Lesson: Statistics

Exercise 14.1 (9)

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?

Solution:

First, find the class mark for each interval. For

i

x

this, the following relation should be used.

i

Upper limit Lower limit

x

2

Now, calculate and .

i

x

i i

x f

Number of

plants

Number of

houses ( )

i

f

i

x

i i

x f

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

Total

20

162

From the table,

and

i

f 20

i i

f x 162

Now, find the value of mean.

Mean

i i

i

f x

x

f

162

20

8.1

Therefore, mean number of plants per house is 8.1.

As the values of class marks and are small,

i

x

i

f

so direct method has been used for finding the

mean.

Question: 2

Consider the following distribution of daily wages

of 50 workers of a factory.

Daily wages

(in Rs)

100 120

120 140

140 160

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.

Solution:

First, find the class mark for each interval. For

i

x

this, the following relation should be used.

i

Upper limit Lower limit

x

2

Class size

h 20

Take assumed mean and calculate and

a 150

i

d ,

i

u

as follows.

i i

fu

Daily wages

(in Rs)

Number of

workers

i

f

i

x

i i

d x 150

i

i

d

u ,

h

h 20

i i

fu

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,

and

i

f 50

i i

fu 12

So, mean is given by,

x

i i

i

fu

x a h

f

12

150 20

50

12

150 2

5

24

150

5

150 4.8

145.20

Therefore, Rs 145.20 is the mean daily wage of the

workers of the factory.

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

11 13

13 15

15 17

17 19

19 21

21 23

23 25

Number

of

children

7

6

9

13

f

5

4

Solution:

Given: Mean pocket allowance

Rs18

Find the class mark for each interval. For this,

i

x

the following relation should be used.

i

Upper limit Lower limit

x

2

Take assumed mean and calculate and

a 18

i

d

i i

fd

as follows.

Daily pocket

allowance (in Rs)

Number of

children

i

f

Class

mark

xi

i i

d x 18

i i

fd

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

2f

21 23

5

22

4

20

23 25

4

24

6

24

Total

i

f 44 f

2f 40

From the table,

and

i

f 44 f

i i

fd 2f 40

So, mean is given by,

x

i i

i

fd

x a

f

2f 40

18 18

44 f

2f 40

18 18

44 f

2f 40

0

44 f

2f 40 0

2f 40

f 20

Hence, the missing frequency is .

f

20

Question: 4

Thirty women were examined in a hospital by a

doctor and the number of heart beats per minute

were recorded and summarised as follows. Find 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

Solution:

Find the class mark for each interval. For this,

i

x

the following relation should be used.

i

Upper limit Lower limit

x

2

Class size

h 3

Take assumed mean and calculate

a 75.5

i

d ,

i

u

and as follows.

i i

fu

Number of

heart beats

per minute

Number of

women

i

f

i

x

i i

d x 75.5

i

i

d

u ,h 3

h

i i

fu

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,

and

i

f 30

i i

fu 4

So, mean is given by,

x

i i

i

fu

x a h

f

4

75.5 3

30

4

75.5

10

75.5 0.4

75.9

Hence, mean heart beats per minute for these

women are 75.9 beats per minute.

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?

Solution:

Given table:

Number of mangoes

Number of boxes

50 52

15

53 55

110

56 58

135

59 61

115

62 64

25

The class intervals are not continuous, as there is a

gap of between two class intervals.

1

So, must be added to the upper-class limit and

0.5

must be subtracted from the lower-class limit of

0.5

each interval.

Then, find the class mark for each interval. For

i

x

this, the following relation should be used.

i

Upper limit Lower limit

x

2

Class size

h 3

Take assumed mean and calculate and

a 57

i

d ,

i

u

as follows.

i i

fu

Number of

mangoes

Number

of boxes

i

f

i

x

i i

d x 57

i

i

d

u ,h 3

h

i i

fu

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

From the table,

and

i

f 400

i i

fu 25

Now, find the mean .

x

i i

i

fu

x a h

f

25

57 3

400

1

57 3

16

3

57

16

57 0.1875

57.1875

57.19

Hence, the mean number of mangoes kept in a

packing box is 57.19.

