A food manufacturer uses an extruder that yields revenue for the firm at a rate of $200 per hour when it operates. However, the extruder breaks down an average of 2 times every day it operates. If Y denotes the number of breakdowns per day, the daily revenue generated by the machine is R=1600-50(Y^2). Find the expected daily revenue for the extruder.
I know that E(R) = 1600 – 50(lambda) and the answer says that lambda is 6 and that the revenue is $1300 but I don’t know how to find lambda.
A definition says that lambda is the expected number of occurances in the defined interval. So I think the interval for this question is 1 day. $200/hr is equal to $4800/day.
The lambda of the E(R) equation and the lambda that is the usual parameter of the Poisson distribution do not seem to be the same here.
As you noted, given the average of 2 breakdowns per day, the lambda parameter for the Poisson distribution, call it L, is 2.
That means that the probability of exactly k failures in a day is:
Pr(k, L) = L^k e^(-L) / k!
But for the revenue we want to know the expected value of Y^2, not Y. That is:
E(Y^2) = sum over k of k^2 Pr(k, L)
but we know that this is related to the variance: “The variance is equal to the mean of the squares minus the square of the mean.”
The mean of the squares is what we want. The mean is just L so the square of the mean is L^2, and the variance of the Poisson distribution is also L. (Check the Wikipedia page on the Poisson distribution)
So we have L = E(Y^2) – L^2 or
E(Y^2) = L + L^2 = 2 + 2^2 = 2 + 4 = 6