Class 11 Statistics NCERT Solutions for Chapter 6 2021: Download PDF

NCERT Solutions for Class 11 Statistics Chapter 6

NCERT Solutions for Class 11 Statistics Chapter 6: We are always concerned with the students’ development. NCERT Solutions For Class 11 Statistics Chapter 6 enable students to comprehend each topic in depth, resulting in good knowledge and the development of expertise in that topic, chapter, or subject. The resources were created by a group of subject experts. 

NCERT Solutions for Class 11 Statistics Chapter 6 PDF

NCERT Solutions For Statistics Class 11 Ch 6

 


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NCERT Solutions for Class 11 Statistics Chapter 6: Overview

Measures of Dispersion

 Introduction

The NCERT Solutions for class 11th chapter 6 teach a new topic of dispersion to the students in this section. The students have already learned several statistical topics such as data collecting, data summarization, tools, and so on. The kids will study about dispersion in this lesson. Our NCERT Solutions explain it with illustrations.

 Measures Based upon the Spread of Values

Students will learn how to use a variety of measures to calculate dispersion based on values in this section. They are the set’s range, defined as the difference between the set’s largest and lowest values. The quartile dispersion is the following one. It is the technique of partitioning the entire set into four equal sections in order to eliminate the maximum and minimum values that cause dispersion.

Measures of Dispersion from Average

The measures of dispersion in the Statistics Class 11 NCERT Solutions describe the remaining two measures in this part with enough examples. We’d utilise range and quartile dispersion if we had a set of data.

If the average needs to be calculated, standard deviation and mean deviation should be used instead. The mean deviation can be calculated in a variety of ways, including determining the median using both grouped and ungrouped data. Each concept is also supported by two or more instances.

Absolute and Relative Dispersion of Measures

Class 11 Statistics NCERT Solutions Chapter 6 explains another topic. The meaning and explanation of the Coefficient of Mean Deviation were discussed by NCERT Solutions. It also defines the significance of each term in the formula and explains the derived formula.

Lorenz Curve 

NCERT Class 11 Statistics Chapter 6 Measures of Dispersion introduces students to a new concept known as the Lawrence curve. The students have learnt how to measure dispersion using sets, averages, deviations, and other methods. The Lorenz curve is the graph paper curve that represents all of these measurements. The Lorenz Curve must be thoroughly studied by students.

The building of the Lorenz curve was also covered in NCERT Solutions for Class 11 Statistics Chapter 6. Students can understand it as a step-by-step approach involving numerous things to study because it plays such an important function in the field of statistics. These Solutions also include a detailed graphic of these steps of the Lorenz curve to aid understanding.

Conclusion

This is the ending part of the chapter. Here the NCERT Solutions Class 11 Statistics Chapter 6 explains the importance of dispersion and the measures, different cases of dispersion like deviations, average, etc with proper examples and exercises also.

An exercise with the 12 Long answer type questions

NCERT Solutions provide a free download for PDF. A group of language experts are available to clarify the doubts of students through live chat.

Access NCERT Solutions For Class 11 Statistics Chapter 6

Question 1:

A measure of dispersion is a good supplement to the central value in understanding a frequency distribution. Comment.  

ANSWER:

The study of the averages is only one sided distribution story. In order to understand the frequency distribution fully, it is essential to study the variability of the observations. The average measures center of the data whereas the quantum of the variation is measured by the measures of dispersion like range, quartile deviation, mean deviation and Standard Deviation.

For example, if a country has very high income group people and very low income group people, then we can say that the country has large income disparity.

Question 2:

Which measure of dispersion is the best and how?

ANSWER:

Standard Deviation is the best measure of dispersion as it satisfies the most essentials of the good measure of dispersion. The following points make Standard Deviation the best measure of dispersion:

  1. Most of the statistical theory is based on Standard Deviation. It helps to make comparison between variability of two or more sets of data. Also, Standard Deviation helps in testing the significance of random samples and in regression and correlation analysis.
  2. It is based on the values of all the observations. In other words, Standard Deviation makes use of every item in a particular distribution.
  3. Standard Deviation has a precise value and is a well-defined and definite measure of dispersion. That is, it is rigidly defined.
  4. It is independent of the origin.
  5. It is widely used measure of dispersion as all data distribution is nearer to the normal distribution.
  6. It enables algebraic treatment. It has correct mathematical processes in comparison to range, quartile deviation and mean deviation.

