Question

At a certain coffee shop, all the customers buy a cup of coffee; some also buy a doughnut. The shop owner believes that the number of cup s 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.
A) The shop is open every day but Sunday. Assuming day-to-day sales are independent, what's the probability he'll sell more than 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:

a) 0.2514

b) No

c) 0.2611

Step-by-step explanation:

For coffee cups sold daily

Mean(μ) = 320 cups

Standard deviation (σ) = 20 cups

For doughnut sold daily

Mean(μ) = 150 doughnut

Standard deviation (σ) = 12 doughnut

Let X be the random variable representing the number of cups.

Let y be the random variable representing the number of doughnuts.

a) The shop is opened everyday except Sunday's. This means that the shop is opened 6 times in a week.

The probability that he will sell more 2000 cups of coffee in a week = Pr(x>2000)

Using normal distribution

Z = (x - μ)/σ

Z = (2000 - 6(320))/6(20)

Z = (2000 - 1920) / 120

Z = 80/120

Z = 0.67

From the normal distribution table, 0.67 = 0.2486

Φ(z) = 0.2486

Recall that if Z is positive,

Pr(x>a) = 0.5 - Φ(z)

Pr(x>2000) = 0.5 - 0.2486

= 0.2514

b) On each cup of coffee, he makes a profit of 50 cents and makes a profit of 40 cents on each doughnut. We have 0.5x + 0.4y

Z = (x - μ) / σ

Z = (300 - (0.5*320 + 0.4*150)) / √0.5^2*20^2 + 0.4^2*12^2

Z = (300 - (160+60)) / √100+23.04

Z = (300 -220)/√123.04

Z = 80/11.09

= 7.02

Since 300 is more than 7 standard deviation from the mean and the value of z cannot be found in the normal distribution table, then he has NO reasonable chance to earn a profit more than 300.

c) The probability that on any given day he will sell a doughnut to more than half of his coffee customers = Pr(y - 0.5x > 0)

From normal distribution,

Z = (x - μ) / σ

Z = (0 - (150 - 0.5*320)) / √12^2 + 0.5^2*20^2

Z = (0 - (150 - 160)) / √144 + 100

Z = (0 -(-10))/√244

Z = 10/15.62

= 0.64

From the normal distribution table, 0.64 = 0.2389

Φ(z) = 0.2389

Recall that if Z is positive,

Pr(x>a) = 0.5 - Φ(z)

Pr(y - 0.5x >0) = 0.5 - 0.2389

= 0.2611