Use the single-server drive-up bank teller operation referred to in Problems 1 and 2 to determine the following operating characteristics for the system: a.) The probability that no customers are in the system. b.) The average number of customers waiting. c.) The average number of customers in the system. d.) The average time a customer spends waiting. e.) The average time a customer spends in the system. f.) The probability that arriving customers will have to wait for service.

This question is incomplete, here's the complete question:

1. Willow Brook National Bank operates a drive-up teller window that allows customers to complete bank transactions without getting out of their cars. On weekday mornings, arrivals to the drive-up teller window occur at random, with an arrival rate of 24 customers per hour or 0.4 customers per minute. 3. Use the single-server drive-up bank teller operation referred to in Problems 1 to determine the following operating characteristics for the system: a.) The probability that no customers are in the system. b.) The average number of customers waiting. c.) The average number of customers in the system. d.) The average time a customer spends waiting. e.) The average time a customer spends in the system. f.) The probability that arriving customers will have to wait for service.

Explanation:

Arrival rate \lambda = 24 customers per hour or 0.4 customers per minute

Service rate \mu = 36 customers per hour or 0.6 customers per minute (from problem 1)

a.) The probability that no customers are in the system , P0 = 1 - \lambda / \mu

= 1 - (24/36) = 1/3 = 0.3333

b.) The average number of customers waiting

Lq = \lambda^2 / [\mu(\mu - \lambda)] = 242 / [36 * (36 - 24)] = 1.33

c.) The average number of customers in the system.

L = Lq + \lambda / \mu = 1.33 + (24/36) = 2

d.) The average time a customer spends waiting.

Wq = \lambda / [\mu(\mu - \lambda)] = 24 / [36 * (36 - 24)] = 0.0555 hr = 3.33 min

e.) The average time a customer spends in the system

W = Wq + 1/\mu = 0.0555 + (1/36) = 0.0833 hr = 5 min

f.) The probability that arriving customers will have to wait for service.

= 1 - P0 = 1 - 0.3333 = 0.6667

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