This Exam P sample reference tests Poisson Distribution. The payment count is (N-1)+. Its expectation is λ-1+exp(−)λ, so multiplying by 10,000 gives 7,231.30, choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BThe value 5,000 replaces the nonnegative covered count by the uncapped expression equal to the count minus one. That expression is negative when no event occurs.
Original practice · fully worked
Original variant: Poisson payment capped at two events
A network experiences a Poisson number N of severe interruptions with mean 2.2 per year. A contract pays 500 per interruption but covers no more than two interruptions. Determine the expected annual payment.
A 767.31
B 332.69
C 500.00
D 900.00
E 1100.00
Variant answer in brief
The covered count is min(N,2). Its expectation is P(N≥1)+P(N≥2), producing payment 767.31, choice A.
Setup
Setup
The contract covers the first two interruptions only, so the covered count is the smaller of the Poisson count and two.
Y=500min(N,2),N∼Poisson(2.2)
Model
Model
The expected capped count equals the probability of at least one interruption plus the probability of at least two interruptions.
E[min(N,2)]=Pr(N≥1)+Pr(N≥2)=2−e−2.2(2+2.2)
Compute
Compute
Evaluating those two Poisson tails and multiplying by 500 gives approximately 767.3134.
E[Y]=500{2−4.2e−2.2}=767.3134
Answer
Answer
The expected capped-interruption payment is approximately 767.31, selecting choice A.
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