NCERT Solutions for Class 10 Maths Chapter 14 Statistics

 

NCERT solutions for class 10 maths chapter 14 Statistics - Today's world is highly dependent on data. Each and every sector has a group of data that represents various information. In the previous classes, you have studied the classification of the given data into grouped and ungrouped frequency distributions and have learnt to represent given data in the form of graphs such as histograms, bar graphs, etc and mean mode and median. Solutions of NCERT class 10 maths chapter 14 Statistics consists of the solutions for the questions based on the study of measures of central tendency, namely, mean, mode and median from the ungrouped data to that of grouped data. CBSE NCERT solutions for class 10 maths chapter 14 Statistics, consists detailed explanation of all 4 exercises comprising 25 questions. Statistics is the branch of applied mathematics that deals with the collection of data and represent it in a meaningful way. In NCERT class 10 maths chapter 14 Statistics you will also get the solutions of cumulative frequency, its distribution and how to draw cumulative frequency curves questions. Students face many real-life problems where the fundamentals of statistics are used to represent data in graphs or in tabular form. For classwise and subjectwise NCERT solutions, you can click on the given link.

Types of questions asked from class 10 maths chapter 14 Statistics

  • A direct method of finding mean
  • Assumed mean method of finding mean
  • Step deviation method of finding mean
  • Question-related to median
  • Questions based on the mode

NCERT solutions for class 10 maths chapter 14 Statistics Excercise: 14.1

Q1 A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

1

Which method did you use for finding the mean, and why?

Answer:

Number of plants

 

Number of houses    

f_i

Classmark

x_i

f_ix_i

0-2

1

1

1

2-4

2

3

6

4-6

1

5

5

6-8

5

7

35

8-10

6

9

54

10-12

2

11

22

12-14

3

13

39

 

\sum f_i 

=20

 

\sum f_ix_i

=162

Mean, 

\overline x = \frac{\sum f_ix_i}{\sum f_i}
= \frac{162}{20} = 8.1
We used the direct method in this as the values of x_i\ and\ f_i are small. 

Q2 Consider the following distribution of daily wages of 50 workers of a factory.
2
Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer:

Let the assumed mean be a = 550

Daily 

Wages   

Number of

workers f_i

Classmark

x_i

d_i = x_i -a

f_id_i

500-520

12

510

-40

-480

520-540

14

530

-20

-280

540-560

8

550

0

0

560-580

6

570

20

120

580-600

10

590

40

400

 

\sum f_i 

=50

 

 

\sum f_ix_i

=-240

Mean, 

\overline x =a + \frac{\sum f_id_i}{\sum f_i}
= 550 + \frac{-240}{50} = 550-4.8 = 545.20
Therefore, the mean daily wages of the workers of the factory is Rs. 545.20

Statistics Excercise: 14.1

Q3 following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

 3

Answer:

Daily pocket

allowance

Number of

children f_i

Classmark

x_i

f_ix_i

11-13

7

12

84

13-15

6

14

84

15-17

9

16

144

17-19

13

18

234

19-21

f

20

20f

21-23

5

22

110

23-25

4

24

96

 

\sum f_i 

=44+f

 

\sum f_ix_i

=752+20f

Mean, 

\overline x = \frac{\sum f_ix_i}{\sum f_i}
\implies 18 = \frac{752+20f}{44+f}

\\ \implies 18(44+f) =( 752+20f) \\ \implies 2f = 40 \\ \implies f = 20
Therefore the missing f = 20

Q4 Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

          

Answer:

Let the assumed mean be a = 75.5

 

No. of heartbeats

per minute   

Number of

women f_i

Classmark

x_i

d_i = x_i -a

f_id_i

65-68

2

66.5

-9

-18

68-71

4

69.5

-6

-24

71-74

3

72.5

-3

-9

74-77

8

75.5

0

0

77-80

7

78.5

3

21

80-83

4

81.5

6

24

83-86

2

84.5

9

18

 

\sum f_i

=30 

 

 

\sum f_ix_i

=12

Mean, 

\overline x =a + \frac{\sum f_id_i}{\sum f_i}
= 75.5 + \frac{12}{30} = 75.5 + 0.4 = 75.9
Therefore, the mean heartbeats per minute of these women are 75.9

Q5 In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers of mangoes. The following was the distribution of mangoes according to the number of boxes.

