Exercise
14.1
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?
To find the class mark (x_{i}) for each interval, the following relation is used.
Class mark (x_{i})
x_{i }and f_{i}x_{i} can be calculated as follows.
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 (x_{i}) and f_{i} are small.
Class mark (x_{i})
x_{i }and f_{i}x_{i} can be calculated as follows.
Number of plants

Number of houses
(f_{i})

x_{i}

f_{i}x_{i}

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

Mean,
Therefore, mean number of plants per house is 8.1.
Here, direct method has been used as the values of class marks (x_{i}) and f_{i} are small.
Question 2:
Consider the following distribution of daily wages of 50 worker of a factory.
Find the mean daily wages of the workers of the factory by using an appropriate method.
Daily wages (in Rs) 
100 − 120

120 − 140

140 −1 60

160 − 180

180 − 200

Number of workers 
12

14

8

6

10

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), d_{i}, u_{i}, and f_{i}u_{i} can be calculated as follows.
From the table, it can be observed that
Therefore, the mean daily wage of the workers of the factory is Rs 145.20.
Class size (h) of this data = 20
Taking 150 as assured mean (a), d_{i}, u_{i}, and f_{i}u_{i} can be calculated as follows.
Daily wages
(in Rs)
 Number of workers (f_{i}) 
x_{i}

d_{i} = x_{i }− 150

f_{i}u_{i}
 
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

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

To find the class mark (x_{i}) for each interval, the following relation is used.
Given that, mean pocket allowance,
Taking 18 as assured mean (a), d_{i} and f_{i}d_{i} are calculated as follows.
From the table, we obtain
Hence, the missing frequency, f, is 20.
Given that, mean pocket allowance,
Taking 18 as assured mean (a), d_{i} and f_{i}d_{i} are calculated as follows.
Daily pocket allowance
(in Rs)

Number of children
f_{i}

Class markx_{i}

d_{i} = x_{i }− 18

f_{i}d_{i}

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

2 f

21 − 23

5

22

4

20

23 − 25

4

24

6

24

Total

2f − 40

Hence, the missing frequency, f, is 20.
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

To find the class mark of each interval (x_{i}), the following relation is used.
Class size, h, of this data = 3
Taking 75.5 as assumed mean (a), di, u_{i}, f_{i}u_{i} are calculated as follows.
From the table, we obtain
Therefore, mean hear beats per minute for these women are 75.9 beats per minute.
It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore 1/2 has to be subtracted from the lower class limit
of each interval.
, has to be added to the upper class limit and
Class mark (x_{i}) can be obtained by using the following relation.
Class size (h) of this data = 3
Taking 57 as assumed mean (a), d_{i}, u_{i}, f_{i}u_{i} are calculated as follows.
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 f_{i, }d_{i} are big and also, there is a common multiple between all d_{i}.
Class size, h, of this data = 3
Taking 75.5 as assumed mean (a), di, u_{i}, f_{i}u_{i} are calculated as follows.
Number of heart beats per minute

Number of women
f_{i}

x_{i}

d_{i} = x_{i }− 75.5

f_{i}u_{i}
 
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

Therefore, mean hear 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?
Number of mangoes

Number of boxes f_{i}

50 − 52

15

53 − 55

110

56 − 58

135

59 − 61

115

62 − 64

25

of each interval.
Class mark (x_{i}) can be obtained by using the following relation.
Class size (h) of this data = 3
Taking 57 as assumed mean (a), d_{i}, u_{i}, f_{i}u_{i} are calculated as follows.
Class interval

f_{i}

x_{i}

d_{i}_{ }= x_{i} − 57

f_{i}u_{i}
 
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

Mean number of mangoes kept in a packing box is 57.19.
Step deviation method is used here as the values of f_{i, }d_{i} are big and also, there is a common multiple between all d_{i}.
Question 6:
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 Rs) 
100 − 150