The method used here is the step deviation method,

as the values of and are large and, also there is a

i

f

i

d

common multiple between all i.e. .

i

d

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.

Solution:

First, find the class mark for each interval. For

i

x

this, the following relation should be used.

i

Upper limit Lower limit

x

2

Class size

h 50

Take assumed mean and calculate and

a 225

i

d ,

i

u

as follows.

i i

fu

Daily

expenditure

(in Rs)

Number of

household

s

i

f

i

x

i i

d x 225

i

i

d

u ,

h

h 50

i i

fu

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,

and

i

f 25

i i

fu 7

Now, find the mean

x

i i

i

fu

x a h

f

7

225 50

25

225 7 2

225 14

211

Hence, mean daily expenditure on food is Rs 211.

Question: 7

To find out the concentration of in the air (in

2

SO

parts per million, i.e., ppm), the data was collected

for 30 localities in a certain city and is presented

below:

Concentration of (in ppm)

2

SO

Frequency

0.00 – 0.04

0.04 – 0.08

0.08 – 0.12

0.12 – 0.16

0.16 – 0.20

0.20 – 0.24

4

9

9

2

4

2

Find the mean concentration of in the air.

2

SO

Solution:

First, find the class mark for each interval. For

i

x

this, the following relation should be used.

i

Upper limit Lower limit

x

2

Class size

h 0.04

Take assumed mean and calculate

a 0.14

i

d ,

i

u

and as follows.

i i

fu

Concentra

tion of

(in

2

SO

ppm)

Frequency

i

f

i

x

i i

d x 0.14

i

i

d

u ,

h

h 0.04

i i

fu

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,

and

i

f 30

i i

fu 31

Now, find the mean .

x

i i

i

fu

x a h

f

31

0.14 0.04

30

0.14 0.04133

0.09867

0.099ppm

Hence, mean concentration of in the air is 0.099

2

SO

ppm.

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

Solution:

First, find the class mark for each interval. For

i

x

this, the following relation should be used.

i

Upper limit Lower limit

x

2

Take assumed mean and calculate and

a 17

i

d

i i

fd

as follows.

Number of

days

Number of

students

i

f

i

x

i i

d x 17

i i

fd

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,

and b

i

f 40

i i

fu 181

Now, find the mean .

x

i i

i

fd

x a

f

181

17

40

17 4.525

12.475

12.48

Hence, the mean number of days for which a

student was absent are 12.48.

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

3

10

11

8

3

cities

Solution:

First, find the class mark for each interval. For

i

x

this, the following relation should be used.

i

Upper limit Lower limit

x

2

Class size

h 10

Take assumed mean and calculate and

a 70

i

d ,

i

u

as follows.

i i

fu

Literacy

rate (in %)

Number

of cities

i

f

i

x

i i

d x 70

i

i

d

u ,

h

h 10

i i

fu

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,

and

i

f 35

i i

fu 2

Now, find the mean .

x

i i

i

fu

x a h

f

2

70 10

35

20

70

35

70 0.57

69.43

Hence, mean literacy rate is 69.43%.

Exercise 14.2 (6)

Question: 1

The following table shows the ages of the patients

admitted in a hospital during a year:

Age (in

years)

5 – 15

15 – 25

25 – 35

35 – 45

45 – 55

55 – 65

Number

of patients

6

11

21

23

14

5

Find the mode and the mean of the data given

above. Compare and interpret the two measures of

central tendency.

Solution:

First, find the class mark for each interval. For

i

x

this, the following relation should be used.

i

Upper limit Lower limit

x

2

Take assumed mean and calculate and

a 30

i

d

i i

fd

as follows.

Age (in

years)

Number of

patients

i

f

i

x

i i

d x 30

i i

fd

5 – 15

6

10

20

120

15 – 25

11

20

10

110

25 – 35

21

30

0

0

35 – 45

23

40

10

230

45 – 55

14

50

20

280

55 – 65

5

60

30

150

Total

80

430

From the table,

and

i

f 80

i i

fu 430

Now, find the mean .

x

i i

i

fd

x a

f

430

30

80

30 5.375

35.375

35.38

So, mean of the given data is 35.38.

This represents that, on an average, the age of a

patient admitted to hospital was years.

35.38

Now, the maximum class frequency is 23 belonging

to class interval .

35 40

Modal class

35 40

Lower limit of modal class

l

35

Frequency of modal class

1

f

23

Class size

h 10

Frequency of class preceding the modal class

0

f

21

Frequency of class succeeding the modal class

2

f

14

Therefore, mode is given by,

1 0

1 0 2

f f

Mode l h

2f f f

23 21

35 10

2 23 21 14

2

35 10

46 35

20

35

11

35 1.81

36.8

Hence, mode is 36.8 and it represents the age of

maximum number of patients admitted in hospital

i.e. 36.8 years.

Question: 2

The following data gives the information on the

observed lifetimes (in hours) of 225 electrical

components:

Lifetimes (in

hours)

0 – 20

20 – 40

40 – 60

60 – 80

80 – 100

100 – 120

Frequency

10

35

52

61

38

29

Determine the modal lifetimes of the components.

Solution:

The maximum class frequency is 61 belonging to

class interval .

60 80

Lower limit of modal class

l

60

Frequency of modal class

1

f

61

Class size

h 20

Frequency of class preceding the modal class

0

f

52

Frequency of class succeeding the modal class

2

f

38

Therefore, mode is given by,

1 0

1 0 2

f f

Mode l h

2f f f

61 52

60 20

2 61 52 38

9

60 20

122 90

9 20

60

32

180

60

32

60 5.625

65.625

Hence, modal lifetime of electrical components is

65.625 hours.

Question: 3

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:

Expenditure (in Rs)

Number of families

1000 – 1500

1500 – 2000

2000 – 2500

2500 – 3000

3000 – 3500

3500 – 4000

4000 – 4500

4500 – 5000

24

40

33

28

30

22

16

7

Solution:

The maximum class frequency is 40 belonging to

class interval .

1500 2000

Lower limit of modal class

l

1500

Frequency of modal class

1

f

40

Class size

h 500

Frequency of class preceding the modal class

0

f

24

Frequency of class succeeding the modal class

2

f

33

Now, find the mode.

1 0

1 0 2

f f

Mode l h

2f f f

40 24

1500 500

2 40 24 33

16

1500 500

80 57

16 500

1500

23

8000

1500

23

1500 347.826

1847.826

1847.83

So, modal monthly expenditure was Rs 1847.83.

Now, find the class mark for each interval. For

i

x

this, the following relation should be used.

i

Upper limit Lower limit

x

2

Take assumed mean and calculate

a 2750

i

d ,

i

u

and as follows.

i i

fu

Expenditur

e (in Rs)

Number

of

families

i

f

i

x

i i

d x 2750

i

i

d

u ,

h

h 500

i i

fu

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

Total

200

35

From the table,

and

i

f 200

i i

fu 35

Now, find the mean .

x

i i

i

fu

x a h

f

35

2750 500

200

35 5

2750

2

175

2750

2

2750 87.5

2662.5

Hence, mean monthly expenditure is Rs 2662.50.

Question: 4

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.

Number of students

per teacher

Number of

states/U.T.

15 – 20

20 – 25

25 – 30

30 – 35

35 – 40

40 – 45

3

8

9

10

3

0

45 – 50

50 – 55

0

2

Solution:

The maximum class frequency is 10 belonging to

class interval .

30 35

Lower limit of modal class

l

30

Frequency of modal class

1

f

10

Class size

h 5

Frequency of class preceding the modal class

0

f

9

Frequency of class succeeding the modal class

2

f

3

Now, find the mode.

1 0

1 0 2

f f

Mode l h

2f f f

10 9

30 5

2 10 9 3

1

30 5

20 12

5

30

8

30 0.625

30.6

So, mode represents that, most of the states/U.T.

have a teacher-student ratio as 30.6.

Now, find the class mark for each interval. For

i

x

this, the following relation should be used.

i

Upper limit Lower limit

x

2

Take assumed mean and calculate

a 32.5

i

d ,

i

u

and as follows.

i i

fu

Number of

students per

teacher

Number of

states/U.T.

i

f

i

x

i i

d x 32.5

i

i

d

u ,

h

h 5

i i

fu

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

Total

35

23

From the table,

and

i

f 35

i i

fu 23

Now, find the mean .

x

i i

i

fu

x a h

f

23

32.5 5

35

23

32.5

7

32.5 3.28

29.2

Hence, mean of data is Rs 29.2 which represents the

average teacher-student ratio.

Question: 5

The given distribution shows the number of runs

scored by some top batsmen of the world in one-

day international cricket matches.

Runs scored

Number of batsmen

3000 – 4000

4000 – 5000

5000 – 6000

6000 – 7000

7000 – 8000

8000 – 9000

9000 – 10000

10000 – 11000

4

18

9

7

6

3

1

1

Find the mode of the data.

Solution:

The maximum class frequency is 18 belonging to

class interval .

4000 5000

Lower limit of modal class

l

4000

Frequency of modal class

1

f

18

Class size

h 1000

Frequency of class preceding the modal class

0

f

4

Frequency of class succeeding the modal class

2

f

9

Now, find the mode.

1 0

1 0 2

f f

Mode l h

2f f f

18 4

4000 1000

2 18 4 9

14

4000 1000

23

4000 608.695

4608.695

Hence, mode of the given data is 4608.695.

Question: 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:

Number

of cars

0 –

10

10 –

20

20 –

30

30 –

40

40 –

50

50 –

60

60 –

70

70 –

80

Frequency

7

14

13

12

20

11

15

8

Solution:

The maximum class frequency is 20 belonging to

class interval .

40 50

Lower limit of modal class

l

40

Frequency of modal class

1

f

20

Class size

h 10

Frequency of class preceding the modal class

0

f

12

Frequency of class succeeding the modal class

2

f

11

Now, find the mode.

1 0

1 0 2

f f

Mode l h

2f f f

20 12

40 10

2 20 12 11

8

40 10

17

80

40

17

40 4.7

44.7

Hence, mode of the given data is 44.7.

Exercise 14.3 (7)

Question: 1

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.

Monthly consumption

(in units)

Number of consumers

65 – 85

85 – 105

105 – 125

125 – 145

145 – 165

165 – 185

185 – 205

4

5

13

20

14

8

4

Solution:

First, find the class mark for each interval. For

i

x

this, the following relation should be used.

i

Upper limit Lower limit

x

2

Take assumed mean and calculate and

a 135

i

d ,

i

u

as follows.

i i

fu

Monthly

consumptio

n (in units)

Number of

consumers

i

f

i

x

i i

d x 135

i

i

d

u ,

h

h 20

i i

fu

65 – 85

4

75

60

3

12

85 – 105

5

95

40

2

10

105 – 125

13

115

20

1

13

125 – 145

20

135

0

0

0

145 – 165

14

155

20

1

14

165 – 185

8

175

40

2

16

185 – 205

4

195

60

3

12

Total

68

7

From the table,

and

i

f 68

i i

fu 7

Now, find the mean .

x

i i

i

fu

x a h

f

7

135 20

68

140

135

68

135 2.058

137.058

So, mean of the given data is 137.058.

Now, the maximum class frequency is 20 belonging

to class interval .

125 145

Modal class

125 145

Lower limit of modal class

l

125

Frequency of modal class

1

f

20

Class size

h 20

Frequency of class preceding the modal class

0

f

13

Frequency of class succeeding the modal class

2

f

14

Therefore, mode is given by,

1 0

1 0 2

f f

Mode l h

2f f f

20 13

125 20

2 20 13 14

7

125 20

13

140

125

13

125 10.76

135.76

Hence, the mode is 135.76.

Now, find the median of the given data, and for

this, cumulative frequency is calculated as follows.

Monthly

consumption (in

units)

Number of

consumers

Cumulative

frequency

65 – 85

85 – 105

105 – 125

125 – 145

145 – 165

165 – 185

185 – 205

4

5

13

20

14

8

4

4

4 5 9

9 13 22

22 20 42

42 14 56

56 8 64

64 4 68

From the table,

n 68

Cumulative frequency just greater than

cf

is 42, belonging to interval .

n 68

34

2 2

125 145

Median class

125 145

Lower limit of median class

l

125

Class size

h 20

Frequency of median class

f

20

Cumulative frequency of class preceding

cf

median class

22

Now, find the median.

n

cf

2

Median l h

f

34 22

125 20

20

12

125 20

20

125 12

137

Therefore, median, mode and mean of the given

data are 137, 135.76 and 137.05 respectively.

The three quantities are approximately the same in

this case.

Question: 2

If the median of the distribution given below is 28.5,

find the values of and .

x

y

Class interval

Frequency

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

5

x

20

15

y

5

Total

60

Solution:

Cumulative frequency is given by,

Class interval

Frequency

Cumulative

frequency

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

5

x

20

15

y

5

5

5 x

25 x

40 x

40 x y

45 x y

Total

n

60

From the table,

n 60

45 x y 60

x y 15 ......(1)

Median of the data is given as , belonging to

28.5

interval .

20 30

Median class

20 30

Lower limit of median class

l

20

Class size

h 10

Frequency of median class

f

20

Cumulative frequency of class preceding

cf

median class

5 x

Put the values in the formula of median.

n

cf

2

Median l h

f

60

5 x

2

28.5 20 10

20

30 5 x

28.5 20 10

20

25 x

8.5

2

17 25 x

x 8

Put the value of in equation .

x

1

8 y 15

y 7

Hence, .

x 8,

y 7

Question: 3

A life insurance agent found the following data for

distribution of ages of 100 policy holders. Calculate

the median age, if policies are given only to persons

having age 18 years onwards but less than 60 years.

Age (in years)

Number of policy holders

Below 20

Below 25

Below 30

Below 35

Below 40

Below 45

Below 50

Below 55

Below 60

2

6

24

45

78

89

92

98

100

Solution:

It is not essential to adjust the frequencies

according to class intervals, as class width is not the

same.

According to the question, the frequency table is of

less than type represented with upper class limits.

The policies were given only to persons with age 18

years onwards but less than 60 years.

So, class intervals with their respective cumulative

frequency is given by,

Age (in

years)

Number of policy

holders

Cumulative

frequency

cf

18 – 20

2

2

20 – 25

6 2 4

2 4 6

25 – 30

24 6 18

6 18 24

30 – 35

45 24 21

24 21 45

35 – 40

78 45 33

45 33 78

40 – 45

89 78 11

78 11 89

45 – 50

92 89 3

89 3 92

50 – 55

98 92 6

92 6 98

55 – 60

100 98 2

98 2 100

From the table,

n 100

Cumulative frequency just greater than

cf

is 78, belonging to interval .

n 100

50

2 2

35 40

Median class

35 40

Lower limit of median class

l

35

Class size

h 5

Frequency of median class

f

33

Cumulative frequency of class preceding

cf

median class

45

Now, find the median.

n

cf

2

Median l h

f

50 45

35 5

33

5

35 5

33

25

35

33

35 0.76

35.76

Hence, the median age is 35.76 years.

Question: 4

The lengths of 40 leaves of a plant are measured

correct to the nearest millimetre, and the data

obtained is represented in the following table:

Length (in mm)

Number of leaves

118 – 126

127 – 135

136 – 144

145 – 153

154 – 162

163 – 171

172 – 180

3

5

9

12

5

4

2

Find the median length of the leaves.

Solution:

The class intervals are not continuous, as there is a

gap of between two class intervals.

1

So, must be added to the upper-class limit and

0.5

must be subtracted from the lower-class limit of

0.5

each interval.

So, class intervals with their respective cumulative

frequency is given by,

Length (in mm)

Number of

Cumulative

leaves

frequency

cf

117.5 – 126.5

3

3

126.5 – 135.5

5

3 5 8

135.5 – 144.5

9

8 9 17

144.5 – 153.5

12

17 12 29

153.5 – 162.5

5

29 5 34

162.5 – 171.5

4

34 4 38

171.5 – 180.5

2

38 2 40

From the table,

n 40

Cumulative frequency just greater than

cf

is 29, belonging to interval .

n 40

20

2 2

144.5 153.5

Median class

144.5 153.5

Lower limit of median class

l

144.5

Class size

h 9

Frequency of median class

f

12

Cumulative frequency of class preceding

cf

median class

17

Now, find the median.

n

cf

2

Median l h

f

20 17

144.5 9

12

3

144.5 9

12

9

144.5

4

144.5 2.25

146.75

Hence, the median length of leaves is 146.75 mm.

