Question

The proportion of time Y that an industrial robot is in operation during a 40 hour week is a random variable with probability density function

f(y)= 2y, 0≤y≤1

0, elsewhere

a.) Find E(Y) and V(Y)

b.) For the robot under study, the profit X for a week is given by X=200Y-60. Find E(X) and V(X).

c.)Find an interval in which the profit should lie for at least 75% of the weeks that the robot is in use.

Answer 1

E(Y) is calculated by integrating the probability density function, resulting in 2/3, and V(Y) is found to be 1/18. For the profit X = 200Y - 60, E(X) is 80, and V(X) is 2000/9. An interval for the profit is determined using empirical rules or z-scores tailored to the distribution type.

Step-by-step explanation:

Finding E(Y) and V(Y)

To find the expected value E(Y) and variance V(Y) for the random variable Y with probability density function f(y) = 2y, we use the following formulas:

E(Y) = ∫ y * f(y) dy

V(Y) = ∫ (y - E(Y))^2 * f(y) dy

By integrating from 0 to 1, we get:

E(Y) = ∫ y * 2y dy from 0 to 1 = 2/3

V(Y) = ∫ (y - 2/3)^2 * 2y dy from 0 to 1 = 1/18

Finding E(X) and V(X)

For the profit X = 200Y - 60, the expected value E(X) and variance V(X) can be found using E(Y) and V(Y) and the properties of linear transformations of random variables:

E(X) = 200 * E(Y) - 60 = 80

V(X) = (200^2) * V(Y) = 2000/9

Profit Interval for 75% of Weeks

To find an interval for the profit X where at least 75% of the observations should lie, we consider the properties of the random variable. Since X is a linear transformation of Y, which is uniformly distributed, we rely on the empirical rule or corresponding z-scores for a uniform distribution to estimate such interval.