150 − 200

200 − 250

250 − 300

300 − 350

Number of households 
4

5

12

2

2

To find the class mark (x_{i}) for each interval, the following relation is used.
Class size = 50
Taking 225 as assumed mean (a), d_{i}, u_{i}, f_{i}u_{i} are calculated as follows.
From the table, we obtain
Therefore, mean daily expenditure on food is Rs 211.
Class size = 50
Taking 225 as assumed mean (a), d_{i}, u_{i}, f_{i}u_{i} are calculated as follows.
Daily expenditure (in Rs)

f_{i}

x_{i}

d_{i}_{ }= x_{i} − 225

f_{i}u_{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

Total

25

− 7

Therefore, mean daily expenditure on food is Rs 211.
Question 7:
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:
Find the mean concentration of SO_{2} in the air.
concentration of SO_{2} (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

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), d_{i}, u_{i},_{ }f_{i}u_{i} are calculated as follows.
From the table, we obtain
Therefore, mean concentration of SO_{2} in the air is 0.099 ppm.
Class size of this data = 0.04
Taking 0.14 as assumed mean (a), d_{i}, u_{i},_{ }f_{i}u_{i} are calculated as follows.
Concentration of SO_{2}(in ppm)

Frequency
f_{i}

Class mark
x_{i}

d_{i}_{ }= x_{i }− 0.14

f_{i}u_{i}
 
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

Therefore, mean concentration of SO_{2} in the air is 0.099 ppm.
Exercise 14.2
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.
To find the class marks (x_{i}), the following relation is used.
Taking 30 as assumed mean (a), d_{i} and f_{i}d_{i}are calculated as follows.
From the table, we obtain
Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years.
It can be observed that the maximum class frequency is 23 belonging to class interval 35 − 45.
Modal class = 35 − 45
Lower limit (l) of modal class = 35
Frequency (f_{1}) of modal class = 23
Class size (h) = 10
Frequency (f_{0}) of class preceding the modal class = 21
Frequency (f_{2}) of class succeeding the modal class = 14
Mode =
Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.
Taking 30 as assumed mean (a), d_{i} and f_{i}d_{i}are calculated as follows.
Age (in years)

Number of patients
f_{i}

Class mark
x_{i}

d_{i}_{ }= x_{i} − 30

f_{i}d_{i}

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

Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years.
It can be observed that the maximum class frequency is 23 belonging to class interval 35 − 45.
Modal class = 35 − 45
Lower limit (l) of modal class = 35
Frequency (f_{1}) of modal class = 23
Class size (h) = 10
Frequency (f_{0}) of class preceding the modal class = 21
Frequency (f_{2}) of class succeeding the modal class = 14
Mode =
Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 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.
From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 − 80.
Therefore, modal class = 60 − 80
Lower class limit (l) of modal class = 60
Frequency (f_{1}) of modal class = 61
Frequency (f_{0}) of class preceding the modal class = 52
Frequency (f_{2}) of class succeeding the modal class = 38
Class size (h) = 20
Therefore, modal lifetime of electrical components is 65.625 hours.
Therefore, modal class = 60 − 80
Lower class limit (l) of modal class = 60
Frequency (f_{1}) of modal class = 61
Frequency (f_{0}) of class preceding the modal class = 52
Frequency (f_{2}) of class succeeding the modal class = 38
Class size (h) = 20
Therefore, 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

