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For this case, we must find the inverse of the following function:
To find the inverse we follow the steps below:
We rewrite the denominator:
We exchange variables:
We solve for y:
We multiply on both sides of the equation by (y-7)
We subtract xy on both sides of the equation:
We add 1 to both sides:
We factor for y:
We divide both sides by (-5-x):
So, we have:
Answer:
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) = (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
