PLEASE SHOW EACH AND EVERY STEP! THANK YOU!! Emilio, the owner of Pepe Delgados,

Question

PLEASE SHOW EACH AND EVERY STEP! THANK YOU!!

Emilio, the owner of Pepe Delgados, is interested in taking a more statistical approach to predicting customer load. He begins the process by gathering data on the number of arrivals in five-minute intervals. The data below represent the arrivals in five-intervals between 7:00p ans 8:00p every Saturday for three weeks. Assume that arrivals are Poisson distributed. The questions that follow pertain to the 7:00p to 8:00p time period, (adapted from [Black 2012]) a. Compute lambda using the data from all three weeks as one data set. b. What is the probability that no customers arrive during any five-minute interval? c. What is the probability that six or more customers arrive during any five-minute interval? d. What is the probability that during a 10-minute interval fewer than four customers arrive? e. What is the probability that no customers arrive during any 10-minute interval? Compare with part b. f. What is the probability that exactly eight customers arrive in any 15-minute interval?

Explanation / Answer

Solution

Back-up Theory

If X = number of arrivals in a 5-minute interval, follows a Poisson Distribution with parameter , then

P(X = x) = (e-)(x)/(x!),where is the mean arrival rate per 5 minutes……………..(1)

If X = number of arrivals in a 5-minute interval, follows a Poisson Distribution with parameter , and Y = number of arrivals in a 10-minute interval, then

Y follows a Poisson Distribution with parameter 2 [note: 10 = 2x5]………………………………..(2)

Note: all probabilities are obtained by using Excel Function

Now, to work out the solution,

Part (a)

From the given data, mean arrival per 5 minute interval = 108/36 = 3.

So, = 3 ANSWER

Part (b)

Probability of no customer arrival in any 5-minute interval = P(X = 0) = e-3

= 0.0498 ANSWER

Part (c)

Probability of 6 or more customer arrivals in any 5-minute interval = P(X 6)

= 1 – P(X 5) = 1 – 0.9161 = 0.0839 ANSWER

Part (d)

Probability of fewer than 4 customer arrivals in any 10-minute interval = P(X 3/ = 6) [vide (2) under Back-up Theory]

= 0.2851 ANSWER

Part (e)

Probability of no customer arrival in any 10-minute interval = P(X = 0/ = 6) [vide (2) under Back-up Theory]

= 0.0025 ANSWER

Part (f)

Probability of exactly 8 customer arrivals in any 15-minute interval = P(X = 8/ = 9) [vide (2) under Back-up Theory]

= 0.1318 ANSWER

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