Respuesta :
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 X = number of cups of coffee sold and Y = number of donuts sold.
The random variable X follows a Normal distribution with parameters μ = 320 and σ = 20.
The random variable Y follows a Normal distribution with parameters μ = 150 and σ = 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:
[tex]P(X>2000)=P(\frac{X-\mu}{\sigma}>\frac{2000-(6\times320)}{6\times20})\\=P(Z>0.67)\\=1-P(Z<0.67)\\=1-0.7486\\=0.2514[/tex]
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 = 0.5X + 0.4Y
Compute the probability that the shop owner earns more than $300 as profit as follows:
[tex]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[/tex]
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:
Y > 0.5X
Y - 0.5X > 0
Compute the probability of the event (Y - 0.5X > 0) as follows:
[tex]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<0.64)\\=1-0.7389\\=0.2611[/tex]
Thus, the probability that the shop owner will sell a doughnut to more than half of his coffee customers is 0.2611.
