John is four years older than frank, the sum of their ages is 36

A new shopping mall is considering setting up an information desk manned by one employee. Based upon information obtained from s

quester [9]

Answer:

a) $P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

b) $p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

c) $L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

d) $L_q =\frac{20^2}{30(30-20)}=1.333 people$

e) $W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

f) $W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

Step-by-step explanation:

Notation

P represent the probability that the employee is idle

$p_x$ represent the probability that the employee is busy

$L_s$ represent the average number of people receiving and waiting to receive some information

$L_q$ represent the average number of people waiting in line to get some information

$W_s$ represent the average time a person seeking information spends in the system

$W_q$ represent the expected time a person spends just waiting in line to have a question answered

This an special case of Single channel model

Single Channel Queuing Model. "That division of service channels happen in regards to number of servers that are present at each of the queues that are formed. Poisson distribution determines the number of arrivals on a per unit time basis, where mean arrival rate is denoted by λ".

Part a

Find the probability that the employee is idle

The probability on this case is given by:

In order to find the mean we can do this:

$\mu = \frac{1question}{2minutes}\frac{60minutes}{1hr}=\frac{30 question}{hr}$

And in order to find the probability we can do this:

$P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

Part b

Find the proportion of the time that the employee is busy

This proportion is given by:

$p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

Part c

Find the average number of people receiving and waiting to receive some information

In order to find this average we can use this formula:

$L_s= \frac{\lambda}{\lambda -\mu}$

And replacing we got:

$L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

Part d

Find the average number of people waiting in line to get some information.

For the number of people wiating we can us ethe following formula"

$L_q =\frac{\lambda^2}{\mu(\mu-\lambda)}$

And replacing we got this:

$L_q =\frac{20^2}{30(30-20)}=1.333 people$

Part e

Find the average time a person seeking information spends in the system

For this average we can use the following formula:

$W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

Part f

Find the expected time a person spends just waiting in line to have a question answered (time in the queue).

For this case the waiting time to answer a question we can use this formula:

$W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

6 0

2 years ago

Read 2 more answers

Anyone know what -10 + 9 is?

KonstantinChe [14]

Answer:

Step-by-step explanation:

-1

3 0

3 years ago

Read 2 more answers

What rectangle shows the final image

frozen [14]

Is there an image attached?

6 0

3 years ago

In a high school graduating class of 128 students, 52 are on the honor roll. Of these, 48 are going on to college; of the other

olga nikolaevna [1]

Answer:

a. 0.8125

b. 0.1875

c. 0.03125

Step-by-step explanation:

from the information given, we can come up with the following data

Total No of students =128

Total Nos of students going to college = 48+56=104

Total Nos of students not going to college and on the honor roll= 52-48=4

a. To determine the probability of students going to college we have

P(going to college) = (total number of students going to college)/total number of students

P(going to college) = 104/128

P(going to college) = 0.8125

b. To determine the probability of students not going to college, we use the rule that says total probability is 1, hence

Pr(not going to college)=1-Pr(going to college)

Pr(not going to college)=1-0.8125

Pr(not going to college)=0.1875

c. To determine the probability of students NOT going to college and on the pay roll we have

Pr = (Total Number of students not going to college and on the honor roll)/total number of students

Pr=4/128

Pr=0.03125

7 0

3 years ago

A child walks 5.0 meters north, then 4.0 meters east, and finally 2.0 meters south. What is the magnitude of the resultant displ

Dafna11 [192]

<h2>file:///E:/Kinematics%20Review%208080.pdf</h2><h2></h2><h2>here is your answer</h2>

6 0

2 years ago