln(1.1), whose square root is 1.467886, so choice D is correct.

Setup

For a Poisson count, the mean and variance are equal. Convert the first standard deviation into its mean.

\lambda_1=(1.50)^2=2.25

Model

Express the stated zero-count probability ratio with the Poisson mass at zero.

e^{-\lambda_2}=1.1e^{-\lambda_1}

\lambda_2=\lambda_1-\ln(1.1)

Compute

Calculate the second mean and then take its square root to obtain the standard deviation.

\lambda_2=2.25-\ln(1.1)=2.1546898202\ldots

\operatorname{SD}(N_2)=\sqrt{\lambda_2}=1.4678861741\ldots

Answer

The second standard deviation rounds to 1.47.

\boxed{1.47\quad\text{(D)}}

Error diagnosis

Why the other letters tempt

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

AThis reduces the second Poisson mean far too much by treating the 1.1 probability multiplier as a direct linear change instead of taking its logarithm.
BThis similarly applies the zero-probability ratio to the standard deviation scale rather than to the exponential zero-count probability.
CThis subtracts ln(1.1) directly from the first standard deviation, giving 1.4047, instead of subtracting it from the Poisson mean.
EThis increases the standard deviation even though a larger zero-count probability requires a smaller Poisson mean.

Original practice · fully worked

Original variant: sampled fault reports

A server's number N of fault reports in one cycle is Poisson. The coefficient of variation of N is 0.40. Each report is independently selected for a detailed audit with probability 0.25; let M be the selected count. Calculate the standard deviation of M.

A
0.40
B
0.63
C
0.79
D
1.25
E
2.50

Variant answer in brief

For a Poisson count, CV=1/sqrt(lambda), so CV 0.40 gives lambda=6.25. Independent thinning makes M Poisson with mean 0.25(6.25)=1.5625 and standard deviation 1.25, so choice D is correct.

Setup

Recover the original Poisson mean from its coefficient of variation.

\operatorname{CV}(N)=\frac{\sqrt\lambda}{\lambda}=\frac1{\sqrt\lambda}=0.40

\lambda=\frac1{0.40^2}=6.25

Model

Independent selection thins a Poisson count to another Poisson count.

M\sim\operatorname{Poisson}(0.25\lambda)

E[M]=\operatorname{Var}(M)=1.5625

Compute

Take the square root of the thinned-count variance.

\operatorname{SD}(M)=\sqrt{1.5625}=1.25

Answer

The selected-count standard deviation is 1.25.

\boxed{1.25\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #613 — Verified Solution.” Verified 2026-07-30.

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