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13 March, 23:58

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?

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  1. 14 March, 00:08
    0
    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
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