Question 3:

Some measures of dispersion depend upon the spread of values whereas some calculate the variation of values from a central value. Do you agree?

ANSWER:

Yes, it is true that some measures of dispersion depend upon the spread of values, whereas some calculate the variation of values from the central value. The spread of values is determined by the absolute measures of dispersion like Range, Quartile Mean Deviation, and Standard Deviation.

These measures express dispersion in terms of original unit of the series and it cannot be used for the comparison of statistical data having different units. On the other hand, the relative measures of the dispersion calculate the variability of the values from a central value. The relative measure includes coefficient of Range, Mean Deviation and Variation. It is used when the comparison has to be made between two statistical sets. These measures are free from any units.

Question 4:

In a town, 25% of the persons earned more than Rs 45,000 whereas 75% earned more than 18,000. Calculate the absolute and relative values of dispersion.

ANSWER:

Absolute Value of Dispersion

Relative Value of Dispersion

Question 5:

The yield of wheat and rice per acre for 10 districts of a state is as under:

District 1 2 3 4 5 6 7 8 9 10
Wheat 12 10 15 19 21 16 18 9 25 10
Rice 22 29 12 23 18 15 12 34 18 12


Calculate for each crop,
(i) Range
(ii) Q.D.
(iii) Mean Deviation about Mean
(iv) Mean Deviation about Median
(v) Standard Deviation
(vi) Which crop has greater variation?
(vii) Compare the value of different measures for each crop.

ANSWER:

(i) Range

  1. Wheat

Highest value of distribution (H) = 25

Lowest value of distribution (L) = 9

Range = H – L

          = 25 – 9

          =16

  1. Rice

Highest value of distribution (H) = 34

Lowest value of distribution (L) = 12

Range = H – L

           =34 – 12

= 22

(ii) Quartile Deviation

  1. Wheat

Arranging the production of wheat in increasing order

9, 10, 10, 12, 15, 16, 18, 19, 21, 25

= 2.75th item

=Size of 2th item + 0.75 (size of 3rd item – size of 2nd item)

= 10 + 0.75 (10 – 10)

= 10 + 0.75 × 0

=10

= 8.25th

=Size of 8th item + 0.25 (size of 9th item – size of 8th item)

= 19 + 0.25 (21 – 19)

= 19 + 0.25 × 2

= 19 + 0.50 = 19.50

= 4.75

  1. Rice

Arranging the data of production of rice

12, 12, 12, 15, 18, 18, 22, 23, 29, 34

= 2.75th item

= size of 2nd item + 0.75 (size of 3rd item – size of 2nd item)

= 12 + 0 .75 (12 – 12)

= 12 + 0.75 × 0

= 12

= 8.25th item

= Size of 8th item + 0.25 (size of 9th item – size of 8th item)

= 23 + 0.25 (29 – 23)

= 23 + 0.25 × 6

= 23 + 1.5

= 24.5

= 6.25

 

(iii) Mean Deviation about Mean

 

  1. Wheat

Wheat Production

x

 

9

6

10

5

10

5

12

3

15

0

16

1

18

3

19

4

21

6

25

10

   

 

  1. Rice

Rice Production

 x

 

12

6

12

6

12

6

15

3

18

0

18

0

22

4

23

5

29

11

34

16

   

 

(iv) Mean Deviation about Median

 

  1. Wheat

 

Production of Wheat

x

 

09

6

10

5

10

5

12

3

15

0

16

1

18

3

19

4

21

6

25

10

 

 

 

  1. Rice

Production of rice

d = X–18

12

6

12

6

12

6

15

3

 

0

18

0

22

4

23

5

29

11

34

16

 

 

Since n is even

(v) Standard Deviation

 

