Question

Let E1 and E2 denote the proportions of time that employees I and II actually spent working on their assigned tasks during a working day. The joint density of E1 and E2 is given by f(e1, e2) = e1 e2, 0 ≤ e1 ≤ 1, 0 ≤ e2 ≤ 1 0, elsewhere Employee I has a higher rating than employee II and a measure of total productivity of these employees is 40E1 33E2. Find the expected value and the variance of this measure of productivity.

Answer 1

To calculate the expected value of the measure of productivity, multiply the proportions of time spent by the respective weights. To calculate the variance, find the expectations of the squared proportions of time and evaluate the expression using the formulas for variance.

Step-by-step explanation:

To find the expected value and variance of the measure of productivity, we need to calculate the expectations of the measure and its squared difference from the mean. Let's denote the measure of productivity as P = 40E1 + 33E2. The expected value E(P) can be calculated as:

E(P) = E(40E1 + 33E2) = 40E(E1) + 33E(E2) = 40 * E1 + 33 * E2

To calculate the variance Var(P), we need to calculate E(P^2) first:

E(P^2) = E((40E1 + 33E2)^2) = E(1600E1^2 + 2640E1E2 + 1089E2^2) = 1600E(E1^2) + 2640E(E1E2) + 1089E(E2^2)

Using the joint density function f(e1, e2) = e1 * e2, we can calculate the expectations E(E1^2) and E(E2^2):

E(E1^2) = ∫∫(e1^2 * e2) de1 de2 = ∫[0,1] ∫[0,1] e1^2 * e2 de1 de2 = 1/6

E(E2^2) = ∫∫(e1 * e2^2) de1 de2 = ∫[0,1] ∫[0,1] e1 * e2^2 de1 de2 = 1/6

Finally, we can calculate the variance using the formulas:

Var(P) = E(P^2) - (E(P))^2 = 1600 * 1/6 + 2640 * E(E1E2) + 1089 * 1/6 - (40E1 + 33E2)^2