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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:
Let us solve the problem using direct method

Number of plants


Number of houses (fi)

Class mark (xi)

fi xi

0 - 2

2 - 4

4 - 6

6 - 8

8 - 10

10 - 12

12 - 14

1

2

1

5

6

2

3

1

3

5

7

9

11

13

1

6

5

35

54

22

39

Σ fi = 20

Σ fixi = 162


Mean ( = = = 8.1
Thus the mean number of plants per house = 8.1
The method used here is direct method becau
se the numerical values of xi and fi are small.

Question-2

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

Daily wages
(in
` )

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:
(Direct method)
The following problem is solved using direct method

Daily wages
(in
` )

Number of workers (fi)

Class mark (xi)

fi xi
100 - 120
120 - 140
140 - 160
160 - 180
180 - 200
12
14
8
6
10
110
130
150
170
190
1320
1820
1200
1020
1900

Total

Σ fi = 50

Σ fixi = 7260


The mean of the given data ( = = = 145.2
The mean daily wages of the workers = 
`145.20. 
 

 

Question-3

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

Daily pocket allowance (in
`  )

11 – 13

13 – 15

15 – 17

17 – 19

19 – 21

21 – 23

23 – 25

Number of workers

7

6

9

13

f

5

4


Solution:
The mean pocket allowance = `18  

 Let us solve the problem using assumed mean method
 

Daily pocket allowance
(in
`)

Number of children

xi

di = xi - A
       = xi - 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

2f

21 - 23

 

5

22

4

20

23 – 25

4

24

6

24

 

å fi = 44 + f

   

å fidi = - 40 + 2f


Here fidi = -40 + 2f           fi = 44 + f
Mean = A + = 18
              = 18 + = 18
Þ  
  = 18 - 18
    = 0
   -40 + 2f = 0         2f = 40
     f = = 20.

 

Question-4

Thirty women are 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, clothing a suitable method.
 

Numbers of heart beats per minute

65 - 68

68 - 71

71 - 74

74 - 77

77 - 80

80 - 83

83 - 36
Number of women 2 4 3 8 7 4 2


Solution:
The given problem is solved using assumed mean method
 


Number of heart beats per minute


Number of women (fi)


Class mark (xi)

     di = xi - A
         = xi - 75.5


fidi

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

 


Σ fi = 30

   


Σ fidi = 12


Mean ( x) =  A +  where A = 75.5
                = 75.5 +
                = 75.5 + 0.4
           
(x) = 75.9
∴ The mean heart beats per minute = 75.9.

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


Solution:
 


Question-6

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

Daily Expenditure
(in
`)

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:
 

Daily expenditure (in ` )


Number of households (fi)


Class Mark (xi)

            ui =


fiui


100 - 150


4


125


-2


-8


150 - 200


5


175


-1


-5


200 - 250


12


225


0


0


250 - 300


2


275


1


2


300 - 350


2


325


2


4

 

fi = 25

   

fiui = -7


Mean = A +
              = 225 +
              = 225 - 14
              = 211


∴ The mean of daily expenditure on food = `211.  

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:
Find the mean concentration of SO2 in the air.

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


Solution:
Let us solve the problem by Step-deviation method

Concentration of SO2(in ppm)

Frequency (fi)

xi

di = xi - 0.1

ui =

fiui
0.0 - 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
0.02
0.06
0.1
0.14
0.18
0.22
-0.08
-0.04
0
0.04
0.08
0.12
-2
-1
0
1
2
3
-8
-9
0
2
8
6

Total

             
Σ f = 30

    
Σ fiui = -1

Mean (x) = A + = 0.1 +
            = 0.1 - (0.00133)
            = 0.09867


∴ The mean concentration of SO2 in the air is 0.09867 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:
 

Number of day

Number of students (fi)

xi

di = xi – A
= xi – 17

Ui = di/2

fiui

0 – 6

11

3

-14

-7

-77

6 – 10

10

8

-9

-4.5

-45

10 – 14

7

12

-5

-2.5

-17.5

14 – 20

4

17

0

0

0

20 – 28

4

24

7

3.5

14

28 – 38

3

33

16

8

24

38 – 40

1

39

22

11

11

Σ fi = 40

   Σ  fiui = -90.5


Mean (x) = A + × h = 17 + 

              = 17 – 4.525
              = 12.475
The mean number of days a student was absent = 12.48 days.

