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Your team has been hired by a new pretzel shop, Twisted Pretzel. They are trying

ID: 3301894 • Letter: Y

Question

Your team has been hired by a new pretzel shop, Twisted Pretzel. They are trying to determine the number of employees they should schedule. Twisted Pretzel says each employee can handle 15 arrivals per hour. You decide the first step is to know the probability of the number of customers arriving at the store in a given hour. You conducted an experiment and recorded the number of arrivals per hour for 7 days. Assume the arrivals follow a Poisson Probability Distribution.

Deliverables:

Create a report to be submitted to Twisted Pretzel. It should include at least the following all presented in one document:

At least one page presenting your results.

The probability of each possible number of arrivals from 0 to 50 per hour.

The probability of more than 15 arrivals and the probability of more than 30 arrivals.

How many employees do you think you should have working?

Include any charts and tables needed to convey your results. Be sure to follow all of the data visualization rules.

Day

Time

Arrivals

1

10-11

24

1

11-12

18

1

12-1

22

1

1-2

27

1

2-3

28

1

3-4

20

1

4-5

13

1

5-6

25

1

6-7

20

1

7-8

19

2

10-11

16

2

11-12

21

2

12-1

33

2

1-2

23

2

2-3

21

2

3-4

20

2

4-5

18

2

5-6

21

2

6-7

23

2

7-8

27

3

10-11

19

3

11-12

14

3

12-1

21

3

1-2

23

3

2-3

17

3

3-4

23

3

4-5

20

3

5-6

31

3

6-7

23

3

7-8

26

4

10-11

22

4

11-12

26

4

12-1

37

4

1-2

29

4

2-3

27

4

3-4

19

4

4-5

24

4

5-6

19

4

6-7

20

4

7-8

28

5

10-11

24

5

11-12

18

5

12-1

27

5

1-2

33

5

2-3

20

5

3-4

24

5

4-5

18

5

5-6

21

5

6-7

21

5

7-8

23

6

10-11

27

6

11-12

19

6

12-1

29

6

1-2

19

6

2-3

24

6

3-4

22

6

4-5

32

6

5-6

23

6

6-7

22

6

7-8

21

7

10-11

20

7

11-12

28

7

12-1

17

7

1-2

20

7

2-3

29

7

3-4

20

7

4-5

18

7

5-6

23

7

6-7

22

7

7-8

25

Day

Time

Arrivals

1

10-11

24

1

11-12

18

1

12-1

22

1

1-2

27

1

2-3

28

1

3-4

20

1

4-5

13

1

5-6

25

1

6-7

20

1

7-8

19

2

10-11

16

2

11-12

21

2

12-1

33

2

1-2

23

2

2-3

21

2

3-4

20

2

4-5

18

2

5-6

21

2

6-7

23

2

7-8

27

3

10-11

19

3

11-12

14

3

12-1

21

3

1-2

23

3

2-3

17

3

3-4

23

3

4-5

20

3

5-6

31

3

6-7

23

3

7-8

26

4

10-11

22

4

11-12

26

4

12-1

37

4

1-2

29

4

2-3

27

4

3-4

19

4

4-5

24

4

5-6

19

4

6-7

20

4

7-8

28

5

10-11

24

5

11-12

18

5

12-1

27

5

1-2

33

5

2-3

20

5

3-4

24

5

4-5

18

5

5-6

21

5

6-7

21

5

7-8

23

6

10-11

27

6

11-12

19

6

12-1

29

6

1-2

19

6

2-3

24

6

3-4

22

6

4-5

32

6

5-6

23

6

6-7

22

6

7-8

21

7

10-11

20

7

11-12

28

7

12-1

17

7

1-2

20

7

2-3

29

7

3-4

20

7

4-5

18

7

5-6

23

7

6-7

22

7

7-8

25

Explanation / Answer

Solution

Let X = number of arrivals per hour. We are given that X ~ Poisson ().