24

1500 − 2000

40

2000 − 2500

33

2500 − 3000

28

3000 − 3500

30

3500 − 4000

22

4000 − 4500

16

4500 − 5000

7

It can be observed from the given data that the maximum class frequency is 40, belonging to 1500 − 2000 intervals.
Therefore, modal class = 1500 − 2000
Lower limit (l) of modal class = 1500
Frequency (f_{1}) of modal class = 40
Frequency (f_{0}) of class preceding modal class = 24
Frequency (f_{2}) of class succeeding modal class = 33
Class size (h) = 500
Therefore, modal monthly expenditure was Rs 1847.83.
To find the class mark, the following relation is used.
Class size (h) of the given data = 500
Taking 2750 as assumed mean (a), d_{i}, u_{i}, and f_{i}u_{i}are calculated as follows.
From the table, we obtain
Therefore, mean monthly expenditure was Rs 2662.50.
Therefore, modal class = 1500 − 2000
Lower limit (l) of modal class = 1500
Frequency (f_{1}) of modal class = 40
Frequency (f_{0}) of class preceding modal class = 24
Frequency (f_{2}) of class succeeding modal class = 33
Class size (h) = 500
Therefore, modal monthly expenditure was Rs 1847.83.
To find the class mark, the following relation is used.
Class size (h) of the given data = 500
Taking 2750 as assumed mean (a), d_{i}, u_{i}, and f_{i}u_{i}are calculated as follows.
Expenditure (in Rs)

Number of families
f_{i}

x_{i}

d_{i} = x_{i} − 2750

f_{i}u_{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

Total

200

− 35

Therefore, mean monthly expenditure was Rs 2662.50.
Question 4:
The following distribution gives the statewise teacherstudent 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

3

20 − 25

8

25 − 30

9

30 − 35

10

35 − 40

3

40 − 45

0

45 − 50

0

50 − 55

2

It can be observed from the given data that the maximum class frequency is 10 belonging to class interval 30 − 35.
Therefore, modal class = 30 − 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f_{1}) of modal class = 10
Frequency (f_{0}) of class preceding modal class = 9
Frequency (f_{2}) of class succeeding modal class = 3
It represents that most of the states/U.T have a teacherstudent ratio as 30.6.
To find the class marks, the following relation is used.
Taking 32.5 as assumed mean (a), d_{i}, u_{i}, and f_{i}u_{i} are calculated as follows.
Therefore, mean of the data is 29.2.
It represents that on an average, teacher−student ratio was 29.2.
Therefore, modal class = 30 − 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f_{1}) of modal class = 10
Frequency (f_{0}) of class preceding modal class = 9
Frequency (f_{2}) of class succeeding modal class = 3
It represents that most of the states/U.T have a teacherstudent ratio as 30.6.
To find the class marks, the following relation is used.
Taking 32.5 as assumed mean (a), d_{i}, u_{i}, and f_{i}u_{i} are calculated as follows.
Number of students per teacher

Number of states/U.T
(f_{i})

x_{i}

d_{i} = x_{i} − 32.5

f_{i}u_{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

Total

35

− 23

Therefore, mean of the data is 29.2.
It represents that on an average, teacher−student ratio was 29.2.
Question 5:
The given distribution shows the number of runs scored by some top batsmen of the world in oneday international cricket matches.
Runs scored

Number of batsmen

3000 − 4000

4

4000 − 5000

18

5000 − 6000

9

6000 − 7000

7

7000 − 8000

6

8000 − 9000

3

9000 − 10000

1

10000 − 11000

1

Find the mode of the data.
From the given data, it can be observed that the maximum class frequency is 18, belonging to class interval 4000 − 5000.
Therefore, modal class = 4000 − 5000
Lower limit (l) of modal class = 4000
Frequency (f_{1}) of modal class = 18
Frequency (f_{0}) of class preceding modal class = 4
Frequency (f_{2}) of class succeeding modal class = 9
Class size (h) = 1000
Therefore, mode of the given data is 4608.7 runs.
Therefore, modal class = 4000 − 5000
Lower limit (l) of modal class = 4000
Frequency (f_{1}) of modal class = 18
Frequency (f_{0}) of class preceding modal class = 4
Frequency (f_{2}) of class succeeding modal class = 9
Class size (h) = 1000
Therefore, mode of the given data is 4608.7 runs.
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

From the given data, it can be observed that the maximum class frequency is 20, belonging to 40 − 50 class intervals.
Therefore, modal class = 40 − 50
Lower limit (l) of modal class = 40
Frequency (f_{1}) of modal class = 20
Frequency (f_{0}) of class preceding modal class = 12
Frequency (f_{2}) of class succeeding modal class = 11
Class size = 10
Therefore, mode of this data is 44.7 cars.
Therefore, modal class = 40 − 50
Lower limit (l) of modal class = 40
Frequency (f_{1}) of modal class = 20
Frequency (f_{0}) of class preceding modal class = 12
Frequency (f_{2}) of class succeeding modal class = 11
Class size = 10
Therefore, mode of this data is 44.7 cars.
Exercise 14.3
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

