Worked solution · Exam P
This Exam P sample reference tests Poisson Aggregation. Independent monthly Poisson counts from July through November add to a Poisson count with mean 3.5. Its probability of two is 6.125 exp(−3.5)=0.185, choice B.
Independent solution
Setup
Independent Poisson counts add to another Poisson count whose mean is the sum of the component means.
Model
Adding the five monthly means gives an aggregate mean of 3.5.
Compute
Use the Poisson mass at two with mean 3.5, which evaluates to approximately 0.184959.
Answer
The five-month total equals two with probability about 0.185.
Test the method on unseen problems
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.
Original practice · fully worked
Independent service-call counts have mean 0.8 on each of two daytime shifts and mean 0.3 on each of four night shifts. Find the probability of exactly one call over all six shifts.
The aggregate count is Poisson with mean 2(0.8)+4(0.3)=2.8. Thus P(T=1)=2.8 exp(−2.8)=0.17027, choice B.
Setup
Because the six shift counts are independent Poisson variables, their total is Poisson with mean equal to the sum of the shift means.
Model
The two daytime shifts contribute 1.6 and the four night shifts contribute 1.2, for aggregate mean 2.8.
Compute
The Poisson mass at one with mean 2.8 is approximately 0.170268.
Answer
The exactly-one probability is about 0.1703.
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