Willow Brook National Bank operates a drive-up teller window that allows custome

Question

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 30 customers per hour or 0.5 customers per minute. What is the mean or expected number of customers that will arrive in a ten-minute period? lambda = per ten minute period Assume that the Poisson probability distribution can be used to describe the arrival process. Use the arrival rate in part (a) and compute the probabilities that exactly 0, 1, 2, and 3 customers will arrive during a ten-minute period. If required, round your answers to four decimal places. Delays are expected if more than three customers arrive during any ten-minute period. What is the probability that delays will occur? If required, round your answer to four decimal places. P(Delay Problems) =

Explanation / Answer

a. Mean = arrival rate of customers per minute*10 minute period

= 0.5 customers per minute*10

= 5

b. Using the formula: P(x,µ) = (e^-µ)*(µ^x)/x!

µ = 5, e = 2.71828

(i) x = 0

probability = (2.71828^-5)*(5^0)/1

= 0.0067380

(ii) x = 1

probability = (2.71828^-5)*(5^1)/1!

= 0.00674*5/1 = 0.03369

(iii) x = 2

probability =  (2.71828^-5)*(5^2)/2!

= 0.00674*12.5 = 0.08422

(iv) x = 3

probability =  (2.71828^-5)*(5^3)/3!

= 0.00674*20.83333 = 0.14037

c. probability of 3 or less than 3 customers = sum of all values calculated in "b" avove = 0.0067380+0.03369+ 0.08422+0.14037

= 0.26502

Thus probability of delay = 1-0.26502 (1 - probability of 3 or less than 3)

= 0.73498

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