R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive

Chapter 9 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive MCQ

Solution 1

Correct option: (d)

Range is not a measure of central tendency.

Solution 2

Correct option: (a)

Mean cannot be determined graphically.

Solution 3

Correct option: (a)

Since mean is the average of all observations, it is influenced by extreme values.

Solution 4

Correct option: (c)

Mode can be obtained graphically from a histogram.

Solution 5

Correct option: (d)

Ogives are used to determine the median of a frequency distribution.

Solution 6

Correct option: (b)

The cumulative frequency table is useful in determining the median.

Solution 7

Correct option: (b)

Median is given by the abscissa of the point of intersection of the Less than Type and More than Type cumulative frequency curves.

Solution 8

Solution 9

Solution 10

Correct option: (c)

di's are the deviations from A of midpoints of the classes.

Solution 11

Correct option: (b)

While computing the mean of the grouped data, we assume that the frequencies are centred at the class marks of the classes.

Solution 12

Correct option: (b)

Mode = (3 x median) - (2 x mean)

Solution 13

Correct option: (c)

Since the abscissa of the point of intersection of both the ogives gives the median, we have median = 20.5

Solution 14

  

Solution 15

Correct option: (c)

Class having maximum frequency is the modal class.

Here, maximum frequency = 30

Hence, the modal class is 30 - 40.

Solution 16

Solution 17

Solution 18

Correct option: (c)

Mean = 8.9

Median = 9

Mode = 3Median - 2Mean

= 3 x 9 - 2 x 8.9

= 27 - 17.8

= 9.2

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Correct option: (c)

For a symmetrical distribution, we have

Mean = mode = median

Solution 24

Correct option: (c)

Number of families having income more than Rs. 20000 = 50

Number of families having income more than Rs. 25000 = 37

Hence, number of families having income range 20000 to 25000 = 50 - 37 = 13

Solution 25

Solution 26

Correct option: (d)

Mean of 20 numbers = 0

Hence, sum of 20 numbers = 0 x 20 = 0

Now, the mean can be zero if

sum of 10 numbers is (S) and the sum of remaining 10 numbers is (-S),

sum of 11 numbers is (S) and the sum of remaining 9 numbers is (-S),

…….

sum of 19 numbers is (S) and the 20th number is (-S), then their sum is zero.

So, at the most, 19 numbers can be greater than zero.

Solution 27

Solution 28

Solution 29

(a) - (s)

The most frequent value in a data is known as mode.

 

(b) - (r)

Mean cannot be determined graphically.

 

(c) - (q)

An ogive is used to determine median.

 

(d) - (p)

Standard deviation is not a measure of central tendency.

Solution 30

Solution 31

Chapter 9 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive FA

Solution 1

Correct option: (b)

The cumulative frequency table is useful in determining the median.

Solution 2

Correct option: (c)

Mean = 27

Median = 33

Mode = 3Median - 2Mean

= 3 x 33 - 2 x 27

= 99 - 54

= 45

Solution 3

Solution 4

Solution 5

Solution 6

Number of athletes who completed the race in less than 14.6 seconds

= 2 + 4 + 15 + 54

= 75

Solution 7

Solution 8

 

The frequency table is as follows:

 

Classes

Profit (in lakhs Rs.)

Frequency

Number of shops

5 - 10

2

10 - 15

12

15 - 20

2

20 - 25

4

25 - 30

3

30 - 35

4

35 - 40

3

 

The frequency corresponding to the class 20 - 25 is 4.

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

 

Marks

Less than 10

Less than 20

Less than 30

Less than 40

Less than 50

Number of students

3

11

28

48

70

 

We plot the points (10, 3), (20, 11), (30, 28), (40, 48) and (50, 70) to get the cumulative frequency curve as follows:

 

  

Solution 14

Solution 15

 

Marks obtained

Less than 20

Less than 30

Less than 40

Less than 50

Less than 60

Less than 70

Less than 80

Less than 90

Less than 100

Number of students

2

7

17

40

60

82

85

90

100

 

We plot the points (20, 2), (30, 7), (40, 17), (50, 40), (60, 60), (70, 82), (80, 85), (90, 90) and (100, 100) to get the cumulative frequency curve as follows:

 

At y = 50, affix A.

 

Through A, draw a horizontal line meeting the curve at P.

 

Through P, a vertical line is drawn which meets OX at M.

 

OM = 56.

