Question

(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?

7.1.55 Question Help 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 130 doughnuts and a standard deviation of 15. Complete parts a) through c).

Answer 1

Let X_i; i = 1,2,3,4,5,6; denote the number of cups of coffee sold on the six days of a week (except Sunday). Moreover let W be the number of cups of coffee sold in a week (except Sunday when the shop is closed). Clearly, W = X₁ + X₂ + ... + X₆

Now, we are given that the number of cups sold each day is normally distributed with a mean of 310 cups and a standard deviation of 24 cups. Moreover, we are given that day-to-day sales are independent, which implies that:

XiN (310, 24) i = 1, ...,6; X's independent

Thus, the distribution of W = X₁ + X₂ + ... + X₆ is given by:

(A)

W = X1 + X2 + ... + X6 ~ Nu = 310 + 310 + ... + 310, 62 = 242 +242 + ... +242) [Since X's are independent] W ~ N(n = 6 * 310, o2 = 6 * 242) →W ~ N(n = 1860, o2 = 3456) → Z-W-_W - 1860 *2= o = 3456 ~ N(0,1); the standard normal distribution. ~

The probability​ he'll sell over 2000 cups of coffee in a​ week is given by:

2000 – 1860 P(W > 2000) =P (W - o 3456 = P(Z > 2.381448) Using the N(0.1) tables, we get = 0.008622 ANSWER

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