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Class 10 SELINA Solutions Maths Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode)

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise TEST YOURSELF

Solution 1(a)

Correct option: (i)

Given A.P. is 2, 4, 6, 8, ….., 40

Here, first term, a = 2 and common difference, d = 2

Last term, 40 = 2 + (n – 1)(2)

38 = (n – 1)(2)

n = 20

Solution 1(b)

Correct option: (i) 12

Data arranged in ascending order,

8, 8, 9, 10, 12, 12, 12, 13, 15

Number of observations, n = 9 (odd)

Then, median =

Solution 1(c)

Correct option: (iv) 21

Data arranged in ascending order,

10, 12, 14, 16, 17, x

Number of observations, n = 6 (even)

Median =

Now, Mean = Median

Solution 1(d)

Correct option: (iii) 6

First six prime numbers are as follows:

2, 3, 5, 7, 11, 13

Number of observations, n = 6 (even)

Median =

Solution 1(e)

Correct option: (iii) 44

n = 80 (even)

From an ogive,

Inter-quartile range = Q3 – Q1 = 54 – 10 = 44

Solution 2

Mean of 1, 7, 5, 3, 4 and 4 =

m=4

Now, mean of 3, 2, 4, 2, 3, 3 and p = m-1 = 4-1 = 3

Therefore, 17+p = 3 x n …. Where n = 7

17+p = 21

p = 4

Arranging in ascending order:

2, 2, 3, 3, 3, 3, 4, 4

Mean = 4th term = 3

Therefore, q = 3

Solution 3

 

Date

Number

C.f.

1

5

5

2

12

17

3

20

37

4

27

64

5

46

110

6

30

140

7

31

171

8

18

189

9

11

200

10

5

205

11

0

205

12

1

206

 

(i) Mode = 5th July as it has maximum frequencies.

(ii) Total number of terms = 206

Upper quartile =

Lower quartile =

Solution 4

 

Marks

No. of students

c.f.

0-10

5

5

10-20

9

14

20-30

16

30

30-40

22

52

40-50

26

78

50-60

18

96

60-70

11

107

70-80

6

113

80-90

4

117

90-100

3

120

 

(i)

Through mark 60.5, draw a parallel line to x-axis which meets the curve at A, From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = 43

(ii) Number of students who obtained up to 75% marks in the test = 110

Number of students who obtained more than 75% marks in the test = 120 - 110 = 10

(iii) Number of students who obtained less than 40% marks in the test = 52 (from the graph; x=40, y=52)

(iv) Lower quartile = Q1 =

Solution 5

 

Weight

Frequency

C. f.

40-45

5

5

45-50

17

22

50-55

22

44

55-60

45

89

60-65

51

140

65-70

31

171

70-75

20

191

75-80

9

200

(i) Number of students weighing more than 55 kg = 200-44 = 156

Therefore, percentage of students weighing 55 kg or more

(ii) 30% of students =

Heaviest 60students in weight = 9 + 21 + 30 = 60

weight = 65 kg ( from table)

(iii) (a) underweight students when 55.70 kg is standard = 46 (approx) from graph

(b) overweight students when 55.70 kg is standard = 200- 55.70 = 154 (approx) from graph

Solution 6

 

Marks obtained(x)

No. of students (f)

c.f.

fx

5

3

3

15

6

9

12

54

7

6

18

42

8

4

22

32

9

2

24

18

10

1

25

10

Total

25

 

171

Number of terms = 25

(i) Mean =

(ii)

(iii) Mode = 6 as it has maximum frequencies i.e. 6

Solution 7

 

Monthly Income (thousands)

No. of employees

(f)

Cumulative frequency

6-7

20

20

7-8

45

65

8-9

65

130

9-10

95

225

10-11

60

285

11-12

30

315

12-13

5

320

Total

320

 

 

Number of employees = 320

(i)

Through mark 160, draw a parallel line to x-axis which meets the curve at A, From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = Rs 9.3 thousands

(ii) The number of employees with income below Rs 8500 = 95 (approx from the graph)

(iii) Number of employees with income below Rs 11500 = 305 (approx from the graph)

Therefore number of employees (senior employees) = 320-305 =15

(iv) Upper quartile =

Solution 8

  

Solution 9

  

Solution 10

 

  

Solution 11

Here the number of observations i. e is 10, which is even.'

 

 

So, the given data is 13, 35, 43, 46, 46, 50, 55, 61, 71, 80.

In the given data, 46 occurs most frequently.

Mode = 46 

Solution 12

i. The frequency distribution table is as follows:

 

Class interval

Frequency

0-10

2

10- 20

5

20-30

8

30-40

4

40-50

6

 

 

ii.

 

Class interval

Frequency

(f)

Mean value (x)

fx

0-10

2

5

10

10- 20

5

15

75

20-30

8

25

200

30-40

4

35

140

40-50

6

45

270

 

 Sf = 25 

 

 Sf = 695 

 

 

iii. Here the maximum frequency is 8 which is corresponding to class 20 - 30.