 5

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer:

Let the assumed mean be a = 57

Number of 

mangoes

Number of

boxes f_i

Classmark

x_i

d_i = x_i -a

f_id_i

50-52

15

51

-6

-90

53-55

110

54

-3

-330

56-58

135

57

0

0

59-61

115

60

3

345

62-64

25

63

6

150

 

\sum f_i 

=400

 

 

\sum f_ix_i

=75

Mean, 

\overline x =a + \frac{\sum f_id_i}{\sum f_i}
= 57+ \frac{75}{400} = 57+0.1875 = 57.1875 \approx 57.19
Therefore, the mean number of mangoes kept in a packing box is approx 57.19

Q6 The table below shows the daily expenditure on the food of 25 households in a locality.
6
Find the mean daily expenditure on food by a suitable method.

Answer:

Let the assumed mean be a = 225 and h = 50

Daily

Expenditure  

Number of

households f_i

Classmark

x_i

d_i = x_i -a

u_i = \frac{d_i}{h}

f_iu_i

100-150

4

125

-100

-2

-8

150-200

5

175

-50

-1

-5

200-250

12

225

0

0

0

250-300

2

275

50

1

2

300-350

2

325

100

2

4

 

\sum f_i

=25

 

 

 

\sum f_ix_i

= -7

Mean, 

\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h 
= 225 + \frac{-7}{25}\times50 = 225 -14 = 211
Therefore, the mean daily expenditure on food is Rs. 211

Q7 To find out the concentration of SO_2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
7

Find the mean concentration of SO_2 in the air.

Answer:

Class 

Interval

Frequency 

f_i

Classmark

x_i

f_ix_i

0.00-0.04

4

0.02

0.08

0.04-0.08

9

0.06

0.54

0.08-0.12

9

0.10

0.90

0.12-0.16

2

0.14

0.28

0.16-0.20

4

0.18

0.72

0.20-0.24

2

0.22

0.44

 

\sum f_i 

=30

 

\sum f_ix_i

=2.96

Mean, 

\overline x = \frac{\sum f_ix_i}{\sum f_i}
= \frac{2.96}{30} = 0.099
Therefore, the mean concentration of SO_2 in the air is 0.099 ppm

Q8 A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

     

Answer:

Number of

days

Number of

Students f_i

Classmark

x_i

f_ix_i

0-6

11

3

33

6-10

10

8

80

10-14

7

12

84

14-20

4

17

68

20-28

4

24

96

28-38

3

33

99

38-40

1

39

39

 

 

 

 

 

\sum f_i 

=40

 

\sum f_ix_i

=499

Mean, 

\overline x = \frac{\sum f_ix_i}{\sum f_i}
= \frac{499}{40} = 12.475= \frac{499}{40} = 12.475\approx 12.48
Therefore, the mean number of days a student was absent is 12.48 days.

Q9 The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

       9

Answer:

Let the assumed mean be a = 75 and h = 10

Literacy

rates  

Number of

cities f_i

Classmark

x_i

d_i = x_i -a

u_i = \frac{d_i}{h}

f_iu_i

45-55

3

50

-20

-2

-6

55-65

10

60

-10

-1

-10

65-75

11

70

0

0

0

75-85

8

80

10

1

8

85-95

3

90

20

2

6

 

\sum f_i

= 35

 

 

 

\sum f_ix_i

= -2

Mean, 

\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h 
= 70 + \frac{-2}{35}\times10 = 70 -0.57 = 69.43
Therefore, the mean mean literacy rate is 69.43%

Statistics Excercise: 14.2

Q1 The following table shows the ages of the patients admitted in a hospital during a year:

       2.1

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Answer:

The class having maximum frequency is the modal class.

The maximum frequency is 23 and hence the modal class = 35-45

Lower limit (l) of modal class = 35, class size (h) = 10 

Frequency ( f_1 ) of the modal class = 23, frequency ( f_0 ) of class preceding the modal class = 21, frequency ( f_2 ) of class succeeding the modal class = 14.

Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h

\\ = 35 + \left(\frac{23-21}{2(23)-21-14} \right).10 \\ \\ = 35 + \frac{2}{11}.10

= 36.8

Now,

Age

 

Number of

patients f_i

Classmark

x_i

f_ix_i

5-15

6

10

60

15-25

11

20

220

25-35

21

30

630

35-45

23

40

920

45-55

14

50

700

55-65

5

60

300

 

\sum f_i 

=80

 

\sum f_ix_i

=2830

Mean, 

\overline x = \frac{\sum f_ix_i}{\sum f_i}
= \frac{2830}{80} = 35.37
The maximum number of patients are in the age group of 36.8, whereas the average age of all the patients is 35.37.

Q2 The following data gives information on the observed lifetimes (in hours) of 225 electrical components :

2.2

Determine the modal lifetimes of the components.

Answer:

The class having maximum frequency is the modal class.

The maximum frequency is 61 and hence the modal class = 60-80

Lower limit (l) of modal class = 60, class size (h) = 20

Frequency ( f_1 ) of the modal class = 61 frequency ( f_0 ) of class preceding the modal class = 52, frequency ( f_2 ) of class succeeding the modal class = 38.

Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h

\\ = 60 + \left(\frac{61-52}{2(61)-52-38} \right).20 \\ \\ = 60 + \frac{9}{32}.20

= 65.62

Thus, the modal lifetime of 225 electrical components is 65.62 hours

Q3 The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

       2.3

Answer:

The class having maximum frequency is the modal class.

The maximum frequency is 40 and hence the modal class = 1500-2000

Lower limit (l) of modal class = 1500, class size (h) = 500

Frequency ( f_1 ) of the modal class = 40 frequency ( f_0 ) of class preceding the modal class = 24, frequency ( f_2 ) of class succeeding the modal class = 33.

Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h

\\ = 1500 + \left(\frac{40-24}{2(40)-24-33} \right).500 \\ \\ = 1500 + \frac{16}{23}.500

= 1847.82

Thus, the Mode of the data is Rs. 1847.82

Now,

Let the assumed mean be a = 2750 and h = 500

Expenditure

Number of

families f_i

Classmark

x_i

d_i = x_i -a

u_i = \frac{d_i}{h}

f_iu_i

1000-1500

24

1250

-1500

-3

-72

1500-2000

40

1750

-1000

-2

-80

2000-2500

33

2250

-500

-1

-33

2500-3000

28

2750

0

0

0

3000-3500

30

3250

500

1

30

3500-4000

22

3750

1000

2

44

4000-4500

16

4250

1500

3

48

4500-5000

7

4750

2000

4

28

 

\sum f_i

=200

 

 

 

\sum f_ix_i

= -35

Mean, 

\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h 
= 2750 + \frac{-35}{200}\times500 = 2750 -87.5 = 2662.50

Thus, the Mean monthly expenditure is Rs. 2662.50 

Q4 The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

   2.4

Answer:

The class having maximum frequency is the modal class.

The maximum frequency is 10 and hence the modal class = 30-35

Lower limit (l) of modal class = 30, class size (h) = 5

Frequency ( f_1 ) of the modal class = 10 frequency ( f_0 ) of class preceding the modal class = 9, frequency ( f_2 ) of class succeeding the modal class = 3

Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h

\\ = 30 + \left(\frac{10-9}{2(10)-9-3} \right).5 \\ \\ = 30 + \frac{1}{8}.5

= 30.625

Thus, Mode of the data is 30.625

Now,

Let the assumed mean be a = 32.5 and h = 5

Class

Number of

states f_i

Classmark

x_i

d_i = x_i -a

u_i = \frac{d_i}{h}

f_iu_i

15-20

3

17.5

-15

-3

-9

20-25

8

22.5

-10

-2

-16

25-30

9

27.5

-5

-1

-9

30-35

10

32.5

0

0

0

35-40

3

37.5

5

1

3

40-45

0

42.5

10

2

0

45-50

0

47.5

15

3

0

50-55

2

52.5

20

4

8

 

\sum f_i

=35

 

 

 

\sum f_ix_i

= -23

Mean, 

\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h 
= 32.5 + \frac{-23}{35}\times5= 29.22

Thus, the Mean of the data is 29.22

Q5 The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

2.5

Find the mode of the data.

