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Statistics Practical B.Sc. II

 B.Sc. II Statistics Practical Paper -II     

In this article we see all the practical problem Of  B.Sc. II  in Practical Paper -II

Practical Number 1.


Padmbhushan Vasantraodada Patil Mahavidyalaya, Kavathe Mahankal

Department of Statistics

Title: Model sampling from Discrete Uniform distribution

                                                                                                               

Questions: 

1. Draw a model sample of size 10 from the following Discrete Uniform Distribution

P(X) = 1/8; x=1,2,3,...,8

            = 0 otherwise.

Calculate A.M. and H. M. of your sample.


2. Draw a model sample of size 15 from the following Discrete Uniform Distribution Taking values 10,15,20,25,30,35,40,45,50,55. Find mean deviation from mode of your sample.


3. Draw a model sample of size 8 from the following Discrete Uniform Distribution

P(X) = 1/13; x=1,2,3,...,13

            = 0 otherwise.

Obtain the quartiles of your sample.


4. Draw a model sample of size 10 from the Discrete Uniform Distribution over the set of first 25 natural numbers. Find C.V. of your sample.                                                                            

5. Draw a random sample of size 11 from discrete uniform distribution taking values 1, 2,     3, 4, 5. find coefficient of M.D. about mean of your sample.

6. Draw a model sample of size 10 from the Discrete Uniform Distribution over values -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. find mean and variance of your sample.

                                                                                                               

ANSWERS Practical NO. 1.

Q. 1. Draw a model sample of size 10 from the following Discrete Uniform Distribution

P(X) = 1/8; x=1,2,3,...,8

            = 0 otherwise.

Calculate A.M. and H. M. of your sample.

Answer : 

first we see the discrete uniform distribution.

If  X ~ DU(1,…., n ), Then the p.m.f. of discrete uniform distribution is given  as,     

P(X=x) = 1/n ; x=1,2,3,...,n

            = 0 otherwise.

and it's mean = ( n+1)/2

& variance = (n²-1) /12.

then  we, write the Aim  of the question.

Aim: To Draw a model sample of size 10 from the following Discrete Uniform Distribution  and Calculate A.M. and H. M. of your sample.

 Statistical Formula:

A.M. = ∑Xi / n

H. M.  = n /  ∑(1/ Xi)

Given,

P(X) = 1/8; x=1,2,3,...,8

            = 0 otherwise.

here n = 8

and P(X) = 0.125

 now we draw random sample so we create observation table.

Observation Table: 

Sr. No.

xi

p(xi)

F(xi)

Rand. No.

xi(Random Sample)

1/xi

1

1

0.125

0.125

0.180575

2

0.5

2

2

0.125

0.250

0.005194

1

1

3

3

0.125

0.375

0.091513

1

1

4

4

0.125

0.500

0.228582

2

0.5

5

5

0.125

0.625

0.437781

4

0.25

6

6

0.125

0.75

0.900141

8

0.125

7

7

0.125

0.875

0.001238

1

1

8

8

0.125

1

0.931718

8

0.125

9

 

0.079296

1

1

10

0.892168

8

0.125

Total

36

5.625

note: here P(X) is given for X= 1,2,3,4,5,6,7,8. and it is P(x) = 1/8  = 0.125 is same for all 8  values.

then F(X) is obtained as sum of probability, we known that F(X) = P(X≤ x )

if x =1 then F( 1) =  P(X≤ 1 )  =  P(X =1 )= 0.125

if x = 2  then F( 2 ) =  P(X≤ 2 )  =  P(X =1 )+ P(X = 2) = 0.125+0.125 = 0.250 

if x = 3  then F( 3 ) =  P(X≤ 3. )  =  P(X =1 )+ P(X = 2)+ P(X =3)  = 0.1225+0.125+0.125 = 0.375

and so on

and the Rand .No. means random number that are obtained as i. in MS- Excel write  =RAND()   and enter we get random number.

ii. in scientific Calculator we first press shift button  then Rand# ( . button on  scientific calculator) then = button.

the random number obtained as a value is slightly greater than random number should be seen in the column of F(X) and corresponding value of x in front of F(X) is selected as random sample. 

i.e. first random number is 0.180 then we see the value slightly grater than 0.180 in column of F(X) that is 0.250  and corresponding value of x in front of F(X) is 2 and it is selected as random sample.

that are in column of X. and the next column is created for calculating H.M 

the observation table is contains mainly five columns that are 

xi

p(xi)

F(xi)

Rand. No.

xi

and the  next column is depended on what you are told to find. 