 Hence, the modal class is 20 - 30. 

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(A)

Solution 2

(a) Here n = 9

(b)

 

If marks of each student be increased by 4 then new arithmetic mean will be = 59 + 4 = 63

Solution 3

(a) The mean of 7, 11, 6, 5 and 6

(b)

If we subtract 2 from each number, then the mean will be 7-2 = 5

Solution 4

No. of terms = 5

Mean = 8

Sum of numbers = 8 x 5 = 40 .(i)

But, sum of numbers = 6+4+7+a+10 = 27+a ..(ii)

From (i) and (ii)

27+a = 40

a = 13

Solution 5

No. of terms = 5 and mean = 8

Sum of numbers = 5 x 8 = 40 ..(i)

but sum of numbers = 6+y+7+x+14 = 27+y+x .(ii)

from (i) and (ii)

27 + y + x = 40

x + y = 13

y = 13 - x

Solution 6

Age in yrs

xi

Frequency

(fi)

fixi

12

2

24

13

4

52

14

6

84

15

9

135

16

8

128

17

7

119

18

4

72

Total

40

614

 

Solution 7

No. of terms = 10

Mean = 69.5

Sum of the numbers = 69.5 x 10 = 695 ..........(i)

But sum of numbers = 72+70+x+62+ 50+71+90+64+58+82

= 619 + x ......(ii)

from (i) and (ii)

619 + x = 695

x = 76

Solution 8

Height (cm)

xi

No. of Plants

fi

fixi

50

2

100

55

4

220

58

10

580

60

f

60f

65

5

325

70

4

280

71

3

213

Total

28+f

1718 + 60f

Mean = 60.95

Solution 9

 

Wages

(Rs/day) (x)

No. of Workers

(f)

fx

50

2

100

60

4

240

70

8

560

80

12

960

90

10

900

100

6

600

Total

42

3360

 

(i) Mean remains the same if the number of workers in each category is doubled.

Mean = 80

(ii) Mean will be increased by 60% if the wages per day per worker is increased by 60%

New mean =

(iii) No change in the mean if the number of workers is doubled but if wages per worker is reduced by 40%, then

New mean =

Solution 10

 

No. of matches

(x)

No. of boxes

(f)

fx

35

6

210

36

10

360

37

18

666

38

25

950

39

21

819

40

12

480

41

8

328

Total

100

3813

 

(i)

(ii) In the second case,

New mean = 39 matches

Total contents = 39 x 100 = 3900

But total number of matches already given = 3813

Number of new matches to be added = 3900 - 3813 = 87

Solution 11

  

Solution 12

  

Solution 13

  

Solution 14

  

Solution 1(a)

Correct option: (i) 38

Let the fifth variable be x.

Then,

62 + x = 100

x = 38

Solution 1(b)

Correct option: (ii) 136 cm

Mean height of 5 students = 140 cm

Then, the total height of 5 students = 140 × 5 = 700 cm

Height of one student = 156 cm

Then, the total height of the remaining 4 students = 700 – 156 = 544

Hence, mean height of the remaining 4 students

Solution 1(c)

Correct option: (iii) 16 years

Mean age of 8 boys = 16 years

Then, sum of the ages of 8 boys = 16 × 8 = 128 years

Sum of the ages of two more boys = 18 + 14 = 32 years

Now, total age of 10 boys = 128 + 32 = 160 years

Hence, the resulting mean age

Solution 1(d)

Correct option: (iv) 18

Mean value of 15 numbers = 20

Then, the (incorrect) sum of values of 15 numbers = 20 × 15 = 300

Correct sum = 300 – 45 + 15 = 270

Therefore, correct mean

Solution 1(e)

Correct option: (iii)

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(B)

Solution 2

 

Age in years

C.I.

xi

Number of students (fi)

xifi

16 - 18

17

2

34

18 - 20

19

7

133

20 - 22

21

21

441

22 - 24

23

17

391

24 - 26

25

3

75

Total

 

50

1074

Solution 3

(i) Short - cut method

Marks

No. of boys (fi)

Mid-value xi

A = 65

di=x-A

fidi

30 - 40

10

35

-30

-300

40 - 50

12

45

-20

-240

50 - 60

14

55

-10

-140

60 - 70

12

A = 65

0

0

70 - 80

9

75

10

90

80 - 90

7

85

20

140

90 - 100

6

95

30

180

Total

70

-270

(ii) Step - deviation method

Marks

No. of boys (fi)

Mid-value xi

A = 65

fiui

30 - 40

10

35

-3

-30

40 - 50

12

45

-2

-24

50 - 60

14

55

-1

-14

60 - 70

12

A = 65

0

0

70 - 80

9

75

1

9

80 - 90

7

85

2

14

90 - 100

6

95

3

18

Total

70

-27

Here A = 65 and h = 10

Solution 4

C. I.

Frequency (fi)