 

Hence, median = 56  

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Less Than Series:

Class interval

Frequency

Less than 10

2

Less than 15

14

Less than 20

16

Less than 25

20

Less than 30

23

Less than 35

27

Less than 40

30

 

We plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27) and (40, 30) to get less than ogive.

 

More Than Series:

Class interval

Frequency

More than 5

30

More than 10

28

More than 15

16

More than 20

14

More than 25

10

More than 30

7

More than 35

3

 

We plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3) to get more than ogive.

 

The two curves intersect at L. Draw LM OX.

Thus, median = OM = 16 

Solution 21

Less Than Series:

Class interval

Frequency

Less than 45

1

Less than 50

10

Less than 55

25

Less than 60

43

Less than 65

83

Less than 70

109

Less than 75

125

Less than 80

139

Less than 85

149

 

We plot the points (45, 1), (50, 10), (55, 25), (60, 43), (65, 83), (35, 27), (70, 109), (75, 125), (80, 139) and (85, 149) to get less than ogive.

 

More Than Series:

Class interval

Frequency

More than 40

149

More than 45

148

More than 50

139

More than 55

124

More than 60

106

More than 65

66

More than 70

40

More than 75

24

More than 80

10

 

We plot the points (40, 149), (45, 148), (50, 139), (55, 124), (60, 106), (65, 66), (70, 40), (75, 24) and (80, 10) to get more than ogive.

 

Solution 22

  

Chapter 9 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Ex. 18F

Solution 1

Solution 2

Class having maximum frequency is the modal class.

Here, maximum frequency = 27

Hence, the modal class is 40 - 50.

Thus, the lower limit of the modal class is 40.

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Chapter 9 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Ex. 18D

Solution 1

Let assumed mean be 35, h = 10, now we have

Class

Frequency

 

Mid-value

 

 

C.F

 

0-10

10-20

20-30

30-40

40-50

50-60

60-70

5

10

18

30

20

12

5

5

15

25

35 = A

45

55

65

-3

-2

-1

0

1

2

3

5

15

33

63

83

95

100

-15

-20

-18

0

20

24

15

 

N = 100

     

 

 

(i)Mean

 

(ii)N = 100,

Cumulative frequency just after 50 is 63

Median class is 30 - 40

l = 30, h = 10, N = 100, c = 33, f = 30

 

 

(iii)Mode = 3 × median - 2 × mean

= 3 × 35.67 - 2 × 35.6 = 107.01 - 71.2

= 35.81

Thus, Mean = 35.6, Median = 35.67 and Mode = 35.81

Solution 2

Solution 3

Solution 4

Solution 5

Let the assumed mean A be 145.Class interval h = 10.

Class

Frequency

Mid-Value

C.F.

120-130

130-140

140-150

150-160

160-170

2

8

12

20

8

125

135

145=A

155

165

-2

-1

0

1

2

-4

-8

0

20

16

2

10

22

42

50

 

N = 50

   

 

(i)Mean

 

(ii)N = 50,

Cumulative frequency just after 25 is 42

Corresponding median class is 150 - 160

Cumulative frequency before median class, c = 22

Median class frequency f = 20

(iii)Mode = 3 median - 2 mean

= 3 151.5 - 2 149.8 = 454.5 - 299.6

= 154.9

Thus, Mean = 149.8, Median = 151.5, Mode = 154.9

Solution 6

Class

Frequency

Mid-value

 

 

 

C.F.

100-120

120-140

140-160

160-180

180-200

12

14

8

6

10

110

130

150= A

170

190

-2

-1

0

1

2

-24

-14

0

6

20

12

26

34

40

50

 

N = 50

   

 

 

 

     Let assumed mean A = 150 and h = 20

(i)Mean

 

(ii) 

Cumulative frequency just after 25 is 26

Corresponding frequency median class is 120 - 140

So, l = 120, f = 14, h = 20, c = 12

                              

(iii)Mode = 3 Median - 2 Mode

= 3 138.6 - 2 145.2

= 415.8 - 190.4

= 125.4

Hence, Mean = 145.2, Median = 138.6 and Mode = 125.4

Solution 7

 

Class

Frequency

Mid-value

 

 

 

C.F.

100-150

150-200

200-250

250-300

300-350

6

7

12

3

2

125

175

225

275

325

-2

-1

0

1

2

-12

-7

0

3

4

6

13

25

28

30

 

N = 30

   

 

 

 

Let assumed mean = 225 and h = 50

(i)Mean =

(ii) 

Cumulative frequency just after 15 is 25

corresponding class interval is 200 - 250

Median class is 200 - 250

Cumulative frequency c just before this class = 13

 


Hence, Mean = 205 and Median = 208.33

Chapter 9 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Ex. 18A

Solution 1

Solution 2

Solution 3

Solution 4

We have

Class

Frequency

Mid Value

 

 

0-10

10-20

20-30

30-40

40-50

50-60

7

5

6

12

8

2

5

15

25

35

45

55

35

75

150

420

360

110

 

 

Mean

Solution 5

We have

Class

Frequency

 

Mid - value

 

25 - 35

35 - 45

45 - 55

55 - 65

65 - 75

6

10

8

12

4

30

40

50

60

70

180

400

400

720

280

 

 

Mean,

Solution 6

We have

Class

Frequency

 

Mid Value

 

0 - 100

100 - 200

200 - 300

300 - 400

400 - 500

6

9

15

12

8

50

150

250

350

450

300

1350

3750

4200

3600

 

= 50

 

Mean,

Solution 7

Solution 8

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

We have

 

Class

Frequency

 

Mid Value

 

0 - 20

7

10

70

20 - 40

 

30

30

40 - 60

12

50

600

60 - 80

=18 -

70

1260 - 70

80 - 100

8

90

720

100 - 120

5

110

550

 

= 50

 

 

Solution 17

We have, Let A = 25 be the assumed mean

Marks

Frequency

 

Mid value

 

Deviation

 

0 - 10

10 - 20

20 - 30

30 - 40

40 - 50

50 - 60

12

18

27

20

17

6

5

15

25 = A

35

45

55

-20

-10

0

10

20

30

-240

-180

0

200

340

180

 

= 100

   

= 300

Hence mean = 28

Solution 16

Let the assumed mean be 150, h = 20

Marks

Frequency

 

Mid value

 

Deviation

di = - 150

  di

100 - 120

120 - 140

140 - 160

160 - 180

180 - 200

10

20

30

15

5

110

130

150=A

170

190

-40

-20

0

20

40

-400

-400

0

300

200

 

= 80

   

 

di=-300

Hence, Mean = 146.25

Solution 19

Let A = 50 be the assumed mean, we have

Marks

Frequency

 

Mid value

 

Deviation

 

0 - 20

20 - 40

40 - 60

60 - 80

80 - 100

100 - 120

20

35

52

44

38

31

10

30

50 = A

70

90

110

-40

-20

0

20

40

60

-800

-700

0

880

1520

1860

 

= 220

   

Solution 18

We have h = 14 and let A = 35 be the assumed mean.

For calculating the mean, we prepare the table given below:

Marks

Frequency

 

Mid value

 

 

0 - 14

14 - 28

28 - 42

42 - 56

56 - 70

7

21

35

11

16

7

21

35 = A

49

63

-2

-1

0

1

2

-14

-21

0

11

32

 

? = 90

   

Thus, A = 35, = 90, h = 14 and = 8

Hence, Mean = 36.24

Solution 19

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Let h = 20 and assume mean = 550, we prepare the table given below:

Age

Frequency

 

Mid value

 

 

500 - 520

520 - 540

540 - 560

560 - 580

580 - 600

600 - 620

14

9

5

4

3

5

510

530

550 = A

570

590

610

-2

-1

0

1

2

3

-27

-9

0

4

6

15

 

= 40

   

Thus, A = 550, h = 20, and = 40,

Hence the mean of the frequency distribution is 544

Solution 26

The given series is an inclusive series, making it an exclusive series, we have

Class

Frequency

 

Mid value

 

 

24.5 - 29.5

29.5 - 34.5

34.5 - 39.5

39.5 - 44.5

44.5 - 49.5

49.5 - 54.5

54.5 - 59.5

4

14

22

16

6

5

3

27

32

37

42 = A

47

52

57

-3

-2

-1

0

1

2

3

-12

-28

-22

0

6

10

9

 

= 70

   

Thus, A = 42, h = 5, = 70 and

Hence, Mean = 39.36 years

Solution 27

The given series is an inclusive series making it an exclusive series,we get

class

Frequency

 

Mid value

 

 

4.5 - 14.5

14.5 - 24.5

24.5 - 34.5

34.5 - 44.5

44.5 - 54.5

54.5 - 64.5

6

11

21

23

14

5

9.5

19.5

29.5=A

39.5

49.5

59.5

-2

-1

0

1

2

3

-12

-11

0

23

28

15

 

= 80

   

Thus, A = 29.5, h = 10, = 80 and

Hence, Mean = 34.87 years

Solution 28

Solution 29

Chapter 9 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Ex. 18B

Solution 1

Solution 2

We prepare the frequency table, given below

Marks

No. of students

C.F.

