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:
![P(X>2000)=P(\frac{X-\mu}{\sigma}>\frac{2000-(6\times320)}{6\times20})\\=P(Z>0.67)\\=1-P(Z](https://tex.z-dn.net/?f=P%28X%3E2000%29%3DP%28%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cfrac%7B2000-%286%5Ctimes320%29%7D%7B6%5Ctimes20%7D%29%5C%5C%3DP%28Z%3E0.67%29%5C%5C%3D1-P%28Z%3C0.67%29%5C%5C%3D1-0.7486%5C%5C%3D0.2514)
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:
![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](https://tex.z-dn.net/?f=P%28Profit%3E300%29%3DP%28%5Cfrac%7BProfit-%5Cmu%7D%7B%5Csigma%7D%3E%5Cfrac%7B300-%28%280.5%5Ctimes320%29%2B%280.4%5Ctimes150%29%29%7D%7B%5Csqrt%7B0.5%5E%7B2%7D%2820%29%5E%7B2%7D%2B0.4%5E%7B2%7D%2812%29%5E%7B2%7D%7D%7D%29%5C%5C%3DP%28Z%3E7.21%29%5C%5C%5Capprox0)
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:
![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](https://tex.z-dn.net/?f=P%28Y%20-%200.5X%20%3E%200%29%3DP%28%5Cfrac%7B%28Y%20-%200.5X%29%20-%5Cmu%7D%7B%5Csigma%7D%3E%5Cfrac%7B0-%28150-%280.5%5Ctimes320%7D%7B%5Csqrt%7B12%5E%7B2%7D%2B0.5%5E%7B2%7D20%5E%7B2%7D%7D%7D%29%5C%5C%3DP%28Z%3E0.64%29%5C%5C%3D1-P%28Z%3C0.64%29%5C%5C%3D1-0.7389%5C%5C%3D0.2611)
Thus, the probability that the shop owner will sell a doughnut to more than half of his coffee customers is 0.2611.