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 310 cups and a standard deviation of 24 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 120 doughnuts and a standard deviation of 17. 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. 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 8 standard deviations above the mean.

C. No. $300 is more than 7 standard deviations above the mean.

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

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

(Round to three decimal places as needed.)

Explanation / Answer

Answer: 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 310 cups and a standard deviation of 24 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 120 doughnuts and a standard deviation of 17. 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.)

For a week, mean µ= 6*310 =1860,

= sd(6x) = 6sd(x) = 6*24 =144

z value for 2000, z =(2000-1860)/144 =0.97

P( x >2000) = P( z >0.97)

=0.834

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. 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 8 standard deviations above the mean.

Answer: C. No. $300 is more than 7 standard deviations above the mean.

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

Daily profit mean = 0.5*310+0.4*120 =203

Daily profit variance = 0.5^2*24^2+0.4^2*17^2 = 190.24

Daily profit standard deviation =sqrt(190.24)= 13.7928

Z value for 300 , z =(300-190.24)/13.7928 =7.96

$300 is more than 7 SD away from the mean

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

(Round to three decimal places as needed.)

Define a random variable Y = D –½ C. Find the probability that the random variable is greater than zero.

= 120 –½ (310) = -35

Var(Y) = Var(D) + Var(0.5C) =24^2+0.25*17^2 = 648.25

Sd(Y) = sqrt(648.25) =25.4608

Z value for 0, z=(0-(-35))/25.4608 =35/25.4608 =1.37

P( x >0) = P( z >1.37)

=0.085

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