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
Answer and calculation
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.
767.31(A) 