  1. Wheat

Reduction of Wheat

x

Ax = 15

dx – Ax

d2

9

–6

36

10

–5

25

10

–5

25

12

–3

9

15

0

0

16

1

1

18

3

9

19

4

16

21

6

36

25

10

100

 

   

 

  1. Rice

Production of rice

 x

Ax = 18

d = x – A x

d2

12

–6

36

12

–6

36

12

–6

36

15

–3

9

18

0

0

22

4

16

23

5

25

29

11

121

34

16

256

 

   

 

(vi) Coefficient of Variation

  1. Wheat
  2. Rice

 

(vi) Rice crop has greater variation as the coefficient of variation is higher for rice as compared to that of wheat.

(vii) Rice crop has higher Range, Quartile Deviation, Mean Deviation about mean, Mean Deviation about median, Standard Deviation and Coefficient of Variation.

Question 6:

In the previous question, calculate the relative measures of variation and indicate the value which, in your opinion, is more reliable.

ANSWER:

(i) Coefficient of Range

  1. a) Wheat
  2. b) Rice

(ii) Coefficient of Quartile Deviation

  1. a) Wheat
  2. b) Rice

Q1 = 12, Q3 = 24.5

(iii) Coefficient of Mean Deviation from mean

  1. a) Wheat
  2. b) Rice

(iv) Coefficient of Variation

  1. a) Wheat
  2. b) Rice

The coefficient of variation is more reliable than all other measures.

Question 7:

A batsman is to be selected for a cricket team. The choice is between X and Y on the basis of their scores in five previous tests which are:
 

X

25

85

40

80

120

Y

50

70

65

45

80


Which batsman should be selected if we want,
(i) a higher run getter, or
(ii) a more reliable batsman in the team?

ANSWER:

Batsman X

 

X

X –  = x

X – 70

x2

25

– 45

2025

85

+ 15

225

 

– 30

900

80

10

100

120

50

2500

 

 

 
       

 

Batsman Y

 

Y

 

y2

50

– 12

144

70

8

64

65

3

9

45

– 17

289

80

18

324

 

 

 
       

 

(i) Average of Batsman X is higher than that of Batsman Y, so he should be selected if we want to score  higher run.

(ii) The Batsman is more reliable than Batsman Y. This is because the coefficient of variation of Batsman X is higher than that of Batsman Y.

Question 8:

To check the quality of two brands of light bulbs, their life in burning hours was estimated as under for 100 bulbs of each brand.

Life 

 

No. of bulbs

(in hrs)

Brand A

Brand B

0 – 50

 

15

2

50 – 100

 

20

8

100 – 150

 

18

60

150 – 200

 

25

25

200 – 250

 

22

5

 

 

100

100

(i) Which brand gives higher life?

(ii) Which brand is more dependable?

ANSWER:

For Brand A

 

Life

(in hours)

No. of bulbs

M

A = 125

 

d2

fd

fd2

 

f

x

X – A = d

 

 

 

 

0 – 50

15

25

– 100

– 2

4

– 30

60

50 – 100

20

75

– 50

– 1

1

– 20

20

100 – 150

18

125

0

0

0

0

0

150 – 200

25

175

50

1

1

25

25

200 – 250

22

225

100

2

4

44

88

 

 

 

 

 

 

   
                 

 

For Brand B

 

 

Life (in hrs)

No. of bulbs (f)

Mid value m

A = 125

 

d’2

fd’

fd’2

0 – 50

2

25

– 100

– 2

4

– 4

8

50 – 100

8

75

– 50

– 1

1

– 8

8

100 – 150

60

125

0

0

0

0

0

150 – 200

25

175

50

1

1

25

25

200 – 250

5

225

100

2

4

10

20

 

 

 

 

 

 

   
                 

 

(i) The average life of bulb of Brand B has comparatively  higher life than the bulb of Brand A.

(ii) The  bulbs  of Brand B is more dependable as CV of Brand B is lesser than CV of Brand A.

Question 9:

Average daily wage of 50 workers of a factory was Rs 200 with a Standard Deviation of Rs 40. Each worker is given a raise of Rs 20. What is the new average daily wage and Standard Deviation? Have the wages become more or less uniform?

