Question 1

A homemade-ice cream shop owner has noticed that,often,potential customers will stop outside the store,assess the wait in line,and then pass by.The shop has a(n):

Accepted Answer

Customers who decide not to enter a queue because it is too long are considered to be "balking." Since the scenario involves a public place where the number of potential customers is not limited by membership or a specific community, it implies an infinite customer population.

Question 2

Customers arrive according to a Poisson distribution.The average number of customer arrivals per hour is six.The probability that four customers will arrive in the next three hours is:

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The answer of Customers arrive according to a Poisson distribution.The...

Question 3

A bank lobby with six teller windows,each with a separate line,is an example of a:

Accepted Answer

The bank lobby with six teller windows, each with a separate line, is an example of a multiple-channel, single-phase arrangement. Multiple channels refer to the six separate lines, and single-phase means that each customer goes through only one phase (receiving service from one teller).

Question 4

Customers arrive every three minutes governed by a Poisson distribution.A single server is capable of handling twenty-five customers per hour and this service time is exponentially distributed.What is the likelihood that the server is not busy at any given time?

Accepted Answer

The arrival rate of customers is λ= 1/3 per minute, or 20 per hour. 
The service rate is μ = 25 customers per hour. 
The utilization factor ρ = λ/μ = 20/25 = 0.8. 
Using Little's Law, the average time a customer spends in the system R = L/λ, where L is the average number of customers in the system. 
Since the server is capable of handling all customers, L is equivalent to the number of customers in queue, which is given by the Lq= ρ^2 / (1- ρ) = 3.2 customers.
Thus, the expected time a customer spends in the system is R = L/λ = (3.2+0.8)/20 = 0.2 hours. 
The probability that the server is not busy at any given time is equal to the probability that there are no customers in the system, which is equal to the probability of the server being idle. 
The server is idle when there are no customers in the system. Thus, the probability of the server being idle is equal to the probability that no customers arrive during a service time. 
Thus, the probability that the server is not busy at any given time P0= e^(-μT), where T is the length of time we are interested in. 
If we take T = 1 hour, then P0 = e^(-25*1) = 3.8 x 10^-11. 
The likelihood that the server is not busy at any given time is less than or equal to 0.1 since the value is very small. 
Therefore, the correct answer is C.

Question 5

The owner of a desktop publishing company has seven loyal clients who periodically require his services.The owner has:

Accepted Answer

The question mentions that the owner has seven loyal clients who periodically require his services, which indicates a finite customer population. There is no mention of balking (customers who leave without purchasing) or impatience, so options B and D can be eliminated. Therefore, the best choice is C - a finite customer population.

Question 6

Customers arrive every three minutes governed by a Poisson distribution.A single server is capable of handling twenty-five customers per hour and this service time is exponentially distributed.About how long do customers wait in line?

Accepted Answer

The arrival rate is 20 customers per hour (60/3), and the service rate is 25 customers per hour. Using the formula for average wait time in a queue (Wq = λ / μ(μ - λ)), where λ is the arrival rate and μ is the service rate, the wait time is a little over nine minutes.

Question 7

Customers are serviced at a rate of three customers per hour according to an exponential distribution.What is the probability that customer service will require fewer than 10 minutes?

Accepted Answer

We first need to convert the service rate from customers per hour to customers per minute. 
Three customers per hour = 3/60 = 1/20 customers per minute.

Next, we need to find the probability that customer service will require fewer than 10 minutes.

Using the exponential distribution formula: 
P(X < 10) = 1 - e^(-λx) 
where λ is the service rate (1/20) and x is the time in minutes (10).

P(X < 10) = 1 - e^(-1/20 * 10) 
P(X < 10) = 1 - e^(-1/2)

Calculating this out, we get: 
P(X < 10) ≈ 0.393

So the answer is A, less than or equal to 0.40.

Question 8

A super computer-accessory discount store often has customers who leave the checkout line before being served because of excessive waiting times.The store has a(n):

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The answer of A super computer-accessory discount store often has...

Question 9

An automatic,drive-through car wash is an example of a:

Accepted Answer

An automatic,drive-through car wash has only one channel, which is the path that the car follows through the wash. The process is also a single-phase arrangement, as there is only one step to the process (washing the car).

