You cannot copy content of this page

NCERT solution class 10 chapter 15 Statistics exercise 15.1 mathematics

EXERCISE 15.1


Page No 270:

Question 1:

A survey was conducted by a group of students as a part of their environment awareness programme, 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.

Number of plants

0­ − 2

2­ − 4

4 − 6

6 − 8

8 − 10

10 − 12

12 − 14

Number of houses

1

2

1

5

6

2

3

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

Answer:

To find the class mark (xi) for each interval, the following relation is used.

Class mark (xi) = 

xand fixi can be calculated as follows.

Number of plants

Number of houses

(fi)

xi

fixi

0­ − 2

1

1

1 × 1 = 1

2­ − 4

2

3

2 × 3 = 6

4 − 6

1

5

1 × 5 = 5

6 − 8

5

7

5 × 7 = 35

8 − 10

6

9

6 × 9 = 54

10 − 12

2

11

2 ×11 = 22

12 − 14

3

13

3 × 13 = 39

Total

20

162

From the table, it can be observed that

Mean,

Therefore, mean number of plants per house is 8.1.

Here, direct method has been used as the values of class marks (xi) and fi are small.

Video Solution

Question 2:

Consider the following distribution of daily wages of 50 worker of a factory.

Daily wages (in Rs)

100­ − 120

120­ − 140

140 −1 60

160 − 180

180 − 200

Number of workers

12

14

8

6

10

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

Answer:

To find the class mark for each interval, the following relation is used.

Class size (h) of this data = 20

Taking 150 as assured mean (a), diui, and fiui can be calculated as follows.

Daily wages

(in Rs)

Number of workers (fi)

xi

di = x− 150

fiui

100­ −120

12

110

− 40

− 2

− 24

120­ − 140

14

130

− 20

− 1

− 14

140 − 160

8

150

0

0

0

160 −180

6

170

20

1

6

180 − 200

10

190

40

2

20

Total

50

− 12

From the table, it can be observed that

Therefore, the mean daily wage of the workers of the factory is Rs 145.20.

Video Solution

Question 3:

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.

Daily pocket allowance (in Rs)

11­ − 13

13­ − 15

15 −17

17 − 19

19 − 21

21 − 23

23 − 25

Number of workers

7

6

9

13

f

5

4

Answer:

To find the class mark (xi) for each interval, the following relation is used.

Given that, mean pocket allowance, 

Taking 18 as assured mean (a), di and fidi are calculated as follows.

Daily pocket allowance

(in Rs)

Number of children

fi

Class mark xi

di = x− 18

fidi

11­ −13

7

12

− 6

− 42

13 − 15

6

14

− 4

− 24

15 − 17

9

16

− 2

− 18

17 −19

13

18

0

0

19 − 21

f

20

2

f

21 − 23

5

22

4

20

23 − 25

4

24

6

24

Total

2f − 40

From the table, we obtain

Hence, the missing frequency, f, is 20.

Video Solution


Page No 271:

Question 4:

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute

65 − 68

68­ − 71

71 −74

74 − 77

77 − 80

80 − 83

83 − 86

Number of women

2

4

3

8

7

4

2

Answer:

To find the class mark of each interval (xi), the following relation is used.

Class size, h, of this data = 3

Taking 75.5 as assumed mean (a), diuifiui are calculated as follows.

Number of heart beats per minute

Number of women

fi

xi

di = x− 75.5

fiui

65 − 68

2

66.5

− 9

− 3

− 6

68 − 71

4

69.5

− 6

− 2

− 8

71 − 74

3

72.5

− 3

− 1

− 3

74 − 77

8

75.5

0

0

0

77 − 80

7

78.5

3

1

7

80 − 83

4

81.5

6

2

8

83 − 86

2

84.5

9

3

6

Total

30

4

From the table, we obtain

Therefore, mean hear beats per minute for these women are 75.9 beats per minute.

