NCERT solutions for class 10 maths Chapter 13 Statistics

NCERT solutions for class 10 maths Chapter 13 Statistics

Updated on Apr 30, 2025 09:51 AM IST | #CBSE Class 10th

This article is focused on the measurement of central tendencies like median, mode, and mean to analyze the grouped and ungrouped data. For students preparing for exams, the 10th Mathematics Statistics NCERT Answers is a comprehensive guide to solving exercise problems and understanding their applications. The chapter also includes different types of graphs such as pie charts, histograms, line graphs, bar graphs, etc, to visualize data. Students should practice Chapter 13 thoroughly and clear all their doubts to strengthen their command of this chapter.

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  1. NCERT Solutions for Class 10 Maths Chapter 13 Statistics PDF Free Download
  2. NCERT Solutions For Statistics Class 10 Chapter 13 - Important Formulae
  3. NCERT Solutions for Class 10 Maths Chapter 13 Statistics (Exercises)
  4. Statistics Class 10 Solutions Exercise-wise
  5. NCERT Solutions for Class 10 Maths: Chapter Wise
  6. NCERT Books and NCERT Syllabus
  7. NCERT solutions of class 10 subject-wise
  8. NCERT Exemplar solutions: Subject-wise
NCERT solutions for class 10 maths Chapter 13 Statistics
NCERT solutions for class 10 maths Chapter 13 Statistics

The NCERT Solutions of class 10 Maths chapter 13 statistics are created by our subject matter expert keeping in mind the latest syllabus and pattern of CBSE. Students preparing for the Class 10 board exams must go through these solutions. The solutions cover each question of the exercise comprehensively and provide step-by-step solutions. Also, NCERT Chapter 13 Maths Class 10 Notes can be used for deeper knowledge, better understanding, and practice.

NCERT Solutions for Class 10 Maths Chapter 13 Statistics PDF Free Download

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NCERT Solutions For Statistics Class 10 Chapter 13 - 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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$X = \frac{\sum (f_{i}x_i)}{\sum f_i}$

Where '$f_i$' denotes the frequency of the value '$x_i$'.

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

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$X =a+ \frac{\sum (f_{i}d_i)}{\sum f_i}$

Where '$a$' is an assumed mean and '$d_i$' is the deviation of each value '$x_i$' from the assumed mean.

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

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As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

$X =a+ [\frac{\sum (f_{i}u_i)}{\sum f_i}]\times h$

Where '$u_i$' 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 $\frac{(n+1)}2$th term in the ordered set

  • In the case of an even number of observations

Median = average of the values of two middle-terms

Median for grouped data

Median $= l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).h$

Where,

l = lower limit of the median class

c.f = Cumulative frequency of preceding class

f = Frequency of median class

h = class interval

Mode:

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

  • The formula to calculate the mode is:

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

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

Empirical formula

Mode = 3 Median - 2 Mean

NCERT Solutions for Class 10 Maths Chapter 13 Statistics (Exercises)

Class 10 Maths chapter 13 solutions Exercise: 13.1
Page number: 181-183
Total questions: 9


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.

13

Which method did you use to find 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.
13

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

$\sum f_i d_i$

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

13

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,

$\overline x = \frac{\sum f_ix_i}{\sum f_i}$
$⇒18 = \frac{752+20f}{44+f}$
$⇒18(44+f) =( 752+20f)$
$⇒ 2f = 40 $
$⇒ f = 20$
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.

13

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

$\sum f_id_i$

= 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.

13

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

$\sum f_id_i$

= 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 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

$\sum f_iu_i$

= -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:
13

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.

13

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 was 12.48 days.

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

13

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

$\sum f_iu_i$

= -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 literacy rate is 69.43%.

Class 10 Maths chapter 7 solutions Exercise: 13.2
Page number: 186-187
Total questions: 6

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

13

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

Answer:

The class with the 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 :

13

Determine the modal lifetimes of the components.

Answer:

The class with the 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.

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

$\\ = 60 + \left(\frac{61-52}{2(61)-52-38} \right)\times20 \\ \\ = 60 + \frac{9}{32}\times20$

$= 65.62$

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 :

13

Answer:

The class with the 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

$\sum f_iu_i$

= -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.

13

Answer:

The class with the 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, the 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

$\sum f_iu_i$

= -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.

13

Find the mode of the data.

Answer:

The class with the 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, the 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 :

13

Answer:

The class with the 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, the Mode of the data is 44.70.

Class 10 Maths chapter 7 solutions Exercise: 13.3

Page number: 198-200
Total questions: 7

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.

13

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

$\sum f_iu_i$

= 7

MEDIAN:
$N= 68$
$⇒\frac{N}{2} = 34$
$\therefore$ 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 with the 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, the 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.

13

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


$N= 60$
$⇒\frac{N}{2} = 30$

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

Therefore, Median class = 20-30
The 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 aged 18 years onwards but less than 60 years.

13

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


$N= 100$
$⇒\frac{N}{2} = 50$
Therefore, Median class = 35-45
The frequency corresponding to the 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 :

13

Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes to find 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 to find 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


$ N= 40$
$⇒\frac{N}{2} = 20$
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 :

13

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


$N= 400$
$⇒\frac{N}{2} = 200$
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.

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:
13

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 $
$⇒\frac{N}{2} = 50$
$\therefore$ Median class = 7-10; Lower limit, l = 7;

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

$=7+\left(\frac{50-36}{40}\right) \cdot 3$

$= 8.05$

Thus, the median of the data is 8.05

MODE:

The class with the 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.

13

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 $
$⇒\frac{N}{2} = 15$
$\therefore$ 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 students is 56.67 kg.

Statistics Class 10 Solutions Exercise-wise

To ease the learning for students, the below links of the exercises can also be used.

NCERT Solutions for Class 10 Maths: Chapter Wise

NCERT Books and NCERT Syllabus

Here is the latest NCERT syllabus, which is very useful for students before strategizing their study plans.
Also, links to some reference books which are important for further studies.

NCERT solutions of class 10 subject-wise

You can find NCERT Solutions for Maths as well as Science through the given links.

NCERT Exemplar solutions: Subject-wise

You can find NCERT Exemplar Solutions for Maths as well as Science through the given links.

Frequently Asked Questions (FAQs)

1. What are the important topics covered in NCERT Class 10 Maths Chapter 13?

The following topics are covered in NCERT Class 10 Maths Chapter 13:
1. Mean for grouped and ungrouped data.
2. Direct method to find the mean.
3. Method of assumed mean to find the mean.
4. Step deviation method to find the mean.
5. Median for grouped and ungrouped data.
6. Mode for grouped and ungrouped data. 

2. How do you find the mean, median, and mode in Class 10 Statistics?

In Class 10 Statistics, the mean is the average of all values, the median is the middle value when the data is arranged in order, and the mode is the value that appears most frequently.

3. What is the formula for mean deviation in Chapter 13 Statistics?

The formula for mean deviation in Chapter 13 Statistics, depending on whether you're dealing with ungrouped or grouped data, is:
Ungrouped Data:
MD = 1n×∑|xi−x¯|;
Grouped Data:
MD = 1n×∑fi|xi−x¯| 

4. How can I download NCERT Solutions for Class 10 Maths Chapter 13 in PDF?

To download the solutions, click on the following link: https://cache.careers360.mobi/media/uploads/froala_editor/files/ncert-solutions-for-class-10-maths-chapter-14-statistics.pdf

5. What are the types of graphical representations in Statistics Class 10?

In Class 10 Statistics, you'll learn about graphical representations like bar graphs, line graphs, histograms, pie charts, and frequency polygons, each used to visualize different types of data.

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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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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