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 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.

Answer 1

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

Answer:

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:

$E(X) = \sum x p(x)$

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:

$V(X) = E(X^2) - [E(X)]^2\\ E(X^2) = \sum x^2 p(x)\\ E(X^2) = (1^2 *0.3) + (2^2 * 0.5) + (3^2 * 0.1) + (4^2 * 0.1)\\ E(X^2) = 0.3 + 2 + 0.9 + 1.6\\ E(X^2) = 4.8\\V(X) = 4.8 - 2^2\\V(X) = 0.8$

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

Answer 2

The question is incomplete. Here is the 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 5 the number of batches ordered by a randomly chosen customer, and suppose that X has the following pmf.

 x      1         2        3       4

P(x)  0.3     0.5     0.1     0.1

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.

Answer: E(X) = 2

V(X) = 0.8

Expected number of pounds left = 90

Variance of the number of pounds left = 20

Step-by-step explanation: Since this is a probability distribution, to determine E(X), which is a measure of central tendency:

E(X) = ∑[x*P(x)]

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

E(X) = 2

V(X) is the variance for this type of distribution and is calculated as:

V(X) = ∑{[x - E(x)]²*P(x)} = (1-2)²*0.3+(2-2)²*0.5+(3-2)²*0.1+(4-2)²*0.1

V(X) = 0.8

The number of pounds left after the next costumer's order is shipped is given by the equation: Y = 100 - 5X

The expected number is equal to E(Y), which is this case is:

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

One property of E(X) is that if a and b are constants:

E(aX+b) = aE(X) + b

Taking that a = - 5 and b = 100:

E(100 - 5X) =  (-5).2 + 100

E(Y) = 90

The variance of the number of pounds left is given by:

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

For variance, the property is: if a and b are constants:

V(aX+b) = a²V(X)

V(100 - 5X) = (-5)²*0.8

V(Y) = 20