Waiting Line Models:Movies tonight is a typical video and dvd movie rental outlet for home-viewing customers. During the weeknight evening, customers arrive at Movies Tonight with an arrival rate of 1.25 customers per minute. The checkout clerk has a service rate of 2 customers per minute/ Assume Poisson arrivals and exponential service times. a. What is the probability that no customers are in the system?b. What is the average number of customers waiting for service?c. What is the average time a customer waits for service to begin?d. What is the probability that an arriving customer will have to wait for service to begin?e. Do the operating charachteristics indicate that the one-clerk checkout system provides and acceptable level of service?

Step-by-step explanation:

Given:

Arrival rate λ = 1.25 min

Mean μ = 2

Computation:

(a) Probability [No customers in system]

Probability [No customers in system] = 1-[λ/μ]

Probability [No customers in system] = 1-[1.25/2]

Probability [No customers in system] = 0.375

(b) Customers waiting for service

Customers waiting for service = λ²/ [μ(μ-λ)]

Customers waiting for service = 1.25²/ [2(2-1.25)]

Customers waiting for service = 25/24

(c) Average time customer wait

Average time customer wait = λ / [μ(μ-λ)]

Average time customer wait = 1.25/ [2(2-1.25)]

Average time customer wait = 1.25/1.5

(d) May be wait because Customers waiting for service = 25/24

(e) Per customer average time

Per customer average time = 1/(μ-λ)

Per customer average time = 1/(2-1.25)

Per customer average time = 1.33 (Approx)

step-by-step explanation:

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.