stats project report

8/8/2019 Stats Project Report

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Average Arrival Rate of Customers and

Average Phone Calls Received At Front Office

Submitted By

1. Anup Agarwal(13)2. Harish Kharthik(36)3.

Nidhi Singh

4. Sandhya Pai5. Umesh Soni


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CONTENTS

SYNOPSIS 3

INTRODUCTION 4

Poisson Distribution 5 - 8

SURVEY 9 - 10

ANALYSYS 11 15

CONCLUSION 16


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SYNOPSIS Through this project we were asked to find out the average arrival rate of

customer or average service rate and average rate in terms of phone calls the

front office person receives in any of the mobile service provider shop.

The service provider chosen is Vodafone. Vodafone Essar is a cellular operator

in India that covers 23 telecom circles in India. It is the second largest mobi leoperator in terms of revenue behind Bharti Airtel and third largest in terms of

customers. Data is collected by individuals personally visiting Vodafone store

and collecting information regarding the same. This data is then tabulated and

the average arrival rate of customers and average phone calls received are

computed and various aspects are analysed using Poisson distribution.


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INTRODUCTION The purpose of this project is to find out the average arrival rate of

customer and average rate in terms of phone calls the front office person

receives.

The main objective was to collect data of customer arrival rate in the

store by personally visiting the Vodafone store at different timings on different

days for specific time intervals and analyse the data.


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POISSON DISTRIBUTION

The Poisson distribution is named for SIMEON DENIS POISSON (1781-1840), aFrench mathematician who developed the distribution from studies during the

latter part of his lifetime.

In probability theory and statistics, the Poisson distribution (or Poisson law of

small numbers) is a discrete probability distribution that expresses the

probability of a number of events occurring in a fixed period of time if these

events occur with a known average rate and independently of the time since

the last event. (The Poisson distribution can also be used for the number of

events in other specified intervals such as distance, area or volume.)

P(X) = X e-

X!

e is the base of the natural logarithm (e = 2.71828...)

x is the number of occurrences of an event - the probability of which is given by

the function

x! is the factorial of x

is a positive real number, equal to the expected number of occurrences that

occur during the given interval. For instance, if the events occur on average 4

times per minute, and you are interested in probability for k times of events

occurring in a 10 minute interval, you would use as your model a Poisson

distribution with = 104 = 40.


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POISSON DISTRIBUTION CURVES:

The following is the plot of the Poisson probability density function for four

values of .

ATTRIBUTES OF POISSON EXPERIMENT:

A Poisson experiment is a statistical experiment that has the following

properties:

1.The experiment results in outcomes that can be classified as successesor failures.

2.The average number of successes () that occurs in a specified region isknown.

3.The probability that a success will occur is proportional to the size of theregion.

4.The probability that a success will occur in an extremely small region isvirtually zero.

5.The specified region could take many forms. For instance, it could be alength, an area, a volume, a period of time, etc.


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P

SS

DISTRIBUTION AND BINOMIAL DISTRIBUTION:

Th Poi

ondi

ib

ion canb a reasonable approxi ationo the bino ial

b t onl under certain conditions.These conditions occur when nis large and

p is small that is, when the number o trials is large and the binomialprobability o success is small.

The rule most o ten used b y statisticians is that the Poisson is a good

approximationo the binomial when nis greater thanor equal to20 and pis

less thanor equal to0.05.

In cases that meet these conditions, we can substitute the mean o the

binomialdistribution (np)inplace the meano the Poissondistribution, so that

the formula becomes

P(x) = (np)X *e-np

X!

The above diagram shows the comparison of the Poisson distribution (black

dots) and the binomial distribution with n=10 (red line), n=20 (blue line),

n=1000 (green line). All distributions have a mean of 5. The horiontal axis

shows the number of events k. Notice that as n gets larger, the Poisson

distribution becomes an increasingly better approximation for the binomial

distribution with the same mean.


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POISSION DISTRIBUTION AND PROJECT REPORT:

In our project we are taking the help of Poisson distribution in order to

calculate the required rates. We have discrete variables in the form of number

of phone calls and number of phone calls. Based on the data collected, we havecalculated various rates and formed questions in order to facilitate our study

using the application of Poisson distribution.


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SURVEY

The data collected on the arrival of customers at the Vodafone store is as

follows

Time (7:30 - 8:30)

No. ofCustomers

No. ofCalls

7:30 - 7:40 2 1

7:40 - 7:50 1 3

7:50 - 8:00 4 2

8:00 - 8:10 1 1

8:10 - 8:20 2 2

8:20 - 8:30 1 0

Time (6:00 - 7:00)

No. of

Customers

No. of

Calls

6:00 - 6:10 0 1

6:10 - 6:20 3 3

6:20 - 6:30 1 2

6:30 - 6:40 0 1

6:40 - 6:50 2 2

6:50 - 7:00 2 0

Time (11:00 - 12:00)

No. ofCustomers No. ofCalls

11:00 - 11:10 1 1

11:10 - 11:20 0 1

11:20 - 11:30 1 2

11:30 - 11:40 2 0

11:40 - 11:50 0 2

11:50 - 12:00 2 0


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Time (4:30 - 5:30)

No. of

Customers

No. of

Calls

4:30 - 4:40 1 1

4:40 - 4:50 3 2

4:50 - 5:00 2 2

5:00 - 5:10 1 1

5:10 - 5:20 1 1

5:20 - 5:30 1 0

Time (2:00 - 3:00)

No. of

Customers

No. of

Calls

2:00 - 2:10 0 1

2:10 - 2:20 1 0

2:20 - 2:30 2 1

2:30 - 2:40 0 1

2:40 - 2:50 0 0

2:50 - 3:00 1 0

Average Arrival Rate of customers per hour

= Total Number of customers

Total number of Hours

= 38

5

= 6.8

Average Arrival Rate of calls per hour

= Total Number of customers

Total number of Hours

= 34

5

= 6.8


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ANALYSIS

1. It is assumed that if there are less than 2 customers for any hour the

shopkeeper will bear a loss for that hour. So, calculate the probability for

which the shop keeper will not bear a loss.

