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NCERT solutions for class 10 maths Chapter 14 Statistics

NCERT solutions for class 10 maths Chapter 14 Statistics

Edited By Ramraj Saini | Updated on Sep 09, 2023 12:11 PM IST | #CBSE Class 10th

NCERT Solutions For Statistics Class 10 Chapter 14

NCERT Solutions for Class 10 Maths Chapter 14 Statistics are provided here. These NCERT Solutions are created by expert team at Careers360 keeping in mind latest syllabus and pattern of CBSE 2023-24. Students preparing for the Class 10 board exams must go through the Statistics Class 10 Maths chapter 14 NCERT solutions. This is one of the most important chapters in NCERT books for Class 10 Maths. The NCERT solutions for Class 10 Maths Chapter 14 Statistics covers each and every question of the exercise comprehensively and provide step by step solutions.

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NCERT Solutions For Statistics Class 10 Chapter 14 - Important Formulae

Measures of Central Tendency - Mean, Median, and Mode:

>> Mean:

  • Direct Method: The mean, represented as X, can be calculated directly using the formula:
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X = ∑(fixi) / ∑fi

Where 'fi' denotes the frequency of the value 'xi'.

  • Assumed Mean Method: Alternatively, the mean can be calculated using the Assumed Mean Method:

X = a + ∑(fidi) / ∑fi

Where 'a' is an assumed mean and 'di' is the deviation of each value 'xi' from the assumed mean.

  • Step Deviation Method: Another approach is the Step Deviation Method:

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X = a + [∑(fiui) / ∑fi]h

Where 'ui' represents the step deviations and 'h' is the class interval.

>> Median:

  • The median is the central value in a set of observations, and its calculation depends on the number of observations.

  • For an odd number of observations,

Median = [value of the (n+1)/2]th term in the ordered set

  • In the case of an even number of observations

Median = average of the middle two values

>> Mode:

  • The mode represents the value that appears most frequently in a dataset.

  • The formula to calculate mode is:

Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] h

Where,

l = lower limit of the modal class

f1 = frequency of the modal class

f0 = frequency of the class before the modal class

f2 = frequency of the class after the modal class

H = Class interval

Free download NCERT Solutions for Class 10 Maths Chapter 14 Statistics PDF for CBSE Exam.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics (Intext Questions and Exercise)

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. Find the mean daily wages of the workers of the factory by using an appropriate method.
2

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

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.

1636093324773

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. Find the mean daily expenditure on food by a suitable method.

Daily expenditure in rupees

100-150150-200200-250250-300300-350
Number of households451222


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.

1636093576677

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%

NCERT solutions for class 10 maths chapter 14 Statistics Excercise: 14.2

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

1640759472805

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 :

1640759523694

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 :

1640759588002

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.

1640759627719

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.

1640759659792

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 :

1640759687964

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


Statistics Class 10 Maths Chapter 14 Solution 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.

1640759732201

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.

1640759760891

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.

1640759806014

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 :

1640759831563

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 :

1640759970074 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:

1640759938199

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.

1640760021634

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


1

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

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,

1640760151886

Q2

1640760212254

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,

1640760282843

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. 1640760343873 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,

1640760369574

Statistics Class 10 Maths - Important Topics

  • A direct method of finding mean

  • Assumed mean method of finding mean

  • Step deviation method of finding mean

  • Median

  • Mode

NCERT Solutions for Class 10 subject wise

NCERT solutions for Class 10 maths - Chapter Wise


Key Features of NCERT Statistics Class 10 Maths Solutions Chapter 14

Valuable Student Resources: These materials offer invaluable guidance to students studying statistics in Class 10.

Enhanced Academic Performance: By utilizing class 10 math chapter 14 resources, students can significantly improve their performance on questions related to chapter 14 class 10 maths, ensuring they achieve high marks.

Versatile Study Aid: Whether for exam preparation or chapter review, this resource proves to be a versatile tool for students.

Expertly Crafted Solutions: Professional subject specialists have meticulously answered the statistics questions, providing accurate and clear explanations.

CBSE Aligned: The content within these materials is in complete alignment with the CBSE Syllabus (2022-23) and follows the guidelines set forth by the board.

Also, check - NCERT Solutions for Class 10

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

  • First of all, go through NCERT Textbook to have an understanding of Statistics Class 10 chapter.

  • 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 Statistics Class 10 chapter.

  • While practising, if you get stuck anywhere then you can take the help of NCERT Solutions for chapter 14 class 10 maths PDF Download, which cab be used offline.

NCERT Exemplar solutions - Subject Wise

NCERT Books and NCERT Syllabus

Frequently Asked Question (FAQs)

1. What is the weightage of the chapter statistics for CBSE board exam?

Two chapters - statistics and probability combined have 11 marks weightage in the CBSE class 10 board exam. Refer to NCERT exemplar for more problems based on Class 10 Mathematics NCERT syllabus. Also practice the CBSE Class 10 Mathematics previous year papers for good score in the exam. Class 10 Maths chapter 14 examples are discussed above in this article, go and check to become familiar with concepts. 

2. Will the NCERT Solutions for Class 10 Maths Chapter 14 aid in passing the CBSE exams?

Yes, Statistics class 10 solutions is a crucial chapter for Class 10 Maths and includes various math tricks and shortcuts for efficient calculations. Statistics math can aid in understanding and passing the math subject in the CBSE exams. Students should practice problems from the statistics class 10 pdf to do well in the exam. 

3. How do the Chapter 14 statistics class 10 ncert solutions assist in understanding the areas and volumes of geometrical shapes?

The NCERT Solutions for class 10 maths statistics are designed in an interactive format to make it easy for students to grasp the concepts. The PDF of solutions can be used as a primary study material to improve problem-solving skills and accuracy. Interested students can use NCERT solutions class 10 maths chapter 14 pdf download from the Careers360 website and study both offline and online mode.

4. How many questions are present in NCERT Solutions for Class 10 Maths Chapter 14?

The chapter 14 statistics 10th class has a total of 25 questions across 4 exercises. The statistics class 10 exercise 14.1 has 9 questions, the second has 6, the third has 7, and the fourth has 3 questions. Practice the class 10 maths chapter 14 solutions pdf download to score well in the exam. 

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

  • Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

  • Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.


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Dear aspirant !

Hope you are doing well ! The class 10 Hindi mp board sample paper can be found on the given link below . Set your time and give your best in the sample paper it will lead to great results ,many students are already doing so .

https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://school.careers360.com/download/sample-papers/mp-board-10th-hindi-model-paper&ved=2ahUKEwjO3YvJu5KEAxWAR2wGHSLpAiQQFnoECBMQAQ&usg=AOvVaw2qFFjVeuiZZJsx0b35oL1x .

Hope you get it !

Thanking you

Hello aspirant,

The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.

To get the complete datesheet, you can visit our website by clicking on the link given below.

https://school.careers360.com/boards/cbse/cbse-date-sheet

Thank you

Hope this information helps you.

Hello aspirant,

The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.

You don't need to worry. The class 7th paper will be simple and made by your own school teachers.

Thank you

Hope it helps you.

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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