Calculation:
for calculating A. M. i.e. arithmetic mean we first find  ∑Xi = total of Xi 
and for H.. Harmonic Mean we first find    ∑(1/ Xi) = total of (1/ Xi) both totals are calculating in observation table

A.M. = ∑Xi / n ;  HERE Xi = total of Xi = 36 and  n = total number of observations = 10

A.M. =  36/10

A.M. = 3.6

H. M.  = n /  ∑(1/ Xi)

H. M. = 10 / 5.625

H. m. = 1.7777

Result : Random sample of size 10 is 

Sample

2

1

1

2

4

8

1

8

1

8

A.M. of sample is 3.6

H. M. Of sample is 1.7777

                                                                                                               

Q. 2. Draw a model sample of size 15 from the following Discrete Uniform Distribution Taking values 10,15,20,25,30,35,40,45,50,55. Find mean deviation from mode of your sample.

Answer:

first we see the discrete uniform distribution.

If  X ~ DU(1,…., n ), Then the p.m.f. of discrete uniform distribution is given  as,     

and it's mean = ( n+1)/2

& variance = (n²-1) /12.

here x takes values 10, 15, 20, 25, 30, 35, 40, 45, 50, 55.

then total number of observation is 10, i.e. n=10

therefore the p.m.f. of discrete uniform distribution is 

P(X=x) = 1/10 = 0.1; x=10, 15, 20, 25, 30, 35, 40, 45, 50, 55.

             = 0 otherwise.

then  we, write the Aim  of the question.

Aim: Draw a model sample of size 15 from the following Discrete Uniform Distribution. and Find mean deviation from mode of your sample. 

Statistical Formula:

Mean Deviation from mode  = M.D. = (1/n) ∑| Xi  - Mode|

therefore the p.m.f. of discrete uniform distribution is 

P(X=x) = 1/10 = 0.1; x=10, 15, 20, 25, 30, 35, 40, 45, 50, 55.

             = 0 otherwise.

now we draw random sample so we create observation table.

Observation Table: 

Sr. No.

xi

p(xi)

F(xi)

Rand. No.

Random Sample

| Xi  - Mode|

1

10

0.1

0.1

0.235467

20

15

2

15

0.1

0.2

0.230688

20

15

3

20

0.1

0.3

0.585224

35

0

4

25

0.1

0.4

0.144258

15

20

5

30

0.1

0.5

0.574728

35

0

6

35

0.1

0.6

0.094317

10

25

7

40

0.1

0.7

0.445045

30

5

8

45

0.1

0.8

0.125616

15

20

9

50

0.1

0.9

0.872183

50

15

10

55

0.1

1

0.776099

45

10

11

 

0.024739

10

25

12

0.975556

55

20

13

0.569214

35

0

14

0.621989

40

5

15

0.460018

30

5

Total

180

mode is 35 because it repeated 3 times. 35 has maximum frequency.

Calculation: 

Mean Deviation from mode  = M.D. = (1/n) ∑| Xi  - Mode|

M.D. = (1/15) x 180

M. D. = 12

Result: Random sample of size 15 is 

Random Sample

20

20

35

15

35

10

30

15

50

45

10

55

35

40

30

Mean Deviation from mode  =12.

                                                                                                               

Q. 3. Draw a model sample of size 8 from the following Discrete Uniform Distribution

P(X) = 1/13; x=1,2,3,...,13

            = 0 otherwise.

Obtain the quartiles of your sample.

Answer:

here given that x has discrete uniform distribution

P(X) = 1/13; x=1,2,3,...,13

            = 0 otherwise.

Aim: To Draw a model sample of size 8 and find the quartiles of your sample.

Statistical Formula:

quartiles = Qi = Size of (i(N+1)/4}th observation, for i = 1, 2, 3.

here given that x has discrete uniform distribution

P(X) = 1/13 = 0.0769; x=1,2,3,...,13

            = 0 otherwise.

now we draw random sample so we create observation table.

