Class 11 Statistics NCERT Solutions for Chapter 5 2023: Download Free PDF

NCERT Solutions for Class 11 Statistics Chapter 5

NCERT Solutions for Class 11 Statistics Chapter 5: We offer students with an excellent understanding of statistics. They offer up-to-date and improved solutions, with content prepared by subject matter specialists with years of experience. With the help of NCERT Solutions For Class 11 Statistics Chapter 5, students can obtain information as well as good grades.

NCERT Solutions for Class 11 Statistics Chapter 5 PDF

NCERT Solutions For Statistics Class 11 Ch 5

 


Download the Free PDF for NCERT Solutions for Class 11 Statistics Chapter 5

Click Here

NCERT Solutions for Class 11 Statistics Chapter 5: Overview

Class 11 Statistics NCERT Solutions Chapter 5 is free to download in PDF format from our official website. Students should take advantage of this opportunity to learn more. It is up to the student whether to use a soft copy or a hard copy. If they are willing to study from a physical copy, they will be able to learn without having to worry about their internet connection. It also helps to prepare for competitive exams and other board exams.

Measures of Central Tendency

  1. Introduction 

Class 11 Statistics NCERT Solutions As the students have learned how to gather data and portray data, Chapter 5 aims to introduce the measures available for the central tendency in Statistics. The following step is to analyze and interpret the data. We can use three different measures of central trends to assess the data: arithmetic means, median, and mode.

  1. Arithmetic Mean

The arithmetic mean is a widely used metric for determining the average of a set of data. The mean for grouped data and the mean for ungrouped data are two different versions explained in NCERT Solutions for Class 11 Statistics Chapter.

  1. The arithmetic mean is the sum The NCERT Solutions Class 11 Statistics Chapter 5 clearly explains available on our official website with an illustration. Again the grouped data has two kinds of series namely-discrete series and continuous series.

There are three methods for computing the mean for ungrouped and grouped data, according to the Measures of Central Tendency in Statistics Class 11 NCERT Solutions. The three possibilities are a direct methodology, an assumed method, and a step division method. Each strategy may alter depending on whether the data is grouped or ungrouped.

  1. Median

The students are urged to master another measure of central tendency, the median, in this section. The median is a straightforward concept, according to the NCERT Class 11 Statistics Chapter 5 Measures of Central Tendency. It can be computed by taking the middle value, which is the value that divides both sides of the values. However, this is only conceivable if the series’ length is odd. If students are asked to find the median for a set of even number values, we must first find the average of the two middle words. 

Another topic covered with the students is quartiles. The term quartile refers to the division of a full size into four equal pieces. It can be calculated separately for both series by themselves. Percentiles are a topic for which the whole size is divided into 99 Parts which reaches the cent. 

  1. Mode 

Our NCERT solutions for Statistics Chapter 5 in Class 11 specify that students comprehend each measure of central tendency. Because it is impossible for kids to forget the meaning and formula if they grasp them. Despite the fact that the mode is the last measure of central tendency, it is quite important.

The mode is a French word that refers to the most popular or common value in the set. In simple terms, the mean is the value with the greatest number of occurrences. We can calculate directly using the direct technique by looking at itself. There is a specific formula for determining the value of mode in the continuous approach.

  1. The relative position of arithmetic mean, median, and mode

The placements of three measures of central tendency are also mentioned in the NCERT Solutions For Class 11 Statistics Chapter 5. The median should be in the middle of the magnitude of three measures can be stirred in any direction. The arithmetic means and mode will be at one of the two extremes.

  1. Conclusion

Thus Chapter 5 of measures of central tendency has been explained very well by professional teachers to the students. The material is provided with different examples of all six concepts.

Access NCERT Solutions For Class 11 Statistics Chapter 5

Question 1:

Which average would be suitable in the following cases?

(i) Average size of readymade garments.

(ii) Average intelligence of students in a class.

(iii) Average production in a factory per shift.

(iv) Average wage in an industrial concern.

(v) When the sum of absolute deviations from average is least.

(vi) When quantities of the variable are in ratios.

(vii) In case of open-ended frequency distribution.

ANSWER:

(i) The demand for the average size of any readymade garment is the maximum. As, the modal value represents the value with the highest frequency, so the number of the average size to be produced is given by the Modal value.

