Expected cost and Probability

Question

A store owner has overstocked a certain item and decides to use the following promotion to decrease the supply. The item has a marked price of $100. For each customer purchasing the item during a particular day, the owner will reduce the price by a factor of one-half. Thus, the first customer will pay $50 for the item, the second will pay $25, and so on. Suppose that the number of customers who purchase the item during the day has a Poisson distribution with mean 2. Find the expected cost of the item at the end of the day. The cost at the end of the day is 100(1/2)^y where Y is the number of customers who have purchased the item.

Answer 1

For a Poisson distribution with mean λ, P(x) = e^-λλ^x/x!

λ = 2, so P(x) = e^-22^x/x!

E(f(x) )= ∑f(x)P(x) = ∑100*2^-xe^-22^x/x! =

∑100*e^-2/x! =

100*e^-2∑1/x! =

100*e^-2 * e =

100/e