Question

A gas station earns $2.5 in revenue for each gallon of regular gas it sells, $2.45 for each gallon of midgrade gas, and $2.5 for each gallon of premium gas. Let X1 , X2 , and X3 denote the numbers of gallons of regular, midgrade, and premium gasoline sold in a day. Assume that X1 , X2 , and X3 have means μ1 = 1500, μ2 = 500, and μ3 = 300, and standard deviations σ1 = 180, σ2 = 90, and σ3 = 40, respectively.

Answer 1

Answer:

a) $R= 2.5 X_1 +2.45 X_2 + 2.5 X_3$

And the expected value for R is given by:

$E(R) = 2.5 E(X_1) + 2.45 E(X_2) + 2.5 E(X_3) =2.5*1500 + 2.45*500 + 2.5*300=5725$

b) We have that $Cov (X_i, X_j) =0 , i,j =1,2,3$ since are independent

First we need to find the Variance like this:

$Var(R) = Var(X_1) +Var(X_2)+Var(X_3)$

And replacing we have this:

$Var(R) = Var(2.5 X_1) +Var(2.45 X_2)+ Var(2.5 X_3)$

And using properties of variance w ehave this:

$Var(R) = 2.5^2 Var(X_1) + 2.45^2 Var(X_2) + 2.5^2 Var(X_3) = 2.5^2 *180^2 + 2.45^2 *90^2 + 2.5^5 *40^2 =261120.25$

And the deviation would be:

$Sd(R)= \sqrt{261120.25}= 510.999$

Step-by-step explanation:

We have the following info:

$X_1 \sim N (\mu_1 = 1500, \sigma_1 =180)$

$X_2 \sim N (\mu_2 = 500, \sigma_2 =90)$

$X_3 \sim N (\mu_3 = 300, \sigma_3 =40)$

Assuming the following questions:

a) Find the mean daily revenue

Based on the info given the daily revenue is given by:

$R= 2.5 X_1 +2.45 X_2 + 2.5 X_3$

And the expected value for R is given by:

$E(R) = 2.5 E(X_1) + 2.45 E(X_2) + 2.5 E(X_3) =2.5*1500 + 2.45*500 + 2.5*300=5725$

b) Assuming that X1, X2 and X3 to be independent find the standard deviation for the daily revenue

We have that $Cov (X_i, X_j) =0 , i,j =1,2,3$ since are independent

First we need to find the Variance like this:

$Var(R) = Var(X_1) +Var(X_2)+Var(X_3)$

And replacing we have this:

$Var(R) = Var(2.5 X_1) +Var(2.45 X_2)+ Var(2.5 X_3)$

And using properties of variance w ehave this:

$Var(R) = 2.5^2 Var(X_1) + 2.45^2 Var(X_2) + 2.5^2 Var(X_3) = 2.5^2 *180^2 + 2.45^2 *90^2 + 2.5^5 *40^2 =261120.25$

And the deviation would be:

$Sd(R)= \sqrt{261120.25}= 510.999$