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 <em>X</em> = number of cups of coffee sold and <em>Y</em> = number of donuts sold.
The random variable <em>X</em> follows a Normal distribution with parameters <em>μ</em> = 320 and <em>σ </em>= 20.
The random variable <em>Y</em> follows a Normal distribution with parameters <em>μ</em> = 150 and <em>σ </em>= 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:
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 = 0.5<em>X</em> + 0.4<em>Y</em>
Compute the probability that the shop owner earns more than $300 as profit as follows:
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:
<em>Y</em> > 0.5<em>X</em>
<em>Y</em> - 0.5<em>X</em> > 0
Compute the probability of the event (<em>Y</em> - 0.5<em>X</em> > 0) as follows:
Thus, the probability that the shop owner will sell a doughnut to more than half of his coffee customers is 0.2611.