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


 Solution:


Mean (x) = A + × h
             = 70 +
             = 70 – 0.571 = 69.428
             = 69.43%

The mean literacy rate (in percentage) = 69.43%.

Question-10

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:
 

Mean = A + × h

              = 40 +
              = 40 – 4.625
              = 35.375


Mode = l + 
  × h


Here the maximum class frequency is 23, and the class corresponding frequency is 35 – 45. So the modal class is 35 – 75.

Now, lower limit(l) of modal class = 35
Class size(h) = 10
Frequency (f1) of the modal class = 23
Frequency (f0) of class preceding the model class = 21
Frequency (f2) of class succeeding the model class = 14
Mode = l +  
× h

         = 35 +
         = 35 +
         = 35 +
         = 35 + 1.82
         = 36.82.


Question-11

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components

Lifetime
(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:
                                            

Here maximum frequency is 61, the class corresponding to this 60 – 80. So, the model class is 60 – 80.

Now lower limit of the model class (l) = 60
frequency(f1) of the modal class = 61
frequency(f0) of the preceding the model class = 52
frequency(f2) of class succeeding the model class = 38
class size (h) = 20
Mode = l + × h
        = 60 + × 20
        = 60 + × 20
        = 60 +
        = 60 + 5.625
        = 65.625

Modal lifetime of the components = 65.63 hours.

Question-12

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 `)

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

 
Solution:
 

Mode l + × h


The highest frequency is 40. The modal class is 1500 - 2000.
Lower limit (l) of the modal class = 1500
Class size (h) = 500
Frequency of the modal class (f1) = 40
Frequency of the preceding modal class (f0) = 24
Frequency of the succeeding modal class (f2) = 33
Mode = l + × h
Mode = 1500 + × 500
        = 1500 +
        = 1500 +
        = 1500 + 347.83
        = 1847.83 The model monthly expenditure of the family = 
`1847.83
Mean = A +
        = 3250 -
        = 3250 - 587.50
        = 266.65

Mean monthly expenditure =  `2662.5.

Question-13

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 student 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


Solution:
Let us first find the mode.
In this problem, the maximum frequency is 10, and the class corresponding to this frequency is 30 - 35. So, the modal class is 30 - 35. Now
Lower limit (l) of modal class = 30
Class size (h)= 5
frequency (f1) of model class = 10
frequency (f0) proceeding the modal class = 9
frequency (f2) succeeding the modal class = 3
Mode = l +

Substituting the values above in the given formula,

        = 30 +
        = 30 +
        = 30 + = = 30.625

The mode of the given data is 30.6
Now calculate mean by applying direct method.

Number of students / teacher

Number of states/U.T (fi)

Classmark (xi)

di = xi - A
= xi - 325

fidi

15-20

3

17.5

-15

-45

20-25

8

22.5

-10

-80

25-30

9

27.5

-5

-45

30-35

10

32.5

0

0

35-40

3

37.5

5

15

40-45

0

42.5

10

0

45-50

0

47.5

15

0

50 - 55

2

52.5

20

40
å fi = 35 å fidi = -11.5

= A +
   = 32.5 +
   = 32.5 - 3.29 = 29.21

The mean of the given data = 29.21
Thus we interpret that the most states/ U.T. student teacher ratio of 30.6 and on an average this ratio is 29.2.

Question-14

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 batsman

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.

Solution:
The maximum frequency is 18, and the class corresponding to the maximum frequency is 4000 – 5000.
Now the model class is 4000 – 5000
Class size(h) = 1000
lower limit (f) of the model class = 4000
frequency(fo) of the class proceeding the modal class = 4
frequency(f1) of the modal class = 18
frequency(f2) of the class succeeding the modal class = 9
mode = l +
         = 4000 +
         = 4000 +
         = 4000 +
         = 4000 + 608.7
         = 4608.7

Mode = 4608.7 runs.

Question-15

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:
In this problem the maximum frequency is 20, the class corresponding to the maximum frequency is 40 – 50.
Now,
The model class is 40 – 50
Class size (h) = 10
lower limit (l) of the model class = 40
frequency(f1) of the modal class = 20
frequency(f0 )of the class proceeding the modal class = 12
frequency(f2 )of the class succeeding the modal class = 11
Mode = l +
        = 40 +
        = 40 +
        = 40 +
        = 40 + 4.71
        = 44.71 cars

∴ Mode = 44.71 cars.


Question-16

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


Solution:


n = 68, = 34. Here 125 - 145 is the class whose cumulative frequency 42 is greater than 34.

Here
L - lower limit of median class = 125
= 34 where n = number of observations
c.f - cumulative frequency of class proceeding the median class = 22
f - frequency of median class = 20
h - class size = 20
Median = l +
           = 125 +
           = 125 + 12
           = 137

The median of the given data is 137.

The mean can be calculated using assumed mean method.

Monthly consumption
(in units)

Number of consumer (fi)

Class mark (xi)

di = xi- fi
= xi - 135

fidi

65-85

4

75

-60

-240

85-105

5

95

-40

-200

105-125

13

115

-20

-260

125 - 145

20

135

0

0

145 - 165

14

155

20

280

165 - 185

8

175

40

320

185 - 205

4

195

60

240
fi = 68 fidi = 140

Mean( = A +  
            = 135 +
            = 135 + 2.06 = 137.06

The mean of the given data = 137.06
Let us find the mode as follows.
The highest frequency is 20 which corresponds to the class interval 125 - 145. So the modal class is 125 - 145.
Here
Lower limit (l) of the modal class = 125
Class size (h) = 20
frequency (f1) of the modal class = 20
frequency (f0) of the class succeeding the modal class = 13
frequency (f2) of the class succeeding the modal class = 14
mode = l +
        = 125 +
        = 125 +
        = 125 + 10.77
        = 135.77

The mode of the data = 135.77
The three measures are approximately the same in this case.


Question-17

 

Class Interval

Frequency

0 – 10

5

10 – 20

x

20 – 30

20

30 - 40

15

40 – 50

y

50 – 60

5

Total

60

 

Solution:
 

Class Interval

Frequency (fi)

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

Σ fi = 60

 


Here in this problem = n = 45 + x + y                               

It is given that total frequency is 60, thus n = 60 45 + x + y = 60 x + y = 60 - 45
            = 15(1)


The median is 28.5 which lies in the class 20-30
So,
l (lower limit) of the median class = 20
f (frequency) of the median class = 20
cf(cumulative) frequency of the class = 5 + x

proceeding 20 - 30
h (class size) = 10
median = l +
28.5 = 20 +
28.5 = 20 +
 
28.5 - 20 +  
8.5 = 17.0 = 25 - x x = 25 - 17
      = 8

substitute x = 8 in (1)
x + y = 15
8 + y = 15
y = 15 - 8
   = 7

The value of x and y are 8 and 7.
 

Question-18

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


Solution:
 

Age
(in years)

Number of policy holders

Cumulative frequency

Below 20

2

2

Below 25

6

6

Below 30

24

24

Below 35

45

45

Below 40

78

78

Below 45

89

89

Below 50

92

92

Below 55

98

98

Below 60

100

100



Here n = 100, = = 50

The observation lies in the class 35 – 40

Lower limit (l) = 35

cf – cumulative frequency of the class proceeding the interval 35 – 40 = 45.

frequency (f) of the median class 35 – 40 = 33

class size (h) = 5
median = l +
           = 35 +
           = 35 +
           = 35 + 0.758
           = 35.76

The median age is 35.76 years.

Question-19

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

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.)

