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 360 cups and a standard deviation of 25 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 130 doughnuts and a standard deviation of 12. 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?

(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.

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

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

C. No. $300 is more than 5 standard deviations above the mean.Your answer is correct.

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

c) What's the probability that on any given day he'll sell a doughnut to more than half of his coffee customers?

Explanation / Answer

a) here mean sale of cups for 6 days =6*360=2160

and std deviaiton =25*(6)1/2 =61.237

hence P(X>2000)=1-P(X<2000)=1-P(Z<(2000-2160)/61.237)=1-P(Z<-2.6128)=1-0.004=0.996

b)mean profit =0.5*360+0.4*130=232

and std deviation =((0.5*25)2+(0.4*12)2)1/2=13.39

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

c)here we need to find out P(2Y-X>0) =P(X-2Y<0) where Y is number of doughnut and X =coffee cups

here mean of X-2Y=360-2*130=100

and std deviation =(252+(2*12)2)1/2 =34.66

hence P(X-2Y<0)=P(Z<(0-(100))/34.66)=P(Z<-2.8856)=0.0020

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