This Exam P sample reference tests Poisson Distribution. Scaling the constant Poisson rate from thirty years to ten gives mean and variance 2/3. The standard deviation is the square root of 2/3, approximately 0.816, so the answer is D.
Convert the stated long-window count rate into the requested time window.
λ10=2(3010)=32
Model
Model
A Poisson count has variance equal to its mean.
N10∼Poisson(32)
Var(N10)=32
Compute
Compute
Take the square root of the variance to obtain the standard deviation.
SD(N10)=32=0.8164965809…
Answer
Answer
The requested standard deviation rounds to 0.816.
0.816(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.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis divides the correct standard deviation by the square root of ten, as though averaging ten observations rather than counting events over ten years.
BThis corresponds to using variance 2/9, which scales the thirty-year parameter by an extra factor of one third.
CThis is the ten-year Poisson mean and variance 2/3, not their square root.
EThis is the square root of the full thirty-year mean 2 and ignores the shorter time window.
Original practice · fully worked
Original variant: priority server alerts
Priority alerts arrive at a server as a Poisson process with an average of 3.6 alerts per twelve-hour monitoring window. Calculate the standard deviation of the number of alerts during the next five hours.
A 0.417
B 1.000
C 1.225
D 1.500
E 1.897
Answer and calculation
The five-hour mean and variance are 3.6(5/12) = 1.5. The standard deviation is the square root of 1.5, approximately 1.225, so choice C.
Setup
Setup
Scale the twelve-hour rate to five hours.
λ5=3.6(125)=1.5
Model
Model
The five-hour alert count is Poisson, so its variance is also 1.5.