Video Solution

Question 5:

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

Number of mangoes

50 − 52

53 − 55

56 − 58

59 − 61

62 − 64

Number of boxes

15

110

135

115

25

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

Answer:

Number of mangoes

Number of boxes fi

50 − 52

15

53 − 55

110

56 − 58

135

59 − 61

115

62 − 64

25

It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore, has to be added to the upper class limit and has to be subtracted from the lower class limit of each interval.

Class mark (xi) can be obtained by using the following relation.

Class size (h) of this data = 3

Taking 57 as assumed mean (a), diuifiui are calculated as follows.

Class interval

fi

xi

di = xi − 57

fiui

49.5 − 52.5

15

51

− 6

− 2

− 30

52.5 − 55.5

110

54

− 3

− 1

− 110

55.5 − 58.5

135

57

0

0

0

58.5 − 61.5

115

60

3

1

115

61.5 − 64.5

25

63

6

2

50

Total

400

25

It can be observed that

Mean number of mangoes kept in a packing box is 57.19.

Step deviation method is used here as the values of fi, di are big and also, there is a common multiple between all di.

Video Solution

Question 6:

The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs)

100 − 150

150 − 200

200 − 250

250 − 300

300 − 350

Number of households

4

5

12

2

2

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

Answer:

To find the class mark (xi) for each interval, the following relation is used.

Class size = 50

Taking 225 as assumed mean (a), diuifiui are calculated as follows.

Daily expenditure (in Rs)

fi

xi

di = xi − 225

fiui

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

Total

25

− 7

From the table, we obtain

Therefore, mean daily expenditure on food is Rs 211.

Video Solution

Question 7:

To find out the concentration of SO2 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:

concentration of SO2 (in ppm)

Frequency

0.00 − 0.04

4

0.04 − 0.08

9

0.08 − 0.12

9

0.12 − 0.16

2

0.16 − 0.20

4

0.20 − 0.24

2

Find the mean concentration of SO2 in the air.

Answer:

To find the class marks for each interval, the following relation is used.

Class size of this data = 0.04

Taking 0.14 as assumed mean (a), diuifiui are calculated as follows.

Concentration of SO2 (in ppm)

Frequency

fi

Class mark

xi

di = x− 0.14

fiui

0.00 − 0.04

4

0.02

− 0.12

− 3

− 12

0.04 − 0.08

9

0.06

− 0.08

− 2

− 18

0.08 − 0.12

9

0.10

− 0.04

− 1

− 9

0.12 − 0.16

2

0.14

0

0

0

0.16 − 0.20

4

0.18

0.04

1

4

0.20 − 0.24

2

0.22

0.08

2

4

Total

30

− 31

From the table, we obtain

Therefore, mean concentration of SO2 in the air is 0.099 ppm.

Video Solution

Page No 272:

Question 8:

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.

Number of days

0 − 6

6 − 10

10 − 14

14 − 20

20 − 28

28 − 38

38 − 40

Number of students

11

10

7

4

4

3

1

Answer:

To find the class mark of each interval, the following relation is used.

Taking 17 as assumed mean (a), di and fidi are calculated as follows.

Number of days

Number of students

fi

xi

di = xi − 17

fidi

0 − 6

11

3

− 14

− 154

6 − 10

10

8

− 9

− 90

10 − 14

7

12

− 5

− 35

14 − 20

4

17

0

0

20 − 28

4

24

7

28

28 − 38

3

33

16

48

38 − 40

1

39

22

22

Total

40

− 181

From the table, we obtain

Therefore, the mean number of days is 12.48 days for which a student was absent.


Question 9:

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

Literacy rate (in %)

45 − 55

55 − 65

65 − 75

75 − 85

85 − 95

Number of cities

3

10

11

8

3

Answer:

To find the class marks, the following relation is used.

Class size (h) for this data = 10

Taking 70 as assumed mean (a), diui, and fiui are calculated as follows.

Literacy rate (in %)

Number of cities

fi

xi

di = xi − 70

fiui

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

Total

35

− 2

From the table, we obtain

Therefore, mean literacy rate is 69.43%.


 

Leave a Comment

Your email address will not be published. Required fields are marked *

error: Content is protected !!
Free Web Hosting