Question: 5

The following table gives the distribution of the life

time of 400 neon lamps.

Life time (in hours)

Number of lamps

1500 – 2000

2000 – 2500

2500 – 3000

3000 – 3500

3500 – 4000

4000 – 4500

4500 – 5000

14

56

60

86

74

62

48

Find the median life time of a lamp.

Solution:

Class intervals with their respective cumulative

frequency is given by,

Life time

Number of

lamps

i

f

Cumulative

frequency

cf

1500 – 2000

14

14

2000 – 2500

56

14 56 70

2500 – 3000

60

70 60 130

3000 – 3500

86

130 86 216

3500 – 4000

74

216 74 290

4000 – 4500

62

290 62 352

4500 – 5000

48

352 48 400

Total

n

400

From the table,

n 400

Cumulative frequency just greater than

cf

is 216, belonging to interval

n 400

200

2 2

.

3000 3500

Median class

3000 3500

Lower limit of median class

l

3000

Class size

h 500

Frequency of median class

f

86

Cumulative frequency of class preceding

cf

median class

130

Now, find the median.

n

cf

2

Median l h

f

200 130

3000 500

86

70

3000 500

86

35000

3000

86

3000 406.976

3406.976

Hence, the median lifetime of lamps is

3406.98

hours.

Question: 6

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:

Number of

letters

1 – 4

4 – 7

7 – 10

10 – 13

13 – 16

16 – 19

Number of

surnames

6

30

40

16

4

4

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

Solution:

Class intervals with their respective cumulative

frequency is given by,

Number of

letters

Number of

surnames

i

f

Cumulative

frequency

cf

1 4

6

6

4 7

30

6 30 36

7 10

40

36 40 76

10 13

16

76 16 92

13 16

4

92 4 96

16 19

4

96 4 100

Total

n

100

From the table,

n 100

Cumulative frequency just greater than

cf

is 76, belonging to interval .

n 100

50

2 2

7 10

Median class

7 10

Lower limit of median class

l

7

Class size

h 3

Frequency of median class

f

40

Cumulative frequency of class preceding

cf

median class

36

Now, find the median.

n

cf

2

Median l h

f

50 36

7 3

40

14

7 3

40

42

7

40

7 1.05

8.05

So, the median number of letters is .

8.05

Now, find the class mark for each interval. For

i

x

this, the following relation should be used.

i

Upper limit Lower limit

x

2

Take assumed mean and calculate

a 11.5

i

d ,

i

u

and as follows.

i i

fu

Number of

letters

Number of

surnames

i

f

i

x

i i

d x 11.5

i

i

d

u ,

h

h 3

i i

fu

1 4

6

2.5

9

3

18

4 7

30

5.5

6

2

60

7 10

40

8.5

3

1

40

10 13

16

11.5

0

0

0

13 16

4

14.5

3

1

4

16 19

4

17.5

6

2

8

Total

100

106

From the table,

i i i

f 100and fu 106

Now, find the mean .

x

i i

i

fu

x a h

f

106

11.5 3

100

318

11.5

100

11.5 3.18

8.32

So, mean of the given data is 8.32.

Now, the maximum class frequency is 40 belonging

to class interval .

7 10

Modal class

7 10

Lower limit of modal class

l

7

Frequency of modal class

1

f

40

Class size

h 3

Frequency of class preceding the modal class

0

f

30

Frequency of class succeeding the modal class

2

f

16

Now, find the mode.

1 0

1 0 2

f f

Mode l h

2f f f

40 30

7 3

2 40 30 16

10

7 3

34

30

7

34

7 0.88

7.88

Hence, median number of letters in surnames is

8.05, mean number of letters in surnames is 8.32,

and modal size of surnames is 7.88.

Question: 7

The distribution below gives the weights of 30

students of a class. Find the median weight of the

students.

Weight (in

kg)

40 –

45

45 –

50

50 –

55

55 –

60

60 –

65

65 –

70

70 –

75

Number of

students

2

3

8

6

6

3

2

Solution:

Class intervals with their respective cumulative

frequency is given by,

Number of

letters

Number of

surnames

i

f

Cumulative

frequency

cf

40 45

2

2

45 50

3

2 3 5

50 55

8

5 8 13

55 60

6

13 6 19

60 65

6

19 6 25

65 70

3

25 3 28

70 75

2

28 2 30

Total

n

30

From the table,

n 30

Cumulative frequency just greater than

cf

is 19, belonging to interval .

n 30

15

2 2

55 60

Median class

55 60

Lower limit of median class

l

55

Class size

h 5

Frequency of median class

f

6

Cumulative frequency of class preceding

cf

median class

13

Now, find the median.

n

cf

2

Median l h

f

15 13

55 5

6

2

55 5

6

10

55

6

55 1.67

56.67

Hence, median weight is 56.7.

Exercise 14.4 (3)

Question: 1

The following distribution gives the daily income of

50 workers of a factory.

Daily income

(in Rs)

100 – 120

120 – 140

140 – 160

160 – 180

180 – 200

Number of

workers

12

14

8

6

10

Convert the distribution above to a less than type

cumulative frequency distribution, and draw its

ogive.

Solution:

The frequency distribution table of less than type

along with the cumulative frequency is given by,

Daily income (in Rs)

Upper class limit

Cumulative frequency

Less than 120

12

Less than 140

12 14 26

Less than 160

26 8 34

Less than 180

34 6 40

Less than 200

40 10 50

Now, take upper class limits of class intervals on

and their respective frequencies on ,

x axis

y axis

draw its ogive as follows.

Fig. Exc_14.4_1

Question: 2

During the medical check-up of 35 students of a

class, their weights were recorded as follows:

Weight (in kg)

Number of students

Less than 38

Less than 40

0

3

Less than 42

Less than 44

Less than 46

Less than 48

Less than 50

Less than 52

5

9

14

28

32

35

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.

Solution:

The frequency distribution table of less than type

along with the cumulative frequency is given by,

Weight (in kg)

Upper class limits

Number of students

(Cumulative

frequency)

Less than 38

Less than 40

Less than 42

Less than 44

Less than 46

Less than 48

Less than 50

0

3

5

9

14

28

32

Less than 52

35

Now, take upper class limits of class intervals on

and their respective frequencies on ,

x axis

y axis

draw its ogive as follows.

Fig. Exc_14.4_2 (i)

From the table,

n 35

So,

n 35

17.5

2 2

Now, mark the point A whose coordinate is 17.5

y

and its coordinate is 46.5.

x

Therefore, median of this data is 46.5.

Fig. Exc_14.4_2 (ii)

The difference between two consecutive upper class

limits is 2. The class marks with their respective

frequencies are given by,

Weight is kg

Frequency

i

f

Cumulative

frequency

cf

Less than 38

0

0

38 40

3 0 3

3

40 42

5 3 2

5

42 44

9 5 4

9

44 46

14 9 5

14

46 48

28 14 14

28

48 50

32 28 4

32

50 52

35 32 3

35

Total

n

35

Cumulative frequency just greater than

cf

is 28, belonging to interval .

n 35

17.5

2 2

46 48

Median class

46 48

Lower limit of median class

l

46

Class size

h 2

Frequency of median class

f

14

Cumulative frequency of class preceding

cf

median class

14

Now, find the median.

n

cf

2

Median l h

f

17.5 14

46 2

14

3.5

46 2

14

3.5

46

7

46 0.5

46.5

The median of the given data is 46.5.

Hence verified.

Question: 3

The following table gives production yield per

hectare of wheat of 100 farms of a village.

Production

yield (in kg/ha)

50 – 55

55 – 60

60 – 65

65 – 70

70 – 75

75 – 80

Number of

farms

2

8

12

24

38

16

Change the distribution to a more than type

distribution, and draw its ogive.

Solution:

The frequency distribution table of more than type

along with the cumulative frequency is given by,

Production yield

Lower class limits

Number of farms

(Cumulative frequency)

More than or equal to 50

100

More than or equal to 55

100 2 98

More than or equal to 60

98 8 90

More than or equal to 65

90 12 78

More than or equal to 70

78 24 54

More than or equal to 75

54 38 16

Now, take lower class limits of class intervals on

and their respective frequencies on ,

x axis

y axis

draw its ogive as follows.

Fig. Exc_14.4_3