A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5-lb batches. Let X 5 the number of batches ordered by a randomly chosen customer, and suppose that X has pmf. Compute E(X) and V(X). Then compute the expected number of pounds left after the next customer’s order is shipped and the variance of the number of pounds left.

Complete Question:

A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5-lb batches. Let X = the number of batches ordered by a randomly chosen customer, and suppose that X has the pmf shown in the table below. Compute E(X) and V(X).

Then compute the expected number of pounds left after the next customer’s order is shipped and the variance of the number of pounds left.

NOTE: The pmf is shown in the file attached

Expected number of batches, E(X) = 2.0 batches

Variance of batches, V(X) = 0.8 batches²

Expected number of pounds left, E(Y) = 90 lb

Variance of pounds left, V(Y) = 20 lb²

Step-by-step explanation:

The company has 100 batches and it sells to customers in 5 lb batches. The number of pounds left after each customer is filled will be modeled by the mathematical formula: Y = 100 - 5X

The expected value of X, E(X) can be calculated using the formula:

E(X) = (1*0.3) + (2*0.5) + (3*0.1) + (4*0.1)

E(X) = 0.3 + 1 + 0.3 + 0.4

E(X) = 2.0

Expected number of batches, E(X) = 2.0 batches

The variance of X, V(X) can be calculated using the formula:

Variance of batches, V(X) = 0.8 batches²

Y = 100 - 5X

E(Y) = E(100 - 5X)

E(Y) = 100 - 5E(X)

E(Y) = 100 - 5(2)

E(Y) = 90 lb

Expected number of pounds left, E(Y) = 90 lb

Variance of the number of pounds left:

V(Y) = V(100 - 5X)

V(Y) = V(100) + V(-5X)

V(100) = 0

V(Y) = (-5)² V(X)

V(Y) = 25 * 0.8

V(Y) = 20

Variance of pounds left, V(Y) = 20 lb²

(23*2)-12=x     x=34

step-by-step explanation:

more detail

step-by-step explanation: