NCERT Solutions for CBSE Class 10 Mathematics chapter 14 - Statistics

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Chapter 14 - Statistics Excercise Ex. 14.1

Solution 1
Let us find class marks (xi) for each interval by using the relation.

Now we may compute xi and fixi as following
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 we may observe that


So, mean number of plants per house is 8.1.
We have used here direct method as values of class marks (xi) and fi are small.

Solution 2
Let us find class mark for each interval by using the relation.


Class size (h) of this data = 20
Now taking 150 as assured mean (a) we may calculate di, ui and fiui as following.
Daily wages (in Rs) Number of workers (fi) xi di = xi - 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 we may observe that

So mean daily wages of the workers of the factory is Rs.145.20
Solution 3
We may find class mark (xi) for each interval by using the relation.


Given that mean pocket allowance = Rs.18
Now taking 18 as assured mean (a) we may calculate di and fidi as following.
Daily pocket allowance (in Rs.) Number of children fi Class mark xi di = 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 2 f
21 - 23 5 22 4 20
23 - 25 4 24 6 24
Total     2f - 40

From the table we may obtain

Hence the missing frequency f is 20.
Solution 4
We may find class mark of each interval (xi) by using the relation.

Class size h of this data = 3
Now taking 75.5 as assumed mean (a) we may calculate di, ui, fiui as following.

Number of heart beats per minute Number of women fi xi di = xi -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

Now we may observe from table that

So mean heart beats per minute for these women are 75.9 beats per minute.
Solution 5
Number of mangoes Number of boxes
fi
50 - 52 15
53 - 55 110
56 - 58 135
59 - 61 115
62 - 64 25

We may observe that class intervals are not continuous. There is a gap of 1 between two class intervals. So we have to add to upper class limit and subtract from lower class limit of each interval.
And class mark (xi) may be obtained by using the relation

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculate di, ui, fiui as following -

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


Now we may observe that

 
Clearly, mean number of mangoes kept in a packing box is 57.19.
We have chosen step deviation method here as values of fi, di are big and also there is a common multiple between all di.

Solution 6

We may calculate cla

mark (xi) for each interval by using the relation


Class size = 50

Now taking 225 as assumed mean (a) we may calculate di, ui, fiui as following

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

Now we may observe that -

Solution 7
Concentration of SO2 (in ppm) Frequency Class mark xi di = xi - 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

Solution 8
We may find class mark of each interval by using the relation

 

Now taking 16 as assumed mean (a) we may calculate di and fidi as following 


Number of days

Number of students
fi

xi

di = xi - 16

fidi

0 - 6

11

3

-13

-143

6 -10

10

8

-8

-80

10 - 14

7

12

-4

-28

14 - 20

4

16

0

0

20 - 28

4

24

8

32

28 - 38

3

33

17

51

38 - 40

1

39

23

23

Total

40





-145



Now we may observe that

 

 
So, mean number of days is 12.38 days, for which a student was absent.

Solution 9

We may find class marks by using the relation



 

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) we may calculate di, ui, and fiui as following


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


Now we may observe that

So, mean literacy rate is 69.43%.

Chapter 14 - Statistics Excercise Ex. 14.2

Solution 1

We may compute class marks (xi) as per the relation



 
Now taking 30 as assumed mean (a) we may calculate di and fidi as following.


Age (in years)

Number of patients
fi

class mark
xi

di = xi - 30

fidi

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 we may observe that


Clearly, 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.
As we may observe that maximum class frequency is 23 belonging to class interval 35 - 45.
So, modal class = 35 -  45
Lower limit (l) of modal class = 35
Frequency (f1) of modal class = 23
Class size (h) = 10
Frequency (f0) of class preceding the modal class = 21
Frequency (f2) of class succeeding the modal class = 14


Clearly mode is 36.8.It represents that maximum number of patients admitted in hospital were of 36.8 years.
Solution 2
From the data given as above we may observe that maximum class frequency is 61 belonging to class interval 60 - 80.
So, modal class = 60 - 80
Lower class limit (l) of modal class = 60
Frequency (f1) of modal class = 61
Frequency (f0) of class preceding the modal class = 52
Frequency (f2) of class succeeding the modal class = 38
Class size (h) = 20


 
So, modal lifetime of electrical components is 65.625 hours.
Solution 3
We may observe from the given data that maximum class frequency is 40 belonging to 1500 - 2000 intervals.
So, modal class = 1500 - 2000
Lower limit (l) of modal class = 1500
Frequency (f1) of modal class = 40
Frequency (f0) of class preceding modal class = 24
Frequency (f2) of class succeeding modal class = 33
Class size (h) = 500

 

Now we may find class mark as


Class size (h) of give data = 500
Now taking 2750 as assumed mean (a) we may calculate di, ui and fiui as following



Expenditure
(in Rs)

Number of families

fi


xi

di = xi - 2750



fiui

 


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


Now from table may observe that


So, mean monthly expenditure was Rs.2662.50.
Solution 4
We may observe from the given data that maximum class frequency is 10 belonging to class interval 30 - 35.
So, modal class = 30 - 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f1) of modal class = 10
Frequency (f0) of class preceding modal class = 9
Frequency (f2) of class succeeding modal class = 3


It represents that most of states/U.T have a teacher  student ratio as 30.6

Now we may find class marks by using the relation


Now taking 32.5 as assumed mean (a) we may calculate di, ui and fiui as following.


Number of students
per teacher

Number of states/U.T
(fi)

 

xi

 

di = xi - 32.5

 

 

fiui


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




So mean of data is 29.2
It represents that on an average teacher - student ratio was 29.2.
Solution 5
From the given data we may observe that maximum class frequency is 18 belonging to class interval 4000 - 5000.
So, modal class = 4000 - 5000
Lower limit (l) of modal class = 4000
Frequency (f1) of modal class = 18
Frequency (f0) of class preceding modal class = 4
Frequency (f2) of class succeeding modal class = 9
Class size (h) = 1000

So mode of given data is 4608.7 runs.
Solution 6
From the given data we may observe that maximum class frequency is 20 belonging to 40 - 50 class intervals.
So, modal class = 40 - 50
Lower limit (l) of modal class = 40
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 12
Frequency (f2) of class succeeding modal class = 11
Class size = 10

So mode of this data is 44.7 cars.

Chapter 14 - Statistics Excercise Ex. 14.3

Solution 1
We may find class marks by using the relation


Taking 135 as assumed mean (a) we may find di, ui, fiui, according to step deviation method as following


Monthly consumption
(in units)

Number of consumers (f i)

xi class mark

di = xi - 135

 


 

fiui 

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 we may observe that

 


Now from table it is clear that 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 (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 13
Frequency (f2) of class succeeding the modal class = 14


We know that
3 median = mode + 2 mean
= 135.76 + 2 (137.058)
= 135.76 + 274.116
= 409.876
Median = 136.625
So median, mode, mean of given data is 136.625, 135.76, 137.05 respectively.

Solution 2
We may find cumulative frequency for the given data as following


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 - 5

y

40 + x + y

50 - 60

5

45 + x + y

Total (n)

60




It is clear that n = 60

45 + x + y = 60

x + y = 15        (1)
Median of data is given as 28.5 which lies in interval 20 - 30.
So, 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 values of x and y are 8 and 7 respectively.

Solution 3
Here class width is not same. There is no need to adjust the frequencies according to class intervals. Now given frequency table is of less than type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years, we can define class intervals with their respective cumulative frequency as below


Age (in years)

Number of policy holders (fi)

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)

 





Now from table we may observe that n = 100.

Cumulative frequency (cf) just greater than  is 78 belonging to interval 35 - 40
So, 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


So, median age is 35.76 years.
Solution 4
The given data is not having continuous class intervals. We can observe that difference between two class intervals is 1. So, we have to add and subtract

  to upper class limits and lower class limits.
Now continuous class intervals with respective cumulative frequencies can be represented as below 



Length (in mm)

Number or leaves fi

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


From the table we may observe that cumulative frequency just greater then
  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



So, median length of leaves is 146.75 mm.
Solution 5
We can find cumulative frequencies with their respective class intervals as below -


Life time

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

Total (n)

400



Now we may observe that 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


So, median life time of lamps is 3406.98 hours.
Solution 6
We can find cumulative frequencies with their respective class intervals as below 


Number of letters

Frequency (fi)

 


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




Now we may observe that 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

Now we can find class marks of given class intervals by using relation


Taking 11.5 as assumed mean (a) we can find di, ui and fiui according to step deviation method as below.

Number of
letters

Number of
surnames

xi

xi - a

 

fiui

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


fiui = - 106
fi = 100


We know that
3 median = mode + 2 mean
3(8.05) = mode + 2(8.32)
24.15 - 16.64 = mode
7.51 = mode
So, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.51.

Solution 7
We may find cumulative frequencies with their respective class intervals as below


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



So, median weight is 56.67 kg.

Chapter 14 - Statistics Excercise Ex. 14.4

Solution 1
We can find frequency distribution table of less than type as following -


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


Now taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis we can draw its ogive as following -
 
Solution 2
The given cumulative frequency distributions of less than type is -


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


Now taking upper class limits on x-axis and their respective cumulative frequency on y-axis we may draw its ogive as following -

>

Now mark the point A whose ordinate is 17.5 its x-coordinate is 46.5. So median of this data is 46.5.
We may observe that difference between two consecutive upper class limits is 2. Now we may obtain class marks with their respective frequencies 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




Now 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


So median of this data is 46.5
Hence, value of median is verified.

Solution 3
We can obtain cumulative frequency distribution of more than type as following -


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


Now taking lower class limits on x-axis and their respective cumulative frequencies on y-axis we can obtain its ogive as following.