0 - 7

7 - 14

14 - 21

21 - 28

28 - 35

35 - 42

42 - 49

3

4

7

11

0

16

9

3

7

14

25

25

41

50

 

N = = 50

 

Now,

 

The cumulative frequency is 25 and corresponding class is 21 - 28.

Thus, the median class is 21 - 28

l = 21, h = 7, f = 11, c = C.F. preceding class 21 - 28 is 14 and  = 25

 


Hence the median is 28.

Solution 3

We prepare the frequency table given below:

Daily wages

Frequency

C.F.

0 - 100

100 - 200

200 - 300

300 - 400

400 - 500

40

32

48

22

8

40

72

120

142

150

 

N = = 150

 

Now, N = 150, therefore

The cumulative frequency just greater than 75 is 120 and corresponding class is 200 - 300.

Thus, the median class is 200 - 300

l = 200, h = 100, f = 48

c = C.F. preceding median class = 72 and

 

Hence the median of daily wages is Rs. 206.25.

Hence the median is 28.
Solution 4

We prepare the frequency table, given below:

Class

Frequency

 

C.F

5 - 10

10 - 15

15 - 20

20 - 25

25 - 30

30 - 35

35-40

40 - 45

5

6

15

10

5

4

2

2

5

11

26

36

41

45

47

49

 

= 49

 

Now, N = 49

The cumulative frequency just greater than 24.5 is 26 and corresponding class is 15 - 20.

Thus, the median class is 15 - 20

l = 15, h = 5, f = 15

c = CF preceding median class = 11 and

 

Median of frequency distribution is 19.5

Solution 5

We prepare the cumulative frequency table as given below:

Consumption

Frequency

 

C.F

65 - 85

85 - 105

105 - 125

125 - 145

145 - 165

165 - 185

185 - 205

4

5

13

20

14

7

4

4

9

22

42

56

63

67

 

N = = 67

 

Now, N = 67

The cumulative frequency just greater than 33.5 is 42 and the corresponding class 125 - 145.

Thus, the median class is 125 - 145

l = 125, h = 20, and c = CF preceding the median class = 22,  = 33.5

 

Hence median of electricity consumed is 136.5

Solution 6

Frequency table is given below:

Height

Frequency

 

C.F

135 - 140

140 - 145

145 - 150

150 - 155

155 - 160

160 - 165

165 - 170

170 - 175

6

10

18

22

20

15

6

3

6

16

34

56

76

91

97

100

 

N = =100

 

 

N = 100,

The cumulative frequency just greater than 50 is 56 and the corresponding class is 150 - 155

Thus, the median class is 150 - 155

l = 150, h = 5, f = 22, c = C.F.preceding median class = 34

 

Hence, Median = 153.64

Solution 7

The frequency table is given below. Let the missing frequency be x.

Class

Frequency

 

C.F

0 - 10

10 - 20

20 - 30

30 - 40

40 - 50

5

25

x

18

7

5

30

30 + x

48 + x

55 + x

 

Median = 24 Median class is 20 - 30

l = 20, h = 10, f = x, c = C.F. preceding median class = 30

Hence, the missing frequency is 25.

Solution 8

Solution 9

 

Solution 10

Let be the frequencies of class intervals 0 - 10 and 40 - 50

Median is 32.5 which lies in 30 - 40, so the median class is 30 - 40

l = 30, h = 10, f = 12, N = 40 and

Solution 11

The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get

Class

Frequency

C.F

18.5 - 25.5

25.5 - 32.5

32.5 - 39.5

39.5 - 46.5

46.5 - 53.5

53.5 - 60.5

35

96

68

102

35

4

35

131

199

301

336

340

 

fi = N = 340

 

The cumulative frequency just greater than 170 is 199 and the corresponding class is 32.5 - 39.5.

Median class is 32.5 - 39.5

l = 32.5, h = 7, f = 68, c = C.F. of preceding median class = 131

 

Hence median is 36.5 years

Solution 12

Given series is in inclusive form converting it into exclusive form and preparing the cumulative frequency table, we get

Wages per day

(in Rs)

Frequency

 

C.F

60.5 - 70.5

70.5 - 80.5

80.5 - 90.5

90.5 - 100.5

100.5 - 110.5

110.5 - 120.5

5

15

20

30

20

8

5

20

40

70

90

98

 

fi = N =98

 

The cumulative frequency just greater than 49 is 70 and corresponding class is 90.5 - 100.5.

median class is 90.5 - 100.5

l = 90.5, h = 10, f = 30, c = CF preceding median class = 40

 


Hence, Median = Rs 93.50

Solution 13

The given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get

Marks

Frequency

 

C.F

0.5 - 5.5

5.5 - 10.5

10.5 - 15.5

15.5 - 20.5

20.5 - 25.5

25.5 - 30.5

30.5 - 35.5

35.5 - 40.5

40.5 - 45.5

7

10

16

32

24

16

11

5

2

7

17

33

65

89

105

116

121

123

 

fi = N =123

 

The cumulative frequency just greater than 61.5 is 65.

The corresponding median class is 15.5 - 20.5.

Then the median class is 15.5 - 20.5

l = 15.5, h = 5, f = 32, c = C.F. preceding median class = 33

 

Hence, Median = 19.95

Solution 14

 

Marks

Frequency

 

C.F

0 - 10

10 - 20

20 - 30

30 - 40

40 - 50

50 - 60

60 - 70

70 - 80

12

20

25

23

12

24

48

36

12

32

57

80

92

116

164

200

 

N =

 

The cumulative frequency just greater than 100 is 116 and the corresponding class is 50 - 60.

Thus the median class is 50 - 60

l = 50, h = 10, f = 24, c = C.F. preceding median class = 92,  = 100

 

Hence, Median = 53.33

Chapter 9 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Ex. 18C

Solution 1

Solution 2

Solution 3

Solution 4

As the class 26 - 30 has maximum frequency so it is modal class

Hence, mode = 28.5

Solution 5

As the class 1500 - 2000 has maximum frequency, so it os modal class

Hence the average expenditure done by maximum number of workers = Rs. 1820

Solution 6

As the class 5000 - 10000 has maximum frequency, so it is modal class

Hence, mode = Rs. 7727.27

Solution 7

As the class 15 - 20 has maximum frequency so it is modal class.

Hence mode = 17.3 years

Solution 8

As the class 85 - 95 has the maximum frequency it is modal class

Hence, mode = 85.71

Solution 9

The given series is converted from inclusive to exclusive form and on preparing the frequency table, we get

 

Class

Frequency

0.5 - 5.5

5.5- 10.5

10.5 - 15. 5

15.5 - 20.5

20.5 - 25. 5

25.5 - 30.5

30.5 - 35.5

35.5 - 40.5

40.5 - 45.5

45.5 - 50.5

3

8

13

18

28

20

13

8

6

3

As the class 20.5 - 25.5 has maximum frequency, so it is modal class

 

Hence, mode = 23.28

Solution 10

Chapter 9 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Ex. 18E

Solution 1

 

We plot the points (10, 5), (20, 8), (30, 12), (40, 15), (50, 18), (60, 22), (70, 29), (80, 38), (90, 45) and (100, 53) to get the 'less than type' ogive as follows:

 

At y = 26.5, affix A.

 

Through A, draw a horizontal line meeting the curve at P.

 

Through P, a vertical line is drawn which meets OX at M.

 

OM = 68 units

 

Hence, median marks = 68

Solution 2

 

Number of wickets

Less than 15

Less than 30

Less than 45

Less than 60

Less than 75

Less than 90

Less than 105

Less than 120

Number of bowlers

2

5

9

17

39

54

70

80

 

We plot the points (15, 2), (30, 5), (45, 9), (60, 17), (75, 39), (90, 54), (105, 70) and (120, 80) to get the 'less than type' ogive as follows:

 

At y = 40, affix A.

 

Through A, draw a horizontal line meeting the curve at P.

 

Through P, a vertical line is drawn which meets OX at M.

 

OM = 78 units

 

Hence, median number of wickets = 78

Solution 3

 

We plot the points (0, 100), (10, 96), (20, 90), (30, 80), (40, 70), (50, 45), (60, 23) and (70, 5) to get the 'more than type' ogive as follows:

 

At y = 50, affix A.

