This Exam P sample reference tests Poisson Distribution. The payment is 0, 1000, or 2000 according as the Poisson count is 0, 1, or at least 2. Its first two moments give standard deviation about 699, choice B.
A Poisson count with mean 0.6 generates payments of 0, 1000, and 2000 when the count is 0, 1, and at least 2, respectively.
N∼Poisson(0.6)
Y=1000min(N,2)
Model
Model
Obtain the three payment-state probabilities from the Poisson masses at zero and one and the remaining upper-tail probability.
P(N=0)=e−0.6
P(N=1)=0.6e−0.6
P(N≥2)=1−1.6e−0.6
Compute
Compute
The three-state distribution gives mean 573.0897 and second moment 816892.5. Centering the second moment and taking its square root gives 698.8996.
E[Y]=1000P(N=1)+2000P(N≥2)=573.0897
E[Y2]=106P(N=1)+4⋅106P(N≥2)=816892.5
SD(Y)=E[Y2]−E[Y]2=698.8996
Answer
Answer
The payment standard deviation rounds to 699, corresponding to choice B.
699(B)
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.
DThe value 817 is the rounded second moment 816892.5 expressed in thousands. A second moment is not a standard deviation.
EThe value 904 is approximately the square root of the uncentered second moment. It omits subtraction of the squared mean.
Original practice · fully worked
Original variant: service credits capped after two outages
A cloud service experiences N outages in a week, where N is binomial with three trials and outage probability 0.20. It grants an 80-dollar credit per outage but never more than 160 dollars. Find the standard deviation of the weekly credit.
C 44.80 dollars
B 50.60 dollars
A 53.64 dollars
D 58.24 dollars
E 64.00 dollars
Answer and calculation
The credit states 0, 80, and 160 have probabilities 0.512, 0.384, and 0.104. Their moments give a standard deviation of about 53.64 dollars.
Setup
Setup
Capping three-outage credits at 160 creates payment states 0, 80, and 160 from a Binomial(3,0.2) count.
N∼Bin(3,0.2)
C=80min(N,2)
Model
Model
Compute the first two payment-state probabilities directly; the remaining probability 0.104 belongs to two or three outages and hence to the cap.
P(C=0)=0.83=0.512
P(C=80)=3(0.2)(0.8)2=0.384
P(C=160)=0.104
Compute
Compute
The credit has mean 47.36 and second moment 5120, so its centered standard deviation is 53.64 dollars.
E[C]=47.36
E[C2]=5120
SD(C)=5120−47.362=53.64
Answer
Answer
Therefore the weekly-credit standard deviation is 53.64 dollars, corresponding to choice A.