Observation Table: 

Sr. No.

xi

p(xi)

F(xi)

Rand. No.

Random Sample

ascending

1

1

0.0769

0.0769

0.5956

8

2

2

2

0.0769

0.1538

0.9960

13

5

3

3

0.0769

0.2308

0.8944

12

7

4

4

0.0769

0.3077

0.3768

5

8

5

5

0.0769

0.3846

0.7382

10

10

6

6

0.0769

0.4615

0.7611

10

10

7

7

0.0769

0.5385

0.0930

2

12

8

8

0.0769

0.6154

0.4907

7

13

9

9

0.0769

0.6923

 

 

 

10

10

0.0769

0.7692

 

 

 

11

11

0.0769

0.8462

 

 

 

12

12

0.0769

0.9231

 

 

 

13

13

0.0769

1

 

 

 


Calculation:

Q1 = Size of (N+1)/4 th observation

    =Size of (8+1)/4 th observation

    =Size of (2.25) th observation

    =5+0.25*(7-5)

Q1  = 5.5

Q2=Size of 2* (N+1)/4 th observation    

    =Size of 2*(8+1)/4 th observation

    =Size of (4.5) th observation

    =8+0.5*(10-8)

    Q2 =9

Q3 =Size of 3*(N+1)/4 th observation

    =Size of 3*(8+1)/4 th observation

    =Size of (6.75) th observation

    =10+0.75*(12-10)

    Q3 =11.5


Result:  The Quartiles of  sample is Q1  = 5.5, Q2 =9, Q3 =11.5.

                                                                                                               

Q. 4. Draw a model sample of size 10 from the Discrete Uniform Distribution over the set of first 25 natural numbers. Find C.V. of your sample.     

Answer:

here given that x has discrete uniform distribution

and the Discrete Uniform Distribution over the set of first 25 natural numbers.

i.e. x take values 1, 2, 3, 4, ......., 25

P(X) = 1/25; x=1,2,3,...,25

            = 0 otherwise.

Aim: To Draw a model sample of size 10 and Find C.V. of your sample.

Statistical Formula:

C. V. = {(S.D.) / (Mean)} x 100

here Mean = ∑Xi / n

S.D. = √(Variance)

Variance = {∑X²i / n} - (x̄)²

here given that x has discrete uniform distribution

P(X) = 1/25 = 0.04 ; x=1,2,3,...,25

            = 0 otherwise.

now we draw random sample so we create observation table.

Observation Table: 

Sr.No.

xi

p(xi)

F(xi)

Rand.No.

xi(Random sample)

Xi^2

1

1

0.04

0.04

0.594799

15

225

2

2

0.04

0.08

0.25552

7

49

3

3

0.04

0.12

0.472988

12

144

4

4

0.04

0.16

0.62064

16

256

5

5

0.04

0.2

0.899819

23

529

6

6

0.04

0.24

0.104979

3

9

7

7

0.04

0.28

0.902166

23

529

8

8

0.04

0.32

0.178694

5

25

9

9

0.04

0.36

0.379054

10

100

10

10

0.04

0.4

0.102018

3

9

11

11

0.04

0.44

Total

117

1875

12

12

0.04

0.48

13

13

0.04

0.52

14

14

0.04

0.56

15

15

0.04

0.6

16

16

0.04

0.64

17

17

0.04

0.68

18

18

0.04

0.72

19

19

0.04

0.76

20

20

0.04

0.8

21

21

0.04

0.84

22

22

0.04

0.88

23

23

0.04

0.92

24

24

0.04

0.96

25

25

0.04

1

Calculation:

Mean = ∑Xi / n

    = 117/10

 Mean    =11.7


Variance = {∑X²i / n} - (x̄)²

        = (1875/10) -(11.7)²

      Variance   = 50.61

S.D. = √(Variance) 

S.D. = √(50.61)

        S.D.= 7.1141

C. V. = {(S.D.) / (Mean)} x 100

C. V. = {(7.1141) / (11.7)} x 100

C. V. = 60.8040%

Result: 

C. V. of sample is = 60.8040%

                                                                                                               

Q. 5. Draw a random sample of size 11 from discrete uniform distribution taking values 1, 2, 3, 4, 5. find coefficient of M.D. about mean of your sample.

