Respuesta :
Answer:
a) The total gross sales over the next 2 weeks exceeds $5000 is 0.0321.
b) The weekly sales exceed $2000 in at least 2 of the next 3 weeks is 0.9033.
Step-by-step explanation:
Given : The gross weekly sales at a certain restaurant are a normal random variable with mean $2200 and standard deviation $230.
To find : What is the probability that
(a) the total gross sales over the next 2 weeks exceeds $5000;
(b) weekly sales exceed $2000 in at least 2 of the next 3 weeks? What independence assumptions have you made?
Solution :
Let [tex]X_1[/tex] and [tex]X_2[/tex] denote the sales during week 1 and 2 respectively.
a) Let [tex]X=X_1+X_2[/tex]
Assuming that [tex]X_1[/tex] and [tex]X_2[/tex] follows same distribution with same mean and deviation.
[tex]E(X)=E(X_1+X_2)=E(X_1)+E(X_2)[/tex]
[tex]E(X)=2\mu = 2(220)=4400[/tex]
[tex]\sigma_X=\sqrt{var(X_1+X_2)}[/tex]
[tex]\sigma_X=\sqrt{2\sigma^2}[/tex]
[tex]\sigma_X=\sqrt{2}\sigma[/tex]
[tex]\sigma_X=230\sqrt{2}[/tex]
So, [tex]X\sim N(4400,230\sqrt{2})[/tex]
[tex]P(X>5000)=1-P(X\leq5000)[/tex]
[tex]P(X>5000)=1-P(Z\leq\frac{5000-4400}{230\sqrt{2}})[/tex]
[tex]P(X>5000)=1-P(Z\leq1.844)[/tex]
[tex]P(X>5000)=1-0.967[/tex]
[tex]P(X>5000)=0.0321[/tex]
The total gross sales over the next 2 weeks exceeds $5000 is 0.0321.
b) The probability that sales exceed teh 2000 and amount in at least 2 and 3 next week.
We use binomial distribution with n=3.
[tex]P(X>2000)=1-P(X\leq2000)[/tex]
[tex]P(X>2000)=1-P(Z\leq\frac{2000-2200}{230})[/tex]
[tex]P(X>2000)=1-P(Z\leq-0.87)[/tex]
[tex]P(X>2000)=1-0.1922[/tex]
[tex]P(X>2000)=0.808[/tex]
Let Y be the number of weeks in which sales exceed 2000.
Now, [tex]P(Y\geq 2)[/tex]
So, [tex]P(Y\geq 2)=P(Y=2)+P(Y=3)[/tex]
[tex]P(Y\geq 2)=^3C_2(0.8077)^2\cdot (1-0.8077)+^3C_3(0.8077)^3[/tex]
[tex]P(Y\geq 2)=0.37635+0.52692[/tex]
[tex]P(Y\geq 2)=0.90327[/tex]
The weekly sales exceed $2000 in at least 2 of the next 3 weeks is 0.9033.