4

85 − 105

5

105 − 125

13

125 − 145

20

145 − 165

14

165 − 185

8

185 − 205

4

To find the class marks, the following relation is used.
Taking 135 as assumed mean (a), d_{i}, u_{i}, f_{i}u_{i} are calculated according to step deviation method as follows.
From the table, we obtain
From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 − 145.
Modal class = 125 − 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f_{1}) of modal class = 20
Frequency (f_{0}) of class preceding modal class = 13
Frequency (f_{2}) of class succeeding the modal class = 14
To find the median of the given data, cumulative frequency is calculated as follows.
From the table, we obtain
n = 68
Cumulative frequency (cf) just greater than
is 42, belonging to interval 125 − 145.
Therefore, median class = 125 − 145
Lower limit (l) of median class = 125
Class size (h) = 20
Frequency (f) of median class = 20
Cumulative frequency (cf) of class preceding median class = 22
Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively.
The three measures are approximately the same in this case.
From equation (1),
8 + y = 15
y = 7
Hence, the values of x and y are 8 and 7 respectively.
Taking 135 as assumed mean (a), d_{i}, u_{i}, f_{i}u_{i} are calculated according to step deviation method as follows.
Monthly consumption (in units)

Number of consumers (f_{i})

x_{i}class mark

d_{i}=x_{i}− 135
 
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, it can be observed that the maximum class frequency is 20, belonging to class interval 125 − 145.
Modal class = 125 − 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f_{1}) of modal class = 20
Frequency (f_{0}) of class preceding modal class = 13
Frequency (f_{2}) of class succeeding the modal class = 14
To find the median of the given data, cumulative frequency is calculated as follows.
Monthly consumption (in units)

Number of consumers

Cumulative frequency

65 − 85

4

4

85 − 105

5

4 + 5 = 9

105 − 125

13

9 + 13 = 22

125 − 145

20

22 + 20 = 42

145 − 165

14

42 + 14 = 56

165 − 185

8

56 + 8 = 64

185 − 205

4

64 + 4 = 68

n = 68
Cumulative frequency (cf) just greater than
is 42, belonging to interval 125 − 145.
Therefore, median class = 125 − 145
Lower limit (l) of median class = 125
Class size (h) = 20
Frequency (f) of median class = 20
Cumulative frequency (cf) of class preceding median class = 22
Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively.
The three measures are approximately the same in this case.
Question 2:
If the median of the distribution is given below is 28.5, find the values of x and y.
Class interval

Frequency

0 − 10

5

10 − 20

x

20 − 30

20

30 − 40

15

40 − 50

y

50 − 60

5

Total

60

The cumulative frequency for the given data is calculated as follows.
From the table, it can be observed that n = 60
45 + x + y = 60
x + y = 15 (1)
Median of the data is given as 28.5 which lies in interval 20 − 30.
Therefore, median class = 20 − 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
Class interval

Frequency

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

Total (n)

60

45 + x + y = 60
x + y = 15 (1)
Median of the data is given as 28.5 which lies in interval 20 − 30.
Therefore, median class = 20 − 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
From equation (1),
8 + y = 15
y = 7
Hence, the values of x and y are 8 and 7 respectively.
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 year.
Age (in years)

Number of policy holders

Below 20

2

Below 25

6

Below 30

24

Below 35

45

Below 40

78

Below 45

89

Below 50

92

Below 55

98

Below 60

100

Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given 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. Therefore, class intervals with their respective cumulative frequency can be defined as below.
From the table, it can be observed that n = 100.
Cumulative frequency (cf) just greater than
is 78, belonging to interval 35 − 40.
Therefore, median class = 35 − 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45
Therefore, median age is 35.76 years.
Question 2:
Age (in years)