(ii) Median will be the best measure for calculating the average intelligence of students in a class. It is the value that divides the series into two equal parts. So, number of students below and above the average intelligence can easily be estimated by median.

(iii) It is advisable to use mean for calculating the average production in a factory per shift. The average production is best calculated by arithmetic mean.

(iv) Mean will be the most suitable measure. It is calculated by dividing the sum of wages of all the labour by the total number of labours in the industry.

(v) When the sum of absolute deviations from average is the least, then mean could be used to calculate the average. This is an important mathematical property of arithmetic mean. The algebraic sum of the deviations of a set of n values from A.M. is 0.

(vi) Median will be the most suitable measure in case the variables are in ratios. It is least affected by the extreme values.

(vii) In case of open ended frequency distribution, Median is the most suitable measure as it can be easily computed. Moreover, the median value can be estimated even in case of incomplete statistical series.

Question 2(i):

The most suitable average for qualitative measurement is

(a) arithmetic mean

(b) median

(c) mode

(d) geometric mean

(e) none of the above

ANSWER:

Median is the most suitable average for qualitative measurement. This is because Median divides a series in two equal parts.

Question 2(ii):

Which average is affected most by the presence of extreme items?

(a) median

(b) mode

(c) arithmetic mean

(d) geometric mean

(e) harmonic mean

ANSWER:

Arithmetic mean is the most affected by the presence of extreme items. It is one of the prime demerits of the arithmetic mean. It is easily distorted by the extreme values, and also the value of arithmetic mean may not figure out at all in the series.

Page No 72:

Question 2(iii):

The algebraic sum of deviation of a set of n values from A.M. is

(a) n

(b) 0

(c) 1

(d) none of the above

ANSWER:

The algebraic sum of deviation of a set of n values from A.M. is zero. This is one of the mathematical properties of arithmetic mean.

Page No 72:

Question 3:

Comment whether the following statements are true or false.

(i) The sum of deviation of items from median is zero.

(ii) An average alone is not enough to compare series.

(iii) Arithmetic mean is a positional value.

(iv) Upper quartile is the lowest value of top 25% of items.

(v) Median is unduly affected by extreme observations.

ANSWER:

(i) The sum of deviation of items from median is zero. False

The statement is false. This mathematical property applies to the arithmetic mean that states that the sum of the deviation of all items from the mean is zero.

(ii) An average alone is not enough to compare series. True

An average indicates only the behaviour of a particular series. Therefore, in order to measure the extent of divergence of different items from the central tendency is measured by dispersion. So, average is not enough to compare the series.

(iii) Arithmetic mean is a positional value. False

This statement is false as mean is not a positional average, rather the statement holds true for median and mode. The calculation of median and modal values is based on the position of the items in the series, i.e. why these are also termed as positional averages.

(iv) Upper quartile is the lowest value of top 25% of items. True

The value that divides a statistical series into four equal parts, the end value of each part is called quartile. The third quartile or the upper quartile has 75 % of the items below it and 25 % of items above it,

(v) Median is unduly affected by extreme observations. False

This statement is true for Arithmetic mean. Arithmetic mean is most affected by the presence of extreme items. It is one of the prime demerits of the arithmetic mean. It is easily distorted by the extreme values, and also the value of arithmetic mean may not figure out at all in the series.

Page No 72:

Question 4:

If the arithmetic mean of the data given below is 28, find (a) the missing frequency, and (b) the median of the series:

Profit per retail shop (in Rs)

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

Number of retail shops

12

18

27

17

6

 

ANSWER:

(i) Let the missing frequency be f

Arithmetic Mean = 28

 

 

Profit per Retail Shop (in Rs)

No of Retail Shops

Mid Value

 

Class Interval

(f)

(m)

fm

0 – 10

12

5

60

10 – 20

18

15

270

20 – 30

27

25

675

30 – 40

f1

35

35f1

40 – 50

17

45

765

50 – 60

6

55

330

 

 

 

 

 

or, 2240 + 28f1 = 2100 + 35f1

or, 2240 – 2100 = 35f1 – 28f1

or, 140 = 7f1

f1 = 20

 

(ii)

 

Class Interval

Frequency

 

(f)

Cumulative

Frequency

(CF)

 

0 – 10

12

12

 

10 – 20

18

30

 

20 – 30

27

57

 

30 – 40

20

77

 

40 – 50

17

94

 

50 – 60

6

100

 

Total

 

 

 

So, the Median class = Size of  item

                                         = 50th item

50th item lies in the 57th cumulative frequency and the corresponding class interval is 20 – 30.

