A cafeteria serving line has a coffee um from which customers serve themselves.
ID: 370603 • Letter: A
Question
A cafeteria serving line has a coffee um from which customers serve themselves. Arrivals at the um follow a Poisson distribution at the rate of 4.0 per minute. In serving themselves, customers take about 13 seconds, exponentially distributed a. How many customers would you expect to see on the average at the coffee urn? (Do not round intermediate calculations. Round your answer to 2 decimal places.) Average no of customers b. How long would you expect it to take to get a cup of coffee? (Round your answer to 2 decimal places.) Expected time minute(s) c. What percentage of time is the urn being used? (Do not round intermediate calculations. Round your answer to 1 decimal place.) Percentage of time d. What is the probability that three or more people are at the coffee urn? (Do not round intermediate calculations. Round your answer to 1 decimal place.) Probability e. If the cafeteria installs an automatic vendor that dispenses a cup of coffee at a constant time of 13 seconds, how many customers would you expect to see at the coffee urn (waiting and/or pouring coffee)? (Do not round intermediate calculations. Round your answer to 2 decimal places.) Average no of customers f. If the cafeteria installs an automatic vendor that dispenses a cup of coffee at a constant time of 13 seconds, how long would you expect it to take (in minutes) to get a cup of coffee, including waiting time? (Do not round intermediate calculations. Round your answer to 2 decimal places.) Expected time minute(s)Explanation / Answer
PLEASE FIND ANSWERS TO FIRST 4 QUESTIONS :
Given :
Arrival rate = a = 4 / minute
Service rate ( @ 13 seconds per customer ) = 60 / 13
Number of customers at the coffee urn
= Average number of customers waiting + Average number of customers taking service
= a^2/s x ( s -a) + a/s
= 4 x 4 / (60/13) x ( 60/13 – 4) + 4 x 13/60
= 16/4.615x ( 4.615 – 4 ) + 0.866
= 5.637 + 0.866
= 6.503 ( 6.50 rounded to 2 decimal places )
AVERAGE NUMBER OF CUSTOMERS = 6.50
Expected time to get a cup of coffee
= Expected wait time in the queue
= 4/ ( 60/13) x ( 60/13 – 4)
= 4 / 4.615 x 0.615 minutes
= 1.409 minutes ( 1.41 rounded to 2 decimal places )
EXPECTED TIME TO GET A CUP OF COFFEE = 1.41 MINUTES
Percentage of time urn is being used
= a/s x 100
= 4 x 13/60 x 100
= 86.67 %
PERCENTAGE OF TIME URN IS BEING USED = 86.67 %
Probability that there are ZERO people at the coffee urn
= Po = 1 – a/s = 1 – 4 x 13/60 = 8/60 = 0.133
Probability that there are 1 person at the coffee urn
= P1 = ( a/s) x Po =( 4 x 13/60) x 0.133 = 0.8666 x 0.133 = 0.1152
Probability that there are 2 persons at the coffee urn
= P2 = ( a/s)^2 x Po = 0.8666 x 0.8666 x 0.133 = 0.0998
Therefore , probability that there are less than 3 persons at coffee urn
= P0 + P1 +P2
= 0.133 + 0.1152 + 0.0998
= 0.348
Hence, probability that there are 3 or more persons at the coffee urn
= 1 – 0.348
= 0.652
PROBABILITY THAT 3 OR MORE PEOPLE ARE AT THE COFFEE URN = 0.652
AVERAGE NUMBER OF CUSTOMERS = 6.50
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