From the given study data on arrivals for 7 days, is estimated by Xbar which is found to be

= 22.8

With this value of , using Excel Function on Poisson Distribution, probabilities of arrivals 0 (1) 50 are tabulated below:   

=

22.8

x

P(X = x)

0

1.2534E-10

1

2.8577E-09

2

3.2578E-08

3

2.4759E-07

4

1.4113E-06

5

6.4354E-06

6

2.4455E-05

7

7.9652E-05

8

0.00022701

9

0.00057509

10

0.00131121

11

0.00271777

12

0.00516377

13

0.00905646

14

0.01474909

15

0.02241862

16

0.03194654

17

0.04284594

18

0.05427153

19

0.06512583

20

0.07424345

21

0.08060717

22

0.08353834

23

0.08281192

24

0.07867133

25

0.07174825

26

0.0629177

27

0.0531305

28

0.04326341

29

0.03401399

30

0.02585063

31

0.01901272

32

0.01354656

33

0.00935944

34

0.00627633

35

0.00408858

36

0.00258944

37

0.00159565

38

0.00095739

39

0.00055971

40

0.00031903

41

0.00017741

42

9.631E-05

43

5.1067E-05

44

2.6462E-05

45

1.3407E-05

46

6.6454E-06

47

3.2237E-06

48

1.5313E-06

49

7.1251E-07

50

3.249E-07

Total

0.99999974

P(X < 16)

0.05633127

P(X < 31)

0.94131778

P(X > 15)

0.94366873

P(X > 30)

0.05868222

Probability of more than 15 arrivals = P(X > 15) = 0.9437

Probability of more than 30 arrivals = P(X > 30) = 0.0587

Number of employees to be working

Let this number be c. The desirable value of c would depend on the managerial stipulations on queue characteristics like average queue length, average waiting time, average time spent in the system etc.

Since these stipulations are not specified in the question, a definite answer cannot be derived. However, a general methodology is presented here.

We have average arrival rate, = 22.8[computed from the given data] and average service rate, µ = 15 [given].

Then, with number of employees working = c,

Average queue length = E(m) = P0{(µ)(/µ)c}/{(c - 1)!(cµ - )2}…………………………..(1)

Average number of customers in the system = E(n) = E(m) + (/µ)………………………..(2)

Average waiting time = E(w) = E(m)/()…… ……………………………………………..(3)

Average time spent in the system = E(v) = E(w) + (1/µ)..…………………………………..(4)

Percentage idle time of service channel = P0 ………. …………………………………….(5)

P0 = 1/(S1 + S2), where S1 = [0,c - 1](n/n!) and S2 = (c)/[c!{1 - (/c)}]……………….(6)

Using these formulae, and the given stipulations on queue characteristics, the desirable value of c can be found.

=

22.8

x

P(X = x)

0

1.2534E-10

1

2.8577E-09

2

3.2578E-08

3

2.4759E-07

4

1.4113E-06

5

6.4354E-06

6

2.4455E-05

7

7.9652E-05

8

0.00022701

9

0.00057509

10

0.00131121

11

0.00271777

12

0.00516377

13

0.00905646

14

0.01474909

15

0.02241862

16

0.03194654

17

0.04284594

18

0.05427153

19

0.06512583

20

0.07424345

21

0.08060717

22

0.08353834

23

0.08281192

24

0.07867133

25

0.07174825

26

0.0629177

27

0.0531305

28

0.04326341

29

0.03401399

30

0.02585063

31

0.01901272

32

0.01354656

33

0.00935944

34

0.00627633

35

0.00408858

36

0.00258944

37

0.00159565

38

0.00095739

39

0.00055971

40

0.00031903

41

0.00017741

42

9.631E-05

43

5.1067E-05

44

2.6462E-05

45

1.3407E-05

46

6.6454E-06

47

3.2237E-06

48

1.5313E-06

49

7.1251E-07

50

3.249E-07

Total

0.99999974

P(X < 16)

0.05633127

P(X < 31)

0.94131778

P(X > 15)

0.94366873

P(X > 30)

0.05868222