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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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Basic Concepts of Probability and Binomial 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 experiment. an outcome is one of the pos
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Statistical Inference II Notes

Likelihood Ratio Test 
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Measures of Central Tendency :Mean, Median and Mode

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 the whole data and
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Time Series

 Time series  Introduction:-         We see the many variables are changes over period of time that are population (I.e. population are changes over time means population increase day by day), monthly demand of commodity, food production, agriculture production increases and that can be observed over period of times known as time series. Time series is defined as a set of observation arranged according to time is called time series. Or a time Series is a set of statistical observation arnging chronological order. ( Chronological order means it is arrangements of variable according to time) and it gives information about variable.  Also we draw the graph of time series to see the behaviour of variable over time. It can be used of forecasting. The analysis of time series is helpful to economist, business men, also for scientist etc. Because it used to forecasting the future, observing the past behaviour of that variable or items. Also planning for future, here time series use past data h
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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 classify the data into groups.        i. Quantitative Classification :-
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Sequential Analysis: (SPRT)

  Sequential Analysis: We seen that in NP theory of testing hypothesis or in the parametric test n is the sample size and is regarded as fixed and the value of α fixed , we minimize the value of β.  But in the sequential analysis theory invented by A Wald in sequential analysis n is the sample number is not fixed but the both values α and β are fixed as constant. Sequential Probability Ratio Test: (SPRT):
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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. Coefficient of Variation (C.V.)                                                                                                                    I.  Introduction. We have the various measures of central tendency, like Mean, Median & Mode,  it is a single figure that represent the whole data. Now we are interested to study this figure(i.e. measures of central tendency) is proper represe
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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, Regression.  These notes will serve as you
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Statistical Quality Control

 Statistical Quality Control  Statistical quality control (S. Q. C.) is a branch of Statistics it deals with the application of statistical methods to control and improve that quality of product. In this use statistical methods of sampling and test of significance to monitoring and controlling than quality of product during the production process.  The most important word in statistical Quality control is quality  The quality of product is the most important property while purchasing that product the product fulfill or meets the requirements and required specification we say it have good quality or quality product other wise not quality. Quality Control is the powerful technique to diagnosis the lack of quality in material, process of production.  Causes of variation:   When the product are produced in large scale there are variation in the size or composition the variation is inherent and inevitable in the quality of product these variation are classified into two causes.  1) chan
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