Answer:

here given that x has discrete uniform distribution

and the Discrete Uniform Distribution over the set of values 1, 2, 3, 4, 5.

i.e. x take values 1, 2, 3, 4, 5.

P(X) = 1/ 5 = 0.2; x=1,2,3,4, 5.

            = 0 otherwise.

Aim: To Draw a model sample of size 11 and find coefficient of M.D. about mean. 

Statistical Formula:

Mean = ∑Xi / n

Mean Deviation about Mean = M.D.(x̄)= 𝟏 /𝒏 ∑ |𝑿𝒊 − x̄|

Coeff. of  Mean Deviation about Mean = M.D.(x̄) / x̄ 

now we draw random sample so we create observation table.

Observation Table: 

Sr.No.

xi

p(xi)

F(xi)

Rand.No.

xi(Random sample)

| Xi  - Mean|

1

1

0.2

0.2

0.357684

2

0.73

2

2

0.2

0.4

0.199077

1

1.73

3

3

0.2

0.6

0.802135

5

2.27

4

4

0.2

0.8

0.331298

2

0.73

5

5

0.2

1

0.485472

3

0.27

0.372287

2

0.73

0.643617

4

1.27

0.322358

2

0.73

0.570065

3

0.27

0.253577

2

0.73

0.790086

4

1.27

Total

30

10.73
Calculation: 

Mean = ∑Xi / n
            =30/11
           =2.7273
Mean Deviation about Mean = M.D.(x̄)= (𝟏 /𝒏) ∑ |𝑿𝒊 − x̄|
        M.D.(x̄) = (1/11) x (10.73)
        M.D.(x̄) =  0.9755
Coeff. of  Mean Deviation about Mean = M.D.(x̄) / x̄ 
 Coeff. of  Mean Deviation about Mean    = 0.9755 / 2.7273
Coeff. of  Mean Deviation about Mean    = 0.3577

Result: 
Sample of size 11is

Random Sample

2

1

5

2

3

2

4

2

3

2

4
Coeff. of  Mean Deviation about Mean    = 0.3577
                                                                                                               

Q.6. Draw a model sample of size 10 from the Discrete Uniform Distribution over values -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. find mean and variance of your sample.

Answer:

here given that x has discrete uniform distribution

and the Discrete Uniform Distribution over the set of values -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

i.e. x take values -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

P(X) = 1/ 21 = 0.0476; x= -10,-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

            = 0 otherwise.

Aim: To Draw a model sample of size 10 and find mean and variance of your sample. 

Statistical Formula:

Mean = ∑Xi / n

Variance = {∑X²i / n} - (x̄)²

now we draw random sample so we create observation table.

Observation Table: 

Sr. No.

xi

p(xi)

F(xi)

Rand.       No.

xi (Random sample)

Xi^2

1

-10

0.0476

0.0476

0.497454

0

0

2

-9

0.0476

0.0952

0.334415

-3

9

3

-8

0.0476

0.1429

0.978027

10

100

4

-7

0.0476

0.1905

0.919356

9

81

5

-6

0.0476

0.2381

0.63215

3

9

6

-5

0.0476

0.2857

0.066648

-9

81

7

-4

0.0476

0.3333

0.373673

-3

9

8

-3

0.0476

0.381

0.235007

-6

36

9

-2

0.0476

0.4286

0.9047

8

64

10

-1

0.0476

0.4762

0.757458

5

25

11

0

0.0476

0.5238

Total

14

414

12

1

0.0476

0.5714

13

2

0.0476

0.619

14

3

0.0476

0.6667

15

4

0.0476

0.7143

16

5

0.0476

0.7619

17

6

0.0476

0.8095

18

7

0.0476

0.8571

19

8

0.0476

0.9048

20

9

0.0476

0.9524

21

10

0.0476

1

Calculation: 

Mean = ∑Xi / n

        =14/10

Mean =1.4

Variance = {∑X²i / n} - (x̄)²

                ={414/10} - (1.4)²

 Variance =39.44

Result: 

Random Sample is 

Random Sample

0

-3

10

9

3

-9

-3

-6

8

5

 The mean of sample= 1.4 and variance of sample= 39.44 

                                                                                                               


Padmbhushan Vasantraodada Patil Mahavidyalaya, Kavathe Mahankal

Department of Statistics

Title: Model sampling from Binomial distribution

                                                                                                               

1. Draw a model sample of size 10 from the binomial Distribution with parameters n = 8 and p = 0.6 also find A.M., G.M. and H. M. of your sample.

