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 22 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 180 doughnuts and a standard deviation of 15. 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.
A. No. $300 is more than 5 standard deviations above the mean.
B. No. The number of doughnuts he expects to sell plus the number of cups of coffee is less than 600.
C. Yes. $300 is less than 6 standard deviations above the mean.
D. Yes. The number of doughnuts he expects to sell plus the number of cups of coffee is greater than 300.
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So, everything beyond 2 deviation is unusual.
Lets see how much is the expected value at 2 deviation above mean
For coffee, expected mean of the combined distribution = Mean 1 + Mean 2 = .5*310 + .4*180 = $227
For donut, expected mean of the combined distribution = sqrt((.5*22)^2 + (.4*15)^2) = $12.53
So, $300 is (300-227)/12.53 = 5.83 deviations away from mean
So, A is correct. It is unusual to expect $300
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