At a certain coffee shop, all the customers buy a cup of coffee and some also bu

Question

At a certain coffee shop, all the customers buy a cup of coffee and some also buy a doughnut. The shop owner believes that the number of cups he sells each day is normally distributed with a mean of 340 cups and a standard deviation of 23 cups. He also believes that the number of doughnuts he sells each day is independent of the coffee sales and is normally distributed with a mean of 160 doughnuts and a standard deviation of 10. Complete parts a) through c).

a) The shop is open every day but Sunday. Assuming day-to-day sales are independent, what's the probability he'll sell over 2000 cups of coffee in a week? 0.761 (Round to three decimal places as needed.)

b) If he makes a profit of 50 cents on each cup of coffee and 40 cents on each doughnut, can he reasonably expect to have a day's profit of over $300? Explain. by chosing one

A. Yes. The number of doughnuts he expects to sell plus the number of cups of coffee is greater than 300.

B. Yes. $300 is less than 6 standard deviations above the mean.

C. No. The number of doughnuts he expects to sell plus the number of cups of coffee is less than 600.

D. No. $300 is more than 5 standard deviations above the mean.

c. What is the probality that on any given day he will sell a doughnut to more than half of his coffee customers?

Explanation / Answer

Mean number of cups of coffee = 340 cups and standard deviation = 23 cups

Here there are 6 days of operaions so toal number of expected number of coffees = 340 * 6 = 2040

Standard deviation of 6 days of operaions standard deviation of expected number of coffees in 6 days = 23 * sqr(6) = 56.338 coffees

so, if S is the total sales of coffee in 6 days

Here Pr(S > 2000 ; 2040 ; 56.338) =1 - Pr(S < 2000 ; 2040 ; 56.338)

Z = (2000 - 2040)/56.338 = -0.71

Pr(S > 2000 ; 2040 ; 56.338) =1 - Pr(S < 2000 ; 2040 ; 56.338) = 1 - Pr(Z < -0.71) = 1 - 0.239 = 0.761

(2) Expected amount of profit = Profit of one cup of coffe * Expected number of coffee sales + Profit on one doughnut * expected number of doughnut sales = 0.50 * 340 + 0.40 * 160 = $ 234

Here Standard deviation of the profit = 0.50 * STD (Coffee) + 0.40 * STD(doughnut) = sqrt[(0.50 * 23)2 + (0.40 * 10)2] = $ 12.176

so if profit is Pso

Pr( P > $ 300)

Z = (300 - 234)/12.176 = 5.42

so here option D is correct as $ 300 is more than 5 standard deviations above the mean.

(c) here is x is the number of doughnut sales and y is the number of coffee sales

then if Z = X - Y/2

so here E[X - Y/2] = E[X] - E[Y/2] = 160 - 340/2 = 160 - 170 = -10

Var(X - Y/2) = Var(X) + 1/4 * Var(Y) = 100 + 1/4 * 23 * 23 = 232.25

STD(X - Y/2) = 15.24

so,

Now we want to calculate that X - Y/2 > 0

Pr( X- Y/2 > 0 ; -10 ; 15.24) = 1 - Pr(X - Y/2 < - ; -10 ; 15.24)

Z = (0 + 10)/ 15.24 = 0.6562

Pr( X- Y/2 > 0 ; -10 ; 15.24) = 1 - Pr(X - Y/2 < - ; -10 ; 15.24) = 1 - Pr(0.6562) = 1 - 0.7442 = 0.2558

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.