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 21 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 150 doughnuts and a standard deviation of 13. 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. O A. Yes. $300 is less than 6 standard deviations above the mean. OB. No $300 is more than 5 standard deviations above the mean. O C. No. The number of doughnuts he expects to sell plus the number of cups of coffee is less than 600. OD. 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? Round to three decimal places as needed.)

Answer 1

a)

µx = 6µ = 6*340 = 2040                  
σx = √nσ = √6 * 21 =  51.4393
                      
P ( X > 2000   ) = P( (X-µ)/σ ≥ (2000-2040) / 51.4393)              
= P(Z ≥   -0.778   ) = P( Z <   0.778   ) = 0.7816   (answer)

b)

Daily profit mean=340*0.5+150*0.4=230
daily profit variance=(0.5*21)^2+(0.4*13)^2 = 137.29

P(X>300)=P(Z> ((300-230)/√137..29 ) = P(Z>5.97)

so, answer is option B)

c)

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

µ=150-0.5*340 = -20

Var(Y) = Var(D) + Var(0.5C) =13^2+0.25*21^2 = 279.25
P ( X ≥   0   ) = P( (X-µ)/√σ ≥ (0+20) /√279.25)              
= P(Z ≥   1.197   ) = P( Z <   -1.197   ) =    0.1157   (answer)