Number of policy holders (f_{i})

Cumulative frequency (cf)

18 − 20

2

2

20 − 25

6 − 2 = 4

6

25 − 30

24 − 6 = 18

24

30 − 35

45 − 24 = 21

45

35 − 40

78 − 45 = 33

78

40 − 45

89 − 78 = 11

89

45 − 50

92 − 89 = 3

92

50 − 55

98 − 92 = 6

98

55 − 60

100 − 98 = 2

100

Total (n)

Cumulative frequency (cf) just greater than
is 78, belonging to interval 35 − 40.
Therefore, median class = 35 − 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45
Therefore, median age is 35.76 years.
Question 4:
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:
Length (in mm)

Number or leaves f_{i}

118 − 126

3

127 − 135

5

136 − 144

9

145 − 153

12

154 − 162

5

163 − 171

4

172 − 180

2

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)
The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore,1/2=0.5 has to be added and subtracted to upper class limits and lower class limits respectively.
Continuous class intervals with respective cumulative frequencies can be represented as follows.
From the table, it can be observed that the cumulative frequency just greater than
is 29, belonging to class interval 144.5 − 153.5.
Median class = 144.5 − 153.5
Lower limit (l) of median class = 144.5
Class size (h) = 9
Frequency (f) of median class = 12
Cumulative frequency (cf) of class preceding median class = 17
Median
Therefore, median length of leaves is 146.75 mm.
Continuous class intervals with respective cumulative frequencies can be represented as follows.
Length (in mm)

Number or leaves f_{i}

Cumulative frequency

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

is 29, belonging to class interval 144.5 − 153.5.
Median class = 144.5 − 153.5
Lower limit (l) of median class = 144.5
Class size (h) = 9
Frequency (f) of median class = 12
Cumulative frequency (cf) of class preceding median class = 17
Median
Therefore, median length of leaves is 146.75 mm.
Question 5:
Find the following table gives the distribution of the life time of 400 neon lamps:
Life time (in hours)

Number of lamps

1500 − 2000

14

2000 − 2500

56

2500 − 3000

60

3000 − 3500

86

3500 − 4000

74

4000 − 4500

62

4500 − 5000

48

Find the median life time of a lamp.
Thecumulative frequencies with their respective class intervals are as follows.
It can be observed that the cumulative frequency just greater than
is 216, belonging to class interval 3000 − 3500.
Median class = 3000 − 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500
Median
= 3406.976
Therefore, median life time of lamps is 3406.98 hours.
Life time

Number of lamps (f_{i})

Cumulative frequency

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

is 216, belonging to class interval 3000 − 3500.
Median class = 3000 − 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500
Median
= 3406.976
Therefore, median life time 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:
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.
Number of letters

1 − 4

4 − 7

7 − 10

10 − 13

13 − 16

16 − 19

Number of surnames

6

30

40

6

4

4

The cumulative frequencies with their respective class intervals are as follows.
It can be observed that the cumulative frequency just greater than
is 76, belonging to class interval 7 − 10.
Median class = 7 − 10
Lower limit (l) of median class = 7
Cumulative frequency (cf) of class preceding median class = 36
Frequency (f) of median class = 40
Class size (h) = 3
Median
= 8.05
To find the class marks of the given class intervals, the following relation is used.
Taking 11.5 as assumed mean (a), d_{i}, u_{i}, and f_{i}u_{i} are calculated according to step deviation method as follows.
From the table, we obtain
∑f_{i}u_{i} = −106
∑f_{i} = 100
Mean,
= 11.5 − 3.18 = 8.32
The data in the given table can be written as
From the table, it can be observed that the maximum class frequency is 40 belonging to class interval 7 − 10.
Modal class = 7 − 10
Lower limit (l) of modal class = 7
Class size (h) = 3
Frequency (f_{1}) of modal class = 40
Frequency (f_{0}) of class preceding the modal class = 30
Frequency (f_{2}) of class succeeding the modal class = 16
Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88.
Number of letters

