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 320 cups and a standard deviation of 20 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 12. 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? If he makes a profit of 50 cents on each cup of coffee and 40 cents on each doughnut, can he reasonable expect to have a day?s profit of over $300? Explain. 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:

(1) The probability that the shop owner sells over 2000 cups of coffee in a week is 0.2514.

(2) The shop owner has no reasonable chance to expect earning a profit more than $300.

(3) The probability that the shop owner will sell a doughnut to more than half of his coffee customers is 0.2611.

Step-by-step explanation:

Let X = number of cups of coffee sold and Y = number of donuts sold.

The random variable X follows a Normal distribution with parameters μ = 320 and σ = 20.

The random variable Y follows a Normal distribution with parameters μ = 150 and σ = 12.

The shop owner opens the shop 6 days a week.

(1)

Compute the probability that the shop owner sells over 2000 cups of coffee in a week as follows:

$P(X>2000)=P(\frac{X-\mu}{\sigma}>\frac{2000-(6\times320)}{6\times20})\\=P(Z>0.67)\\=1-P(Z$

Thus, the probability that the shop owner sells over 2000 cups of coffee in a week is 0.2514.

(2)

The equation representing the profit earned on selling 1 cup of coffee and 1 doughnut in a day is:

P = 0.5X + 0.4Y

Compute the probability that the shop owner earns more than $300 as profit as follows:

$P(Profit>300)=P(\frac{Profit-\mu}{\sigma}>\frac{300-((0.5\times320)+(0.4\times150))}{\sqrt{0.5^{2}(20)^{2}+0.4^{2}(12)^{2}}})\\=P(Z>7.21)\\\approx0$

The probability of earning a profit more then $300 is approximately 0.

Thus, the shop owner has no reasonable chance to expect earning a profit more than $300.

(3)

The expression representing the statement "he'll sell a doughnut to more than half of his coffee customers" is:

Y > 0.5X

Y - 0.5X > 0

Compute the probability of the event (Y - 0.5X > 0) as follows:

$P(Y - 0.5X > 0)=P(\frac{(Y - 0.5X) -\mu}{\sigma}>\frac{0-(150-(0.5\times320}{\sqrt{12^{2}+0.5^{2}20^{2}}})\\=P(Z>0.64)\\=1-P(Z$

Thus, the probability that the shop owner will sell a doughnut to more than half of his coffee customers is 0.2611.