ANSWER:

N = 50

 = 200

s = 40

So, Total Wages = 200 × 50

= Rs 10,000

Now, increased wage rate = Rs 20

Total raise = 50 × 20 = Rs 1,000

Total Wage after raise = 10,000 + 1,000

=Rs 11,000

= Rs 220

Initial Standard Deviation = Rs 40

So, New Standard Deviation = Rs 40 + Rs 20

= Rs 60

Note: New Standard Deviation will rise by the same amount as the wage of each worker has increased.

Question 10:

If in the previous question, each worker is given a hike of 10% in wages, how are the Mean and Standard Deviation values affected?

ANSWER:

Average wage = Rs 200

Hike in wages = 10% of Rs 200

= Rs 20

Individual raise given to each worker = Rs 20

Total raise in wage = 50 × 20 = Rs 1,000

New Total Wage = Rs 10,000 + Rs 1000

= Rs 11,000

 

Initial Standard Deviation = Rs 40

So, New Standard Deviation = Rs 40 + 20

= Rs 60

Page No 90:

Question 11:

Calculate the Mean Deviation using mean and Standard Deviation for the following distribution.

Classes

Frequencies

20 – 40

3

40 – 80

6

80 – 100

20

100 – 120

12

120 – 140

9

Total

50

 

ANSWER:

Classes

Frequency

m

A = 90

d = X – A

     

fd’

d’2

fd’2

20 – 40

3

30

– 60

60

180

– 6

– 18

36

108

40 – 80

6

60

– 30

30

180

– 3

– 18

9

54

80 – 100

20

90

0

0

0

0

0

0

0

100 – 120

12

110

20

20

240

2

24

4

48

120 – 140

9

130

40

40

360

4

36

16

144

 

 

 

 

 

 

 

 

 

 

Mean Deviation from mean

Question 12:

The sum of 10 values is 100 and the sum of their squares is 1090. Find out the Coefficient of Variation.

ANSWER:

image6914698646333600083

Access Other NCERT Chapters Solutions of Class 11 Statistics

You can download the PDF of NCERT Solutions Class 11 Accountancy for other chapters: 

  • Chapter-7 Correlation
  • Chapter-8 Index Numbers

We have provided all the important above in the article regarding the CBSE NCERT Solutions of class 11 Statistics Chapter 6. If you have any queries, you can mention them in the comment section.

FAQ (Frequently Asked Questions): NCERT Solutions for Class 11 Statistics Chapter 6

What are the steps involved in the construction of the Lorenz Curve?

The procedure involves a set of nine steps.
Determine the class midpoints. Table 2: Col. 2
Multiply the midpoint of the class by the frequency in the class to get the anticipated total income of employees in each c ss. As a result, Col. (4) of Table is obtained.
Calculate the frequency of each class as a proportion of the total frequency. Col. (5) of Table is thus obtained.
Calculate the total income of each class as a percentage (%) of the total income of all classes. Col. (6) of Table 5 was so obtained. 
Prepare a table with less than cumulative frequency and a table with cumulative income.
Col. 6 (2) Employees’ cumulative frequency is shown in Table 6.5.
Table 6.5 displays the total income received by these individuals.
Draw a line from Coordinate(0,0) to Co-ordinate(0,0) (100,100). The line of equal distribution is what it’s called.
Plot the cumulative percentages of employees on the horizontal axis and cumulative income on the vertical axis.

What are the measures used for the Dispersion of Values?

There are four main measures that users can utilize to further disperse scatter values. They’re 
Range
Quartile Dispersion
Mean Deviation
Standard Deviation.

Define mean deviation.

The mean deviation is the average of all the deviations of all the principles taken from an average value (mean, median, mode) of the series while ignoring the signs (+ or -) of the deviation.

What is the standard deviation?

It is the square root of the arithmetic mean of the squared deviations of the items from their mean value.
Define range.
The range is the variance between the lowest and the highest values in a series. 
Range = Highest value in the series – Lowest value in the series
 

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