Question

The Burger King in the ASU-MU has only one person working at the register. Customers arrive 12 per hour according to a Poisson distribution. The Burger King employee on averagecan help 16 customers per hour according to an exponential distributionA. What percentage of the time is the Burger King employee busy?B. What is the average number of customers WAITING to be helped?C. What is the average waiting time for customers in line? (In minute

Answer 1

A) The Burger King employee is busy 75% of the time.

B) The average number of customers waiting to be helped is 2.25.

C) The customers have an average time of 11.25 minutes waiting on the line.

Step-by-step explanation:

let x be the mean arrival rate = 12 customers per hour

let y be the mean service rate = 16 customers per hour

A) the percentage of the time is the Burger King employee busy is given by:

(12/16)*100

= 75%

Therefore, the Burger King employee is busy 75% of the time.

B) the average number of customers WAITING to be helped is given by:

x^2/y(y-x)

= (12^2)/[16(16-12)]

= 2.25 customers

Therefore, the average number of customers waiting to be helped is 2.25.

C) the average waiting time for customers in line is given by:

x/y(y-x)

= 12/[16(16-12)]

= 0.1875 hours

= 11.25 minutes.

Therefore, the customers have an average time of 11.25 minutes waiting on the line.