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3 years ago

13

Questions are easy.................. ؛)

2 answers:

Snowcat [4.5K]3 years ago

5 0

Answer:

x>90

Step-by-step explanation:

x/3>30

(x/3)*3>30*3

x>90

MissTica3 years ago

4 0

Answer:

x>90

Step-by-step explanation:

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The gross weekly sales at a certain restaurant are a normal random variable with mean $2200 and standard deviation $230.What is

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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 $X_1$ and $X_2$ denote the sales during week 1 and 2 respectively.

a) Let $X=X_1+X_2$

Assuming that $X_1$ and $X_2$ follows same distribution with same mean and deviation.

$E(X)=E(X_1+X_2)=E(X_1)+E(X_2)$

$E(X)=2\mu = 2(220)=4400$

$\sigma_X=\sqrt{var(X_1+X_2)}$

$\sigma_X=\sqrt{2\sigma^2}$

$\sigma_X=\sqrt{2}\sigma$

$\sigma_X=230\sqrt{2}$

So, $X\sim N(4400,230\sqrt{2})$

$P(X>5000)=1-P(X\leq5000)$

$P(X>5000)=1-P(Z\leq\frac{5000-4400}{230\sqrt{2}})$

$P(X>5000)=1-P(Z\leq1.844)$

$P(X>5000)=1-0.967$

$P(X>5000)=0.0321$

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.

$P(X>2000)=1-P(X\leq2000)$

$P(X>2000)=1-P(Z\leq\frac{2000-2200}{230})$

$P(X>2000)=1-P(Z\leq-0.87)$

$P(X>2000)=1-0.1922$

$P(X>2000)=0.808$

Let Y be the number of weeks in which sales exceed 2000.

Now, $P(Y\geq 2)$

So, $P(Y\geq 2)=P(Y=2)+P(Y=3)$

$P(Y\geq 2)=^3C_2(0.8077)^2\cdot (1-0.8077)+^3C_3(0.8077)^3$

$P(Y\geq 2)=0.37635+0.52692$

$P(Y\geq 2)=0.90327$

The weekly sales exceed $2000 in at least 2 of the next 3 weeks is 0.9033.

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