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

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

170

doughnuts and a standard deviation of

10

.

​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?

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

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

Answer 1

Answer:

Given,

Mean = 360

Standard deviation = 21

a)

z score = (x - u)/s

substitute values

= (2000 - 6*360)/(6*21)

= - 1.27

P(X > 2000) = P(z > - 1.27)

= 0.8979577 [since from z table]

= 0.8980

= 89.80%

b)

z = (x - u)/s

substitute values

= [300 - 0.5*360 - 0.4*170] / sqrt(0.5^2*21^2 + 0.4^2*10^2)

= 4.63

Here there is no reasonable chance to earn profit of more than 300 due to that 300 > 7 standard deviations from mean.

c)

z = (x - u)/s

substitute values

= [0 - (170 - 0.5*360)] / sqrt(10^2 + 0.5^2*21^2)

= 0.69

P(Y - 0.5X > 0) = 0.2450971 [since from z table]

= 0.2451

= 24.51%

answered by: anonymous