The manager of Pepe Delgados is interested in taking a more statistical approach
ID: 3128871 • Letter: T
Question
The manager 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.
Time frame
Week 1
Week 2
Week 3
7:00-7:05
3
1
5
7:05-7:10
6
2
3
7:10-7:15
4
4
5
7:15-7:20
6
0
3
7:20-7:25
2
2
5
7:25-7:30
3
6
4
7:30-7:35
1
5
7
7:35-7:40
5
4
3
7:40-7:45
1
2
4
7:45-7:50
0
5
8
7:50-7:55
3
3
1
7:55-8:00
3
4
3
Compute lambda using the data from all three weeks as one data set.
What is the probability that no customers arrive during any five-minute interval?
What is the probability that six or more customers arrive during any five-minute interval?
What is the probability that during a 10-minute interval fewer than four customers arrive?
What is the probability that no customers arrive during any 10-minute interval? Compare with part b.
What is the probability that exactly eight customers arrive in any 15-minute interval?
Time frame
Week 1
Week 2
Week 3
7:00-7:05
3
1
5
7:05-7:10
6
2
3
7:10-7:15
4
4
5
7:15-7:20
6
0
3
7:20-7:25
2
2
5
7:25-7:30
3
6
4
7:30-7:35
1
5
7
7:35-7:40
5
4
3
7:40-7:45
1
2
4
7:45-7:50
0
5
8
7:50-7:55
3
3
1
7:55-8:00
3
4
3
Explanation / Answer
For Poisson distribution, lambda is the mean of the data set of values.
Therefore, lambda=(3+6+4=...1+3)/36=3.5.
To compute, probability of no customer arrive at any five time, use the following formula as follows:
P(x,mu)=(e^-mu)(mu^x)/x! where, x is the number of success (x=0) and mu is the average rate (mu=3.5)
P(0,3.5)=(e^-3.5)(3.5^0)/0!=0.030
Probability of six or more customers is
1-P(X<6)=1-0.85=0.15
[P(X<6)=P(0,3.5)+P(1,3.5)+...+P(5,3.5)]=0.85
Probability of less than 4 customers arrives at 10 minute interval, mu changes to 7, use the same formula with new substitution of values is as follows:
P(X<4)=P(0,7)+P(1,7)+P(2,7)+P(3,7)=0.08
P(0,7)=(e^-7)(7^0)/0!=0.001
b.
At 15 minutes interval, mu changes to 10.5. Therefore, P(8,10.5)=0.100
while computing the 10 and 15 minutes interval club the data and divide by 18 and 12 respectively to obtain the mu.
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