 

Page No 72:

Question 5:

The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

Workers

A

B

C

D

E

F

G

H

I

J

Daily Income (in Rs)

120

150

180

200

250

300

220

350

370

260

 

ANSWER:

Workers

Daily Income (in Rs)

(X)

A

120

B

150

C

180

D

200

E

250

F

300

G

220

H

350

I

370

J

260

Total

 

 

N = 10

Arithmetic mean = Rs 240

Page No 72:

Question 6:

Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.

Income (in Rs)

Number of families

More than 75

150

More than 85

140

More than 95

115

More than 105

95

More than 115

70

More than 125

60

More than 135

40

More than 145

25

 

ANSWER:

Income

No. of families

Frequency

Mid Value

fm

Class Interval

(CF)

(f)

(m)

 

75 – 85

150

150 – 140 = 10

80

800

85 – 95

140

140 – 115 = 25

90

2250

95 – 105

115

115 – 95 = 20

100

2000

105 – 115

95

95 – 70 = 25

110

2750

115 – 125

70

70 – 60 = 10

120

1200

125 – 135

60

60 – 40 = 20

130

2600

135 – 145

40

40 – 25 = 15

140

2100

145 – 155

25

25

150

3750

Total

 

 

 

 

 

= Rs 116.33



Page No 73:

Question 7:

The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.

Size of Land Holdings (in acres)

Less than 100

100 – 200

200 – 300

300 – 400

400 and above

Number of families

40

89

148

64

39

 

ANSWER:

Size of Land Holdings

Class Interval

No. of Families

 

(f)

Cumulative Frequency

(CF)

0 – 100

40

40

100 – 200

89

129

200 – 300

148

277

300 – 400

64

341

400 – 500

39

380

Total

 

 

 

So, the Median class = Size of  item = 190th item

 

190th item lies in the 129th cumulative frequency and the corresponding class interval is 200 – 300.

The median size of land holdings = 241.22 acres

Page No 73:

Question 8:

The following series relates to the daily income of workers employed in a firm. Compute (a) highest income of lowest 50% workers (b) minimum income earned by the top 25% workers and (c) maximum income earned by lowest 25% workers.

Daily Income (in Rs)

10 – 14

15 – 19

20 – 24

25 – 29

30 – 34

35 – 39

Number of workers

5

10

15

20

10

5

(Hint: Compute median, lower quartile and upper quartile)

ANSWER:

Daily Income

(in Rs)

Class Interval

No. of Workers

(f)

Cumulative frequency

(CF)

9.5 – 14.5

5

5

14.5 – 19.5

10

15

19.5 – 24.5

15

30

24.5 – 29.5

20

50

29.5 – 34.5

10

60

34.5 – 39.5

5

65

 

 

 

 

(a) Highest income of lowest 50% workers

 

 

32.5th item lies in the 50th cumulative frequency and the corresponding class interval is 24.5 – 29.5.

 

 

(b) Minimum income earned by top 25% workers
In order to calculate the minimum income earned by top 25% workers, we need to ascertain Q3.

48.75th item lies in 50th item and the corresponding class interval is 24.5 – 29.5.

 

 

(c) Maximum  income earned by lowest 25% workers
In order to calculate the maximum income earned by lowest 25% workers, we need to ascertain Q1.

  

16.25th item lies in the 30th cumulative frequency and the corresponding class interval is 19.5 – 24.5

Page No 73:

Question 9:

The following table gives production yield in kg. per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode values.