P(x)= x e- / x!

So,

P(x6) = 1-P(0)-P(1)-P(2)-P(3)-P(4)-P(5)-P(6)

=1-(6.8)0.e-6.8/0! - (6.8)1 e-6.8/1! - (6.8)2 e-6.8/2! - (6.8)3 e-6.8/3! -

(6.8)4 e-6.8/4! - (6.8)5 e-6.8/5! - (6.8)6 e-6.8/6!

=1-10-4[5+38+4365.81+695.045+1056.4689+1338.193]

=0.6357


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The probability of getting more than 6 customers per hour is 0.6357 which

indicates there is a nominal probability of including one more worker in his

shop

3. Draw a graph by computing the arrival rate at x=0,1,2,3.........11,12 per hour

P(0)=(6.8)0.e-6.8/0! = 5*10-4

P(1)=(6.8)1.e-6.8/1! = 38*10-4

P(2)=(6.8)2.e-6.8/2! = 144.4*10-4

P(3)=(6.8)3.e-6.8/3! = 365.81*10-4

P(4)=(6.8)4

.e-6.8

/4! = 695.045*10-4

P(5)=(6.8)5.e-6.8/5! = 1056.4589*10-4

P(6)=(6.8)6.e-6.8/6! = 1339.402*10-4

P(7)=(6.8)7.e-6.8/7! = 1454.208*10-4

P(8)=(6.8)8.e-6.8/8! = 1381.497*10-4

P(9)=(6.8)9.e-6.8/9! = 1166.598*10-4

P(10)=(6.8)10.e-6.8/10! = 886.814*10-4

P(11)=(6.8)11.e-6.8/11! = 612.57*10-4

P(12)=(6.8)12.e-6.8/12! = 387.961*10-4

The probability of the customer arrival rate is maximum for 7 Customers.


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4. Draw a graph by computing the phone call rate at x=0,1,2,3.........11,12 per

hour

P(0)=(6.8)0.e-6.8/0! = 11.137*10-4

P(1)=(6.8)1.e-6.8/1! = 75.736*10-4

P(2)=(6.8)2.e-6.8/2! = 257.504*10-4

P(3)=(6.8)

3

.e

-6.8

/3! = 583.677*10

-4

P(4)=(6.8)4.e-6.8/4! = 992.251*10-4

P(5)=(6.8)5.e-6.8/5! = 1349.462*10-4

P(6)=(6.8)6.e-6.8/6! = 1529.39*10-4

P(7)=(6.8)7.e-6.8/7! = 1485.694*10-4

P(8)=(6.8)8.e-6.8/8! = 1262.839*10-4

P(9)=(6.8)9.e-6.8/9! = 954.145*10-4

P(10)=(6.8)10.e-6.8/10! = 648.819*10-4

P(11)=(6.8)11.e-6.8/11! = 401.088*10-4

P(12)=(6.8)12.e-6.8/12! = 227.283*10-4

0

200

400

600

800

1000

1200

1400

1600

1 2 3 4 5 6 7 8 9 10 11 12 13

No. of Customers

Probability


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The probability of phone call arrival rate is maximum at 6 phone calls.

5. What is the probability that the shopkeeper receives more than four calls per

hour?

P(x>4)= 1-P(0)-P(1)-P(2)-P(3)-P(4)

=1-(6.8)0.e-6.8/0! -(6.8)1.e-6.8/1! -(6.8)2.e-6.8/2! -(6.8)3.e-6.8/3! -(6.8)4.e-6.8/4!

=1-e-6.8[1+6.8+(6.8)2/2+(6.8)3/6+(6.8)4/24]

=1- e-6.8[7.8+23.12+52.4+89.089]

=0.805

The probability of receiving more than four calls is 0.805 which is high.

5. It is assumed that, if the shopkeeper is receiving more than three calls per

hour, Vodafone is having lot of customer problems. So, find the probability for

which the customers are having a lot of problems.

0

200

400

600

800

1000

1200

1400

1600

1800

1 2 3 4 5 6 7 8 9 10 11 12 13

No. of Phone Calls

Pro a ility


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P(x>3) = 1-P(0)-P(1)-P(2)-P(3)

= 1-(6.8)0.e-6.8/0! -(6.8)1.e-6.8/1! -(6.8)2.e-6.8/2! -(6.8)3.e-6.8/3!

=1-e-6.8[1+6.8+(6.8)2/2+(6.8)3/3]

=1-1.11*10-3

[7.8+23.12+52.14]

=1-0.0927

=0.9072

The probability of receiving more than three calls is 0.9072 which is very high.

Hence Vodafone must concentrate more on the reducing the problems faced by

the customers

6. If it is assumed that, receiving zero calls per hour is the condition for

Vodafone is having the best customer satisfaction. Calculate the probability for

which the Vodafone is having best customer satisfaction.

P(x=0) = (6.8)0.e-6.8/0!

= 1.113*10-3

= 0.001113

The probability of receiving zero calls is very meagre which indicates that

customer satisfaction is a concern to Vodafone.


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CONCLUSION

The Probability of number of customers arriving at the Vodafone shop per hour

is maximum for 7 customers which is nominal. The probability of receiving

phone calls is maximum for 6 phone calls which is not a good sign for

Vodafone. Hence Vodafone has to concentrate more on reducing the issues

faced by the customers in the services provided by them thereby increasing the

customer satisfaction.

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