Mckenzie has been saving her earnings from mowing the

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total s

garri49 [273]

Answer

a. The expected total service time for customers = 70 minutes

b. The variance for the total service time = 700 minutes

c. It is not likely that the total service time will exceed 2.5 hours

Step-by-step explanation:

This question is incomplete. I will give the complete version below and proceed with my solution.

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total service time for customers arriving during a 1-hour period. (Assume that a sufficient number of servers are available so that no customer must wait for service.) Is it likely that the total service time will exceed 2.5 hours?

Reference

Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour.

From the information supplied, we denote that

$X=$ Customers that arrive within the hour

and since $X$ follows a Poisson distribution with mean $\alpha$ = 7

Therefore,

$E(X)= 7$

$& V(X)=7$

Let $Y =$ the total service time for customers arriving during the 1 hour period.

Now, since it takes approximately ten minutes to serve each customer,

$Y=10X$

For a random variable $X$ and a constant $c$ ,

$E(cX)=cE(X)\\V(cX)=c^2V(X)$

Thus,

$E(Y)=E(10X)=10E(X)=10*7=70\\V(Y)=V(10X)=100V(X)=100*7=700$

Therefore the expected total service time for customers = 70 minutes

and the variance for serving time = 700 minutes

Also, the probability of the distribution $Y$ is,

$p_Y(y)=p_x(\frac{y}{10} )\frac{dx}{dy} =\frac{\alpha^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-\alpha } \frac{1}{10}\\ =\frac{7^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-7 } \frac{1}{10}$

So the probability that the total service time exceeds 2.5 hrs or 150 minutes is,

$P(Y>150)=\sum^{\infty}_{k=150} {p_Y} (k) =\sum^{\infty}_{k=150} \frac{7^{\frac{k}{10} }}{(\frac{k}{10})! }.e^{-7} .\frac{1}{10} \\=\frac{7^{\frac{150}{10} }}{(\frac{150}{10})! } .e^{-7}.\frac{1}{10} =0.002$

0.002 is small enough, and the function $\frac{7^{\frac{k}{10} }}{(\frac{k}{10} )!} .e^{-7}.\frac{1}{10}$ gets even smaller when $k$ increases. Hence the probability that the total service time exceeds 2.5 hours is not likely to happen.

3 0

3 years ago

4) Choose the correct inequality for this situation.

Ksenya-84 [330]

Answer:

B. 21x + 40 ≤ 124

Step-by-step Explanation:

Maximum amount budgeted = $124 (this means they can't spend more than this)

x = number of people

Cost per head = $21

Given that Mr Walter already spent $40, which is part of the money budgeted, the number of people that can go canoeing cam be expressed with the following inequality:

21x + 40 ≤ 124

(note: the amount total to be spent will either be equal to or greater than $124, because it's the maximum amount budgeted for spending).

6 0

3 years ago

Archimedes drained the water in his tub. 62.5, point, 5 liters of water were drained each minute, and the tub was completely dra

Firlakuza [10]

Answer

its 500 and 8

Step-by-step explanation:

got it from khan academy

5 0

3 years ago

Is h(x)=6 a linear equation, explain

puteri [66]

A linear equation is an equation for a straight line. And yes that is a linear equation. Because<span> it is of the form h(x) = mx + b where m=0 and b = 23 h(x) = mx + 23</span>

8 0

3 years ago

Draw a Triangle whose tangent is 2. Find the values of the other five trigonometric functions

fenix001 [56]

Answer:

you gonna answer it

Step-by-step explanation:

Because that's yours book

4 0

3 years ago