Production yield (kg. per hectare)

50 – 53

53 – 56

56 – 59

59 – 62

62 – 65

65 – 68

68 – 71

71 – 74

74 – 77

Number of farms

3

8

14

30

36

28

16

10

5

 

ANSWER:

(i) Mean

Production Yield

No. of farms

Mid value

A = 63.5

   

50 – 53

3

51.5

–12

–4

–12

53 – 56

8

54.5

–9

–3

–24

56 – 59

14

57.5

–6

–2

–28

59 – 62

30

60.5

–3

–1

–30

62 – 65

36

63.5

0

0

0

65 – 68

28

66.5

+3

+1

28

68 – 71

16

69.5

+6

+2

32

71 – 74

10

72.5

+9

+3

30

74 – 77

5

75.5

+12

+4

20

Total

 

 

 

 

 

 

= 63.5 + 0.32

= 63.82 kg per hectare

 

(ii) Median

 

Class Interval

Frequency

(f)

CF

50 – 53

3

3

53 – 56

8

11

56 – 59

14

25

59 – 62

30

55

62 – 65

36

91

65 – 68

28

119

68 – 71

16

135

71 – 74

10

145

74 – 77

5

150

Total

 

 

 

75th item lies in the 91st cumulative frequency and the corresponding class interval is 62 – 65.

 

 (iii) Mode

 

Grouping Table

 

 

 

Class Interval

I

II

III

IV

V

VI

 

 

 

 

 

 

 

 

 

50 – 53

3

 

 

 

 

 

 

11

22

25

52

   

53 – 56

8

   

 

 

 

 

 

 

56 – 59

14

 

 

 

 

 

 

44

 

 

 

 

 

59 – 62

30

   

 

 

 

 

 

 

62 – 65

 

 

 

 

 

 

 
 

44

 

 

54

 

65 – 68

28

   

 

 

 

 

 

 

68 – 71

16

 

 

 

 

 

 

26

15

31

 

 

 

71 – 74

10

   

 

 

 

 

 

 

74 – 77

5

 

 

 

 

 

 

               

 

  

                                                                                              Analysis Table

 

Column

50 – 53

53 – 56

56 – 59

59 – 62

62 – 65

65 – 68

68 – 71

71 – 74

74 – 77

I

 

 

 

 

 

 

 

 

II

 

 

 

 

 

 

 

III

 

 

 

 

 

 

 

IV

 

 

 

 

 

 

V

 

 

 

 

 

 

VI

 

 

 

 

 

 

Total

1

3

6

3

1

 

Modal class = 62 – 65

 

Access Other Chapters of NCERT Solutions For Class 11 Statistics

You can download the PDF of NCERT Solutions For Class 11 Statistics Other Chapters:

We have provided all the important above in the article regarding the CBSE NCERT Solutions For Class 11 Statistics Chapter 5. If you have any queries, you can mention them in the comment section.

FAQ on NCERT Solutions For Class 11 Statistics Chapter 5

Find the mean, median, and mode for the following list of values: 13, 18, 13, 14, 13, 16, 14, 21, 13.

The mean is the usual average is, 
(13 + 18 + 13 + 14 + 13 + 16 + 14 + 21 + 13) ÷ 9 = 15
The median is the middle value, so first, we have to rewrite the list in numerical order:
13, 13, 13, 13, 14, 14, 16, 18, 21
There are nine numbers in the list, so the size is an odd number. 
13, 13, 13, 13, 14, 14, 16, 18, 21
Hence, the median is 14.
A mode is a number that is repeated more times.
Hence, 13 is the mode.

What is the difference between Mean and Median?

Mean is the average of all the values given in the set. On the contrary, the median is the exact middle number when a list of values will be written in order. So both are completely different and they have unique advantages in the field of Statistics.

What are the purposes of average in the statistical method?

The purposes of average are the statistical method are:
Brief description
Comparison
Formulation of policies
Statistical analysis
One value of all

What are the different kinds of statistical averages?

The different kinds of statistical averages are:
Mathematical average
Positional average

What are the two methods that can calculate the simple arithmetic mean in case 
of an individual series.

The two methods that can calculate the simple arithmetic mean in the case of an individual series are:
 Direct method
 Shortcut method

What are the methods of calculating the simple arithmetic mean?

The methods of calculating the simple arithmetic mean are:
 Individual series
 Discrete series
 Frequency distribution
 

Leave a Comment

Government Jobs for Architects in India 2024 Top 7 Arts and Science Colleges In Coimbatore 2024 Private Agriculture Colleges in India How to Score Above 150 in KCET 2024? Top Summer Research Programs in India 2024