2. Draw a model sample of size 13 from the binomial Distribution with parameter n =10 and p = 0.438 also find C.V. of your sample and compare it with theoretical values.

3. Draw a model sample of size 10 from the binomial Distribution with parameter n = 12 and p = 0.25 also find A.M., Median of your sample.

4. Draw a model sample of size 13 from the binomial Distribution with parameter n = 7 and p = 0.336. Find variance of your sample and compare it with theoretical value.

5. Draw a model sample of size 14 from the binomial Distribution with parameter n = 10 and p = 0.62. Find mean and variance of your sample and compare it with theoretical value.

6. If X has binomial distribution such that P{X = 0} = P{X = 5} and n = 5 then obtain a random sample of size 11 from this distribution. Obtain sample median and sample mode.

7. If x has binomial distribution such that E(X) =2 and V(X) =1.5 then obtain a random sample of size 25 from this distribution. Obtain sample mean and sample variance. Compare these with theoretical mean and variance of the distribution. 

                                                                                                               

ANSWERS Practical NO. 1.

Q1. Draw a model sample of size 10 from the binomial Distribution with parameters n = 8 and p = 0.6 also find A.M., G.M. and H. M. of your sample.

Answer : 








Binomial Distribution PMF Calculator Binomial Distribution PMF

Binomial Distribution PMF

Enter the parameters of the binomial distribution:



Enter the value of k to calculate P(X = k):


PMF for X =

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Changing Color Blog Name  Measures of Central Tendency  I. Introduction. II. Requirements of good measures. III. Mean Definition. IV . Properties  V. Merits and Demerits. VI. Examples VII.  Weighted Arithmetic Mean VIII. Median IX. Quartiles I. Introduction Everybody is familiar with the word Average. and everybody are used the word average in daily life as, average marks, average of bike, average speed etc. In real life the average is used to represent the whole data, or it is a single figure is represent the whole data. the average value is lies around the centre of the data. consider the example if we are interested to measure the height of the all student and remember the heights of all student, in that case there are 2700 students then it is not possible to remember the all 2700 students height so we find out the one value that represent the height of the all 2700 students in college. therefore the single value represent ...
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Business Statistics Notes ( Meaning, Scope, Limitations of statistics and sampling Methods)

  Business Statistics Paper I Notes. Welcome to our comprehensive collection of notes for the Business Statistics!  my aim is to provided you  with the knowledge you need as you begin your journey to comprehend the essential ideas of this subject. Statistics is a science of collecting, Presenting, analyzing, interpreting data to make informed business decisions. It forms the backbone of modern-day business practices, guiding organizations in optimizing processes, identifying trends, and predicting outcomes. I will explore several important topics through these notes, such as: 1. Introduction to Statistics. :  meaning definition and scope of  Statistics. 2. Data collection methods. 3. Sampling techniques. 4. Measures of  central tendency : Mean, Median, Mode. 5. Measures of Dispersion : Relative and Absolute Measures of dispersion,  Range, Q.D., Standard deviation, Variance. coefficient of variation.  6.Analysis of bivariate data: Correlation, Regr...
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Classification, Tabulation, Frequency Distribution, Diagrams & Graphical Presentation.

Business Statistics I    Classification, Tabulation, Frequency Distribution ,  Diagrams & Graphical Presentation. In this section we study the following point : i. Classification and it types. ii. Tabulation. iii. Frequency and Frequency Distribution. iv. Some important concepts. v. Diagrams & Graphical Presentation   I. Classification and it's types:        Classification:- The process of arranging data into different classes or groups according to their common  characteristics is called classification. e.g. we dividing students into age, gender and religion. It is a classification of students into age, gender and religion.  Or  Classification is a method used to categorize data into different groups based on the values of specific variable.  The purpose of classification is to condenses the data, simplifies complexities, it useful to comparison and helps to analysis. The following are some criteria to classi...
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Measures of Dispersion : Range , Quartile Deviation, Standard Deviation and Variance.

