Emilio, the owner of Pepe Delgados, is interested in taking a more statistical a

Question

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 and 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

Let X be the number of arrivals in a 5-minute interval between 7.00PM to 8.00PM. So X follows a Poisson Distribution with parameter , then probability mass function is

P(X = x) = (e-)(x)/(x!),

where is the mean arrival rate per 5 minutes

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

a)

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

So, = 3 Thats the answer

b)

Here x=0,   = 3

Probability of no customer arrival in any 5-minute interval = P(X = 0) = e-3 ................You can use here  =POISSON(0,3,0) in excel you will get probability

P(X = 0)= 0.0498 Thats the answer

c)

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

= 1 – P(X < 5)

=1-P(X<=4) ........................Using Excel =POISSON(4,3,1)

= 1 – 0.8153

= 0.1847 thats the Answer

d)

Probability of fewer than 4 customer arrivals in any 10-minute interval

P(x<4) = P(X 3/ = 6) ............................using excl =POISSON(6,3,1)

= 0.9665 that's the answer

Probability of no customer arrival in any 10-minute interval = P(X = 0/ = 3*2=6)

= 0.0025 that's the answer

f)

Probability of exactly 8 customer arrivals in any 15-minute interval = P(X = 8/ = 3*3=9)

= 0.1318 that's the answer

Hope this will help you. Thanks. Good Luck :)

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