Question 10

Customers arrive every three minutes governed by a Poisson distribution.A single server is capable of handling twenty-five customers per hour and this service time is exponentially distributed.About how many minutes per hour is the server not helping customers?

Accepted Answer

The arrival rate is 20 customers per hour (60/3), and the service rate is 25 customers per hour. The utilization rate is 20/25 = 0.8, meaning the server is busy 80% of the time and idle 20% of the time. Therefore, the server is not helping customers for 0.2 * 60 = 12 minutes per hour.

Question 11

Customers arrive according to a Poisson distribution.The average number of customer arrivals per hour is four.The probability that three customers will arrive in the next two hours is:

Accepted Answer

The average number of customer arrivals in two hours is 8 (since 4 customers arrive per hour on average). Using the Poisson distribution formula $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the average rate (8 for two hours) and $k$ is the number of occurrences (3 customers), we find $P(X=3) = \frac{e^{-8} 8^3}{3!}$. This calculation results in a probability greater than 0.025, making option D correct.

Question 12

Customers arrive according to a Poisson distribution.The average number of customer arrivals per hour is three.The probability that four customers will arrive in the next two hours is:

Accepted Answer

The average rate of arrival is 3 customers per hour, so for two hours, the average is 6. Using the Poisson probability formula, the probability of exactly 4 arrivals in two hours is approximately 0.1338, which falls in the range greater than 0.12 but less than or equal to 0.14.

Question 13

A drive-through system at a fast food restaurant where the first facility takes the order,the second takes the money,and the third provides the food is an example of a:

Accepted Answer

This is a single-channel arrangement since the customer is going through one drive-through lane. However, it is a multiple-phase arrangement since there are three distinct phases to the transaction (order, payment, and food delivery).

Question 14

Customers are serviced at a rate of six customers per hour according to an exponential distribution.What is the probability that customer service will require fewer than 20 minutes?

Accepted Answer

We need to convert the time unit from hour to minutes: 
Six customers per hour $ightarrow$ six customers per 60 minutes. 
Therefore, the rate parameter $\lambda$ of the exponential distribution is $\lambda = 6/60 = 0.1$ customers per minute.

Let X be the time in minutes needed to serve one customer. 
We want to find P(X < 20).

The cumulative distribution function (cdf) of an exponential distribution with rate parameter $\lambda$ is given by 
$F(x) = 1 - e^{-\lambda x}$ for x > 0

Therefore, 
$P(X < 20) = F(20) = 1 - e^{-0.1	imes 20} \approx 0.8187$

So the answer is (C), which corresponds to a probability between 0.80 and 0.90.

Question 15

The single,multiple,and finite waiting line models all assume that the:

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The answer of The single,multiple,and finite waiting line models all...

Question 16

A Laundromat where there are washing machines and dryers is an example of a:

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The answer of A Laundromat where there are washing machines...

Question 17

Customers are serviced at a rate of five customers per hour according to an exponential distribution.What is the probability that customer service will require fewer than 20 minutes?

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The answer of Customers are serviced at a rate of...

Question 18

Customers arrive every three minutes governed by a Poisson distribution.A single server is capable of handling twenty-five customers per hour and this service time is exponentially distributed.How many customers are either receiving service or waiting in line on average?

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The answer of Customers arrive every three minutes governed by...

Question 19

Customers are serviced at a rate of four customers per hour according to an exponential distribution.What is the probability that customer service will require fewer than 30 minutes?

Accepted Answer

The exponential distribution is defined by the parameter λ (lambda), which is the rate of occurrences (customers serviced per hour in this case). Given λ = 4 customers per hour, the mean service time is 1/λ = 1/4 hour or 15 minutes. The probability that a service time is less than a certain value x is given by 1 - e^(-λx). For x = 0.5 hours (30 minutes), this probability is 1 - e^(-4*0.5) = 1 - e^(-2). Since e^(-2) is approximately 0.135, the probability is 1 - 0.135 = 0.865, which is greater than 0.70.

Question 20

Customers arrive according to a Poisson distribution.The average number of customer arrivals per hour is two.The probability that five customers will arrive in the next three hours is:

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