 

Through A, draw a horizontal line meeting the curve at P.

 

Through P, a vertical line is drawn which meets OX at M.

 

OM = 47 units

 

Hence, median marks = 47

Solution 4

 

We plot the points (135, 50), (140, 45), (145, 37), (150, 28), (155, 16) and (160, 2) to get the 'more than type' ogive as follows:

 

At y = 25, affix A.

Through A, draw a horizontal line meeting the curve at P.

 

Through P, a vertical line is drawn which meets OX at M.

 

OM = 151 units

 

Hence, median height = 151 cm

Solution 5

 

We plot the points (140, 156), (160, 153), (180, 145), (200, 130), (220, 90), (240, 40) and (260, 10) to get the 'more than type' ogive as follows:

 

At y = 78, affix A.

 

Through A, draw a horizontal line meeting the curve at P.

 

Through P, a vertical line is drawn which meets OX at M.

 

OM = 226 units

 

Hence, median consumption of electricity = 226 units

Solution 6

 

We plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) to get the 'more than type' ogive as follows:

 

At y = 50, affix A.

Through A, draw a horizontal line meeting the curve at P.

 

Through P, a vertical line is drawn which meets OX at M.

 

OM = 70.5 units

 

Hence, median production yield = 70.5 kg/ha

Solution 7

Less Than Series:

Class interval

Cumulative Frequency

Less than 200

5

Less than 300

11

Less than 400

22

Less than 500

35

Less than 600

40

Less than 700

44

Less than 800

47

Less than 900

49

 

We plot the points (200, 5), (300, 11), (400, 22), (500, 35), (600, 40), (700, 44), (800, 47) and (900, 49) to get 'less than type' ogive.

 

More Than Series:

Class interval

Frequency

More than 100

49

More than 200

44

More than 300

38

More than 400

27

More than 500

14

More than 600

9

More than 700

5

More than 800

2

 

We plot the points (100, 49), (200, 44), (300, 38), (400, 27), (500, 14), (600, 9), (700, 5) and (800, 2) to get more than ogive.

 

Solution 8

More than series


 

We plot the points (400, 230), (450, 210), (500, 175), (550, 135), (600, 103), (650, 79), (700, 52), (750, 34)

 

 

Hence,

Take a point A(0, 115) on the y-axis and draw AP||x-axis meeting the curve at P, Draw PM x-axis intersecting x-axis at M

Then,OM = 590

Hence median = 590

Solution 9

(i) Less than series:


Marks

No. of students

Less than 5

Less than 10

Less than 15

Less than 20

Less than 25

Less than 30

Less than 35

Less than 40

Less than 45

Less than 50

2

7

13

21

31

56

76

94

98

100

Plot the points (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98) and (50, 100)

                    

Join these points free hand to get the curve representing "less than" cumulative curve.

(ii)From the given table we may prepare the 'more than' series as shown below

 

Marks

No. of students

More than 45

More than 40

More than 35

More than 30

More than 25

More than 20

More than 15

More than 10

More than 5

More than 0

2

6

24

44

69

79

87

93

98

100

 

Now, on the same graph paper as above, we plot the point (0, 100), (5, 98), (10, 93), (15, 87), (20, 79), (25, 69), (30, 44), (35, 24) and (40, 6) and (45, 2)

Join these points free hand to get required curve

Here

Two curves intersect at point P(28, 50)

Hence, the median = 28

Solution 10

We may prepare less than series and more than series

(i)Less than series

Height in (cm)

Frequency

Less than 140

Less than 144

Less than 148

Less than 152

Less than 156

Less than 160

Less than 164

Less than 168

Less than 172

Less than 176

Less than 180

0

3

12

36

67

109

173

248

330

416

450

Now on graph paper plot the points (140, 0), (144, 3), (148, 12), (152, 36), (156, 67), (160, 109), (164, 173), (168, 248), (172, 330), (176, 416), (180, 450)

 

(ii)More than series

Height in cm

C.F.

More than 140

More than 144

More than 148

More than 152

More than 156

More than 160

More than 164

More than 168

More than 172

More than 176

More than 180

450

447

438

414

383

341

277

202

120

34

0

Now on the same graph plot the points (140, 450), (144, 447), (148, 438), (152, 414), (156, 383), (160, 341), (164, 277), (168, 202), (172, 120), (176, 34), (180, 0)

 

The curves intersect at (167, 225).

Hence, 167 is the median.