Measures of Dispersion :  I.  Introduction. II. Requirements of good measures. III. Uses of Measures of Dispersion. IV.  Methods Of Studying Dispersion:     i.  Absolute Measures of Dispersions :             i. Range (R)          ii. Quartile Deviation (Q.D.)          iii. Mean Deviation (M.D.)         iv. Standard Deviation (S. D.)         v. Variance    ii.   Relative Measures of Dispersions :              i. Coefficient of Range          ii. Coefficient of Quartile Deviation (Q.D.)          iii. Coefficient of Mean Deviation (M.D.)         iv. Coefficient of Standard Deviation (S. D.)         v. Coefficien...
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Basic Concepts of Probability and Binomial Distribution , Poisson Distribution.

 Probability:  Basic concepts of Probability:  Probability is a way to measure hoe likely something is to happen. Probability is number between 0 and 1, where probability is 0 means is not happen at all and probability is 1 means it will be definitely happen, e.g. if we tossed coin there is a 50% chance to get head and 50% chance to get tail, it can be represented in probability as 0.5 for each outcome to get head and tail. Probability is used to help us taking decision and predicting the likelihood of the event in many areas, that are science, finance and Statistics.  Now we learn the some basic concepts that used in Probability:  i) Random Experiment OR Trail: A Random Experiment is an process that get one or more possible outcomes. examples of random experiment include tossing a coin, rolling a die, drawing  a card from pack of card etc. using this we specify the possible outcomes known as sample pace.  ii)Outcome: An outcome is a result of experi...
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Statistical Inference I ( Theory of estimation : Efficiency)

🔖Statistical Inference I ( Theory of estimation : Efficiency)  In this article we see the  terms:  I. Efficiency. II. Mean Square Error. III. Consistency. 📚 Efficiency:  We know that  two unbiased estimator of parameter gives rise to infinitely many unbiased estimators of parameter. there if one of parameter have two estimators then the problem is to choose one of the best estimator among the class of unbiased estimators. in that case we need to some other criteria to to find out best estimator. therefore, that situation  we check the variability of that estimator, the measure of variability of estimator T around it mean is Var(T). hence If T is an Unbiased estimator of parameter then it's variance gives good precision. the variance is smaller then it give's greater precision. 📑 i. Efficient estimator: An estimator T is said to be an Efficient Estimator of 𝚹, if T is unbiased estimator of    𝛉. and it's variance is less than any other estima...
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The Power of Statistics: A Gateway to Exciting Opportunities

  My Blog The Power of Statistics: A Gateway to Exciting Opportunities     Hey there, future statistician! Ever wondered how Netflix seems to know exactly what shows you'll love, how sports teams break down player performance, or how businesses figure out their pricing strategies? The answer is statistics—a fascinating field that helps us make sense of data in our everyday lives. Let's dive into why choosing statistics for your B.Sc. Part First can lead you to some exciting opportunities.     Why Statistics Matters in Everyday Life     From predicting election outcomes and analyzing social media trends to understanding consumer behavior and optimizing public transport routes, statistics are crucial. It's the backbone of modern decision-making, helping us sift through complex data to uncover meaningful insights that drive innovation and progress.   The Role of Statistics in Future Opportunities ...
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Statistical Inference I ( Theory of Estimation) : Unbiased it's properties and examples

 📚Statistical Inference I Notes The theory of  estimation invented by Prof. R. A. Fisher in a series of fundamental papers in around 1930. Statistical inference is a process of drawing conclusions about a population based on the information gathered from a sample. It involves using statistical techniques to analyse data, estimate parameters, test hypotheses, and quantify uncertainty. In essence, it allows us to make inferences about a larger group (i.e. population) based on the characteristics observed in a smaller subset (i.e. sample) of that group. Notation of parameter: Let x be a random variable having distribution function F or f is a population distribution. the constant of  distribution function of F is known as Parameter. In general the parameter is denoted as any Greek Letters as θ.   now we see the some basic terms :  i. Population : in a statistics, The group of individual under study is called Population. the population is may be a group of obj...
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