Question

Customers at a fast-food restaurant buy both sandwiches and drinks. The mean number of sandwiches is 1.5 with a standard deviation of 0.5. The mean number of drinks is 1.45 with a standard deviation of 0.3. The correlation between the number of sandwiches and drinks purchased by the customer is 0.6. If the profit earned from selling a sandwich is $1.50 and from a drink is $1, what is the expected value and standard deviation of profit made from each customer.

Answer 1

Let X denote the number of sandwiches and Y denote the number of Drinks

Profit from each customer, P = 1.5X + Y

Expected value of P = E(1.5X + Y)

= 1.5*E(X) + E(Y)

= 1.5*1.5 + 1.45 = $3.70

Variance of P = Var(1.5X + Y)

= Var(1.5X) + Var(Y) + 2Cov(1.5X, Y)

= *2

Now, Cov(X,Y) = Corr(X,Y)*SD(X)*SD(Y)

= 0.6*0.5*0.3 = 0.09

Thus, Variance of P =

= 0.9225

Thus, Standard deviation of profit made from each customer = √0.9225

= $0.96