Question

A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece.Let X = # of computers sold, & p(0) =.1, p(1) =.2, p(2) =.3 and p(3) =.4.With h(X) = profit associated with selling X units = revenue –cost of X computers.Find E[h(X)]

Answer 1

$700

Explanation:

Given.

X = Number of Computers Sold

p(0) =.1, p(1) =.2, p(2) =.3 and p(3) =.4.

h(x) = Revenue - Cost

Revenue = 1000X + 200(3 - X)

Cost = 1500

So, h(X) = 1000X+ 200(3 - X) - 1500

h(X) = 1000X + 600 - 200X - 1500

h(X) = 800X - 900

The possible range of Computer sold is 0 to 3 (I.e p(0) to p(3))

Hence, we'll solve for h(0) to h(3)

h(0) = 800(0) - 900

h(1) = 0 - 900

h(0) = -900

h(1) = 800(1) - 900

h(1) = 800 - 900

h(1) = -100

h(2) = 800(2) - 900

h(2) = 1600 - 900

h(2) = 700

h(3) = 800(3) - 900

h(3) = 2400 - 900

h(3) = 1500

Calculating E(h(x))

E(h(x)) = p(0).h(0) + p(1).h(1) + p(2).h(2) + p(3).h(3)

So,

E(h(x)) = 0.1 * -900 + 0.2 * -100 + 0.3 * 700 + 0.4 * 1500

E(h(x)) = -90 - 20 + 210 + 600

E(h(x)) = 700

So, E(h(x)) = $700