Answer:

The class having maximum frequency is the modal class.

The maximum frequency is 18 and hence the modal class = 4000-5000

Lower limit (l) of modal class = 4000, class size (h) = 1000

Frequency ( f_1 ) of the modal class = 18 frequency ( f_0 ) of class preceding the modal class = 4, frequency ( f_2 ) of class succeeding the modal class = 9

Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h

\\ =4000 + \left(\frac{18-4}{2(18)-4-9} \right).1000 \\ \\ = 4000 + \frac{14}{23}.1000

= 4608.70

Thus, Mode of the data is 4608.70

Q6 A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

      2.6

Answer:

The class having maximum frequency is the modal class.

The maximum frequency is 20 and hence the modal class = 40-50

Lower limit (l) of modal class = 40, class size (h) = 10

Frequency ( f_1 ) of the modal class = 20 frequency ( f_0 ) of class preceding the modal class = 12, frequency ( f_2 ) of class succeeding the modal class = 11

Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h

\\ =40+ \left(\frac{20-12}{2(20)-12-11} \right).10 \\ \\ = 40 + \frac{8}{17}.10

= 44.70

Thus, Mode of the data is 44.70


6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :
Edit Q

 

CBSE NCERT solutions for class 10 maths chapter 14 Statistics Excercise: 14.3

Q1 The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

        3.1

Answer:

Let the assumed mean be a = 130 and h = 20

Class

Number of

consumers f_i

Cumulative

Frequency

Classmark

x_i

d_i = x_i -a

u_i = \frac{d_i}{h}

f_iu_i

65-85

4

4

70

-60

-3

-12

85-105

5

9

90

-40

-2

-10

105-125

13

22

110

-20

-1

-13

125-145

20

42

130

0

0

0

145-165

14

56

150

20

1

14

165-185

8

64

170

40

2

16

185-205

4

68

190

60

3

12

 

 

\sum f_i = N

= 68

 

 

 

\sum f_ix_i

= 7

MEDIAN:
N= 68 \implies \frac{N}{2} = 34
\therefore Median class = 125-145; Cumulative Frequency = 42; Lower limit, l = 125; 

c.f. = 22; f = 20; h = 20
Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 125 + \left (\frac{34-22}{20} \right ).20 \\ \\ = 125 + 12

= 137

Thus, the median of the data is 137

MODE:

The class having maximum frequency is the modal class.
The maximum frequency is 20 and hence the modal class = 125-145
Lower limit (l) of modal class = 125, class size (h) = 20
Frequency (f_1) of the modal class = 20; frequency (f_0) of class preceding the modal class = 13, frequency (f_2) of class succeeding the modal class = 14.

Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h

\\ = 125 + \left(\frac{20-13}{2(20)-13-14} \right).20 \\ \\ = 125 + \frac{7}{13}.20

= 135.76

Thus, Mode of the data is 135.76

MEAN:

Mean, 
\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h 
= 130 + \frac{7}{68}\times20 = 137.05

Thus, the Mean of the data is 137.05

Q2 If the median of the distribution given below is 28.5, find the values of x and y.

     3.2

Answer:

Class

Number of

consumers f_i

Cumulative

Frequency

0-10

5

5

10-20

x

5+x

20-30

20

25+x

30-40

15

40+x

40-50

y

40+x+y

50-60

5

45+x+y

 

\sum f_i = N

= 60

 


N= 60 \implies \frac{N}{2} = 30

Now,
Given median = 28.5 which lies in the class 20-30

Therefore, Median class = 20-30
Frequency corresponding to median class, f = 20
Cumulative frequency of the class preceding the median class, c.f. = 5 + x
Lower limit, l = 20; Class height, h = 10

Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ \implies28.5= 20 + \left (\frac{30-5-x}{20} \right ).10 \\ \\ \implies8.5=\frac{25-x}{2} \\ \implies 25-x = 8.5(2) \\ \implies x = 25 - 17 = 8

Also, 

\\ 60 = 45 + x+y \\ \implies x+y = 60-45 = 15 \\ \implies y = 15-x = 15-8 \ \ \ (\because x =8) \\ \implies y = 7

Therefore, the required values are: x=8 and y=7

Q3 A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

3.3

Answer:

Class

Frequency

 f_i

Cumulative

Frequency

15-20

2

2

20-25

4

6

25-30

18

24

30-35

21

45

35-40

33

78

40-45

11

89

45-50

3

92

50-55

6

98

55-60

2

100


N= 100 \implies \frac{N}{2} = 50
Therefore, Median class = 35-45
Frequency corresponding to median class, f = 21
Cumulative frequency of the class preceding the median class, c.f. = 24
Lower limit, l = 35; Class height, h = 10

Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 35 + \left (\frac{50-45}{33} \right ).5 \\ \\

= 35.75

Thus, the median age is 35.75 years.

Q4 The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table :

 3.4

Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to
117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

Answer:

The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes.

Class

Frequency

 f_i

Cumulative

Frequency

117.5-126.5

3

3

126.5-135.5

5

8

135.5-144.5

9

17

144.5-153.5

12

29

153.5-162.5

5

34

162.5-171.5

4

38

171.5-180.5

2

40


\dpi{100} N= 40 \implies \frac{N}{2} = 20
Therefore, Median class = 144.5-153.5

Lower limit, l = 144.5; Class height, h = 9
Frequency corresponding to median class, f = 12
Cumulative frequency of the class preceding the median class, c.f. = 17

Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 144.5 + \left (\frac{20-17}{12} \right ).9 \\ \\

= 146.75

Thus, the median length of the leaves is 146.75 mm

Q5 The following table gives the distribution of the lifetime of 400 neon lamps :
3.5
Find the median lifetime of a lamp.

Answer:

Class

Frequency

 f_i

Cumulative

Frequency

1500-2000

14

14

2000-2500

56

70

2500-3000

60

130

3000-3500

86

216

3500-4000

74

290

4000-4500

62

352

4500-5000

48

400

\dpi{100} N= 400 \implies \frac{N}{2} = 200
Therefore, Median class = 3000-3500

Lower limit, l = 3000; Class height, h = 500
Frequency corresponding to median class, f = 86
Cumulative frequency of the class preceding the median class, c.f. = 130

Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 3000 + \left (\frac{200-130}{86} \right ).500 \\ \\ = 3000+406.97

= 3406.97

Thus, the median lifetime of a lamp is 3406.97 hours

= 146.75

Thus, the median length of the leaves is 146.75 mm

Q6 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

3.6

Determine the median number of letters in the surnames. Find the mean number of
letters in the surnames? Also, find the modal size of the surnames.

Answer:

Class

Number of

surnames f_i

Cumulative

Frequency

Classmark

x_i

f_ix_i

1-4

6

6

2.5

15

4-7

30

36

5.5

165

7-10

40

76

8.5

340

10-13

16

92

11.5

184

13-16

4

96

14.5

51

16-19

4

100

17.5

70

 

 

\sum f_i = N

= 100

 

\sum f_ix_i

= 825

MEDIAN:
N= 100 \implies \frac{N}{2} = 50
\therefore Median class = 7-10; Lower limit, l = 7; 

Cumulative frequency of preceding class, c.f. = 36; f = 40; h = 3
Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 7+ \left (\frac{50-36}{40} \right ).3 \\ \\

= 8.05

Thus, the median of the data is 8.05

MODE:

The class having maximum frequency is the modal class.
The maximum frequency is 40 and hence the modal class = 7-10
Lower limit (l) of modal class = 7, class size (h) = 3
Frequency (f_1) of the modal class = 40; frequency (f_0) of class preceding the modal class = 30, frequency (f_2) of class succeeding the modal class = 16

Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h

\\ = 7 + \left(\frac{40-30}{2(40)-30-16} \right).3 \\ \\ = 125 + \frac{10}{34}.3

= 7.88

Thus, Mode of the data is 7.88

MEAN:

Mean, 
\overline x =\frac{\sum f_ix_i}{\sum f_i} 
= \frac{825}{100} = 8.25

Thus, the Mean of the data is 8.25

Q7 The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

   3.7

Answer:

Class

Number of

students f_i

Cumulative

Frequency

40-45

2

2

45-50

3

5

50-55

8

13

55-60

6

19

60-65

6

25

65-70

3

28

70-75

2

30

MEDIAN:
N= 30 \implies \frac{N}{2} = 15
\therefore Median class = 55-60; Lower limit, l = 55; 

Cumulative frequency of preceding class, c.f. = 13; f = 6; h = 5
Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 55+ \left (\frac{15-13}{6} \right ).5 \\ \\ = 55+\frac{2}{6}.5

= 56.67

Thus, the median weight of the student is 56.67 kg

NCERT solutions for class 10 maths chapter 14 Statistics Excercise: 14.4

Q1 The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Answer:

Daily Income

(Upper-Class Limit)

Cumulative

Frequency

Less than 120

12

Less than 140

26

Less than 160

34

Less than 180

40

Less than 200

50

Now,

Taking upper-class interval on the x-axis and their respective frequencies on the y-axis,

14.4 Ogive 1

Q2 During the medical check-up of 35 students of a class, their weights were recorded as follows:
4.2
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Answer:

Taking upper-class interval on the x-axis and their respective frequencies on the y-axis,

14.4 Ogive 2

N= 35 \implies \frac{N}{2} = 17.5

Marking a point on the curve whose ordinate is 17.5 gives an x-ordinate= 46.5.

Hence, the Median of the data is 46.5

Now,

Weight

(Class)

Frequency

Cumulative

Frequency

>38

0

0

38-40

3

3

40-42

2

5

42-44

4

9

44-46

5

14

46-48

14

28

48-50

4

32

50-52

3

35

N= 35 \implies \frac{N}{2} = 17.5
\therefore Median class = 46-48; Lower limit, l = 46; 

Cumulative frequency of preceding class, c.f. = 14; f = 14; h = 2
Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 46+ \left (\frac{17.5-14}{14} \right ).2 \\ \\ = 46+\frac{7}{14}

= 46.5

Thus, the median using formula is 46.5 which verifies the result.

Q3 The following table gives production yield per hectare of wheat of 100 farms of a village.
4.3
Change the distribution to a more than type distribution, and draw its ogive.

Answer:

Production yield

(Upper-Class Limit)

Cumulative

Frequency

More than or equal to 50

100

More than or equal to 55

98

More than or equal to 60

90

More than or equal to 65

78

More than or equal to 70

54

More than or equal to 75

16

Now,

Taking lower class limit on the x-axis and their respective frequencies on the y-axis,

14.4 Ogive 3

 

NCERT solutions for class 10 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers

Chapter 2

NCERT solutions for class 10 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables

Chapter 4

CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations

Chapter 5

NCERT solutions for class 10 chapter 5 Arithmetic Progressions

Chapter 6

Solutions of NCERT class 10 maths chapter 6 Triangles

Chapter 7

CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry

Chapter 8

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry

Chapter 9

Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry

Chapter 10

CBSE NCERT solutions class 10 maths chapter 10 Circles

Chapter 11

NCERT solutions  for class 10 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles

Chapter 13

CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes

Chapter 14

NCERT solutions for class 10 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 10 maths chapter 15 Probability

 

Solutions of NCERT class 10 - subject wise

NCERT solutions for class 10 maths

Solutions of NCERT for class 10 science

 

How to use NCERT solutions for class 10 maths chapter 14 Statistics?

  • First of all, go through NCERT Textbook to have an understanding of data.

  • Learn the techniques to calculate the mean, median, mode.

  • Once you learn the techniques and formulae, then move to the practice exercises available in the chapter.

  • While practicing, if you get stuck anywhere then you can take the help of NCERT solutions for class 10 maths chapter 14 Statistics.

Keep working hard & happy learning!

 

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