Mid-value xi

A = 87.50

fiui

63 - 70

9

66.50

-3

-27

70 - 77

13

73.50

-2

-26

77 - 84

27

80.50

-1

-27

84 - 91

38

A = 87.50

0

0

91 - 98

32

94.50

1

32

98 - 105

16

101.50

2

32

105 - 112

15

108.50

3

45

Total

150

29

Here A = 87.50 and h = 7

Solution 5

C. I.

frequency

Mid-value (xi)

fixi

0-10

8

5

40

10-20

22

15

330

20-30

31

25

775

30-40

f

35

35f

40-50

2

45

90

Total

63+f

1235+35f

Solution 6

 

C.I.

Frequency

Mid value x

fx

15-25

10

20

200

25-35

20

30

600

35-45

25

40

1000

45-55

15

50

750

55-65

5

60

300

Total

75

 

2850

Solution 7

 

Class

Frequency (f)

Mid Value (x)

fx

0 - 20

7

10

70

20 - 40

p

30

30p

40 - 60

10

50

500

60 - 80

9

70

630

80 - 100

13

90

1170

Total

39 + p

 

2370 + 30p

 

Here mean = 54 ..(ii)

from (i) and (ii)

Solution 8

 

Class

Freq (f)

Mid value

fx

0-20

5

10

50

20-40

f1

30

30f1

40-60

10

50

500

60-80

f2

70

70f2

80-100

7

90

630

100-120

8

110

880

Total

30+f1+f2

 

2060+30f1+70f2

 

Now, and

from (i)

using (i) and (ii)

Solution 9

  

Solution 1(a)

Correct option: (i) 41

C.I.

Class Mark (x)

f

fx

30-40

35

2

70

40-50

45

3

135

 

 

Σf= 5

Σfx = 205

 

Alternate Method:

The mean of the set of observations lies between the range of the given data.

Since, the range of observations is from 30 to 50, only option (i), that is, 41 lies between 30 and 50.

Hence, 41 is the mean.

Solution 1(b)

Correct option: (i) 17

Let the assumed mean, A = 15

Class width, I = 10

C.I.

f

x

u = (x – A)/i

f × u

0-10

5

5

–1

–5

10-20

10

15

0

0

20-30

10

25

1

10

 

Σf = 25

 

 

Σfu = 5

Solution 1(c)

Correct option: () Back answer is 30

x

f

fx

10

10

100

20

a

20a

30

10

300

Note: All four values of ‘a’ given in the options satisfy the equation.

In fact, a can take any value apart from this.

Solution 1(d)

Correct option: (iii) x1 – x2 = 0

x

f

fx

10

x1

10x1

20

20

400

30

x2

30x2

 

Solution 1(e)

Correct option: (ii) 47

C.I.

Class Mark (x)

f

fx

40-50

45

40

1800

50-60

55

10

550

 

 

Σf= 50

Σfx = 2350

Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(C)

Solution 2

Arranging the given data in descending order:

8, 7, 6, 5, 4, 3, 3, 1, 0

The middle term is 4 which is the 5th term.

Median = 4

Solution 3

Arranging the given data in descending order:

28.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5

The middle terms are 24 and 24, 5th and 6th terms

Solution 4

Arranging in ascending order:

22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 21, 32, 32, 33, 35, 35, 36, 36, 37

(i) Middle term is 10th term i.e. 29

Median = 29

(ii) Lower quartile =

(iii) Upper quartile =

(iv) Interquartile range = q3 - q1 =35 - 26 = 9

Solution 5

 

Weight

(kg) x

no. of boys

f

cumulative frequency

37

10

10

38

14

24

39

18

42

40

12

54

41

6

60

Number of terms = 60

(i) median = the mean of the 30th and the 31st terms

(ii) lower quartile (Q1) =

(iii) upper quartile (Q3) =

(iv) Interquartile range = Q3 - Q1 = 40 - 38 = 2

Solution 6

 

Marks

(less than)

Cumulative frequency

10

5

20

24

30

37

40

40

50

42

60

48

70

70

80

77

90

79

100

80

Number of terms = 80

\Median = 40th term.

(i) Median = Through 40th term mark draw a line parallel to the x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

Value of B is the median = 40

(ii) Lower quartile (Q1) = 20th term = 18

(iii) Upper Quartile (Q3) = 60th term = 66

Solution 7

 

Height

(in cm)

No. of

pupils

Cumulative

Frequency

121 - 130

12

12

131 - 140

16

28

141 - 150

30

58

151 - 160

20

78

161 - 170

14

92

171 - 180

8

100

Number of terms = 100

Through 50th term mark draw a line parallel to the x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

Value of B is the median = 148

Median height = 148cm

Solution 8

 

Mode is in 20-30, because in this class there are 20 frequencies.