Solution:

 
Now n = 40, = = 20
This observation lies in the class 144.5 - 153.5
l (lower limit of the classes) = 144.5
cf (cumulative frequency of the class proceeding 144.5 - 153.5 = 17
f (frequency of the median class) = 12
h (class size) = 9
median = l +
           = 144.5 +
           = 144.5 +
           = 144.5 +
           = 144.5 + 2.25
           = 146.75 mm

The median length of the leaves = 146.75 mm.
 

Question-20

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.

Solution:
 

Life time
(in hours)

Number of lamps
(fi)

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

 

Σ fi = 400

 

n = fi = 400
= = 200
Thus the observation lies in the class 3000 - 3500
median = l +

where l - lower limit of the median class = 3000
f = frequency of the median class = 86
cf = cumulative frequency proceeding the median class = 216
h - class size = 500
          median = 3000 +
                    = 3000 +
                    = 3000 + 406.98
                    = 3406.98

Thus the median life time of a lamp is 3406.98 hours.

Question-21

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:
 

Number of letters

Number of surnames

Cumulative frequency

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

9 + 4 = 96

16 - 19

4

96 + 4 = 100

 

100

 

Here n = 100,
Here the observation lies in the class 7 - 10
l - lower limit of the class = 7
f - frequency of the median class (7 - 10) = 40
cf - (cumulative frequency of the class proceeding (7 - 10) = 36
h - class size = 3
median = l +
           = 7 +
           = 7 +
           = 7 + 1.05
           = 8.05

median = 8.05
Median number of letters in the surnames = 8.05
Let us calculate the mean using step deviation method

This mean number of letters in the surname = 8.32
Mean = A +
            = 11.5 +
            = 11.5 - 3.18
            = 8.32

Now we have to find the modal size of the surnames
 


Number of letters


Numbered surnames (f)


1 - 4


6


4 - 7


30


7 - 10


40


10 - 13


16


13 - 16


4


16 - 19


4


Here the maximum frequency is 40. The class corresponding to this frequency is 7 - 10.
 
l = lower limit of the model class = 7
frequency(f1) of the modal class = 40
frequency(f0) of the class proceeding the modal class = 30
frequency(f2) of the class succeeding the modal class = 16
size of the class (h) = 3
mode = l +
         = 7 +
         = 7 +
         = 7 +
         = 7 + = 7 + 0.88
         = 7.88

The model size of the surname = 7.88.

Question-22

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:
 

Weight (in kg)

Number of students
(∑ 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


n = 30,
Here the observation lies in the class 55 – 60
The lower limit (l) = 55
f (frequency of the median class) = 6
cf (cumulative frequency of the class proceeding the median class = 13
Median = 
           =
           = 55 + (5/3)
           = 55 + 1.67
           = 56.67 kg.

Question-23

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

Daily Income
(in
`)

100 -120

120 -140

140 -160

160 -180

180-200

Number of students

12

14

8

6

10

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Solution:
The less than type cumulative frequency distribution is given as follows:

Daily income (in
`)

Number of students

Cumulative frequency

less than 120

12

12

less than 140

14

12 + 14 = 26

less than 160

8

26 + 8 = 34

less than 180

6

34 + 6 = 40

less than 200

10

40 + 10 = 50

Draw the ogive curve by plotting the points(120, 12), (140, 26), (160, 34), (180, 40), (200, 50)
First draw the coordinate axes with upper limits of daily income along the horizontal axis and the cumulative frequency along the vertical axis.



Question-24

During the medical check-up of 35 students of a class, their weights were recorded as follows:
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.

 

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


Solution:
Draw a less than type ogive curve, by plotting the points(38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35).
Here n = 35,

The x-coordinate of the point 17.5 will be the median.


 

Question-25

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 fans

2

8

12

24

38

16


Change the distribution to a more than type distribution, and draw its ogive.

Solution:

Production yield

kg / ha

Number of forms

Cumulative frequency


More than or equal to 50


2


100


More than or equal to 55


8


98


More than or equal to 60


12


90


More than or equal to 65


24


78


More than or equal to 70


38


54


More than or equal to 75


16


16


Now draw the ogive by plotting the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16).





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