Frequency (f_{i})

Cumulative frequency

1 − 4

6

6

4 − 7

30

30 + 6 = 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

is 76, belonging to class interval 7 − 10.
Median class = 7 − 10
Lower limit (l) of median class = 7
Cumulative frequency (cf) of class preceding median class = 36
Frequency (f) of median class = 40
Class size (h) = 3
Median
= 8.05
To find the class marks of the given class intervals, the following relation is used.
Taking 11.5 as assumed mean (a), d_{i}, u_{i}, and f_{i}u_{i} are calculated according to step deviation method as follows.
Number of letters

Number of surnames
f_{i}

x_{i}

d_{i} = x_{i}− 11.5

f_{i}u_{i}
 
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

∑f_{i}u_{i} = −106
∑f_{i} = 100
Mean,
= 11.5 − 3.18 = 8.32
The data in the given table can be written as
Number of letters

Frequency (f_{i})

1 − 4

6

4 − 7

30

7 − 10

40

10 − 13

16

13 − 16

4

16 − 19

4

Total (n)

100

Modal class = 7 − 10
Lower limit (l) of modal class = 7
Class size (h) = 3
Frequency (f_{1}) of modal class = 40
Frequency (f_{0}) of class preceding the modal class = 30
Frequency (f_{2}) of class succeeding the modal class = 16
Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while 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

The cumulative frequencies with their respective class intervals are as follows.
Cumulative frequency just greater than
is 19, belonging to class interval 55 − 60.
Median class = 55 − 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5
Median
= 56.67
Therefore, median weight is 56.67 kg.
Weight (in kg)

Frequency (fi)

Cumulative frequency

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

is 19, belonging to class interval 55 − 60.
Median class = 55 − 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5
Median
= 56.67
Therefore, median weight is 56.67 kg.
Exercise 14.4
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.
The frequency distribution table of less than type is as follows.
Taking upper class limits of class intervals on xaxis and their respective frequencies on yaxis, its ogive can be drawn as follows.
Daily income (in Rs)
(upper class limits)

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

During the medical checkup of 35 students of a class, their weights were recorded as follows:
Weight (in kg)

Number of students

Less than 38

0

Less than 40

3

Less than 42

5

Less than 44

9

Less than 46

14

Less than 48

28

Less than 50

32

Less than 52

35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.
The given cumulative frequency distributions of less than type are
Taking upper class limits on xaxis and their respective cumulative frequencies on yaxis, its ogive can be drawn as follows.
Here, n = 35 So
= 17.5
,
Mark the point A whose ordinate is 17.5 and its xcoordinate is 46.5. Therefore, median of this data is 46.5.
It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below.
The cumulative frequency just greater than
is 28, belonging to class interval 46 − 48.
Median class = 46 − 48
Lower class limit (l) of median class = 46
Frequency (f) of median class = 14
Cumulative frequency (cf) of class preceding median class = 14
Class size (h) = 2
Therefore, median of this data is 46.5.
Hence, the value of median is verified.
Weight (in kg)
upper class limits

Number of students
(cumulative frequency)

Less than 38

0

Less than 40

3

Less than 42

5

Less than 44

9

Less than 46

14

Less than 48

28

Less than 50

32

Less than 52

35

Here, n = 35 So
= 17.5
Mark the point A whose ordinate is 17.5 and its xcoordinate is 46.5. Therefore, median of this data is 46.5.
It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below.
Weight (in kg)

Frequency (f)

Cumulative frequency

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

is 28, belonging to class interval 46 − 48.
Median class = 46 − 48
Lower class limit (l) of median class = 46
Frequency (f) of median class = 14
Cumulative frequency (cf) of class preceding median class = 14
Class size (h) = 2
Therefore, median of this data is 46.5.
Hence, the value of median is 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 ogive.
The cumulative frequency distribution of more than type can be obtained as follows.
Taking the lower class limits on xaxis and their respective cumulative frequencies onyaxis, its ogive can be obtained as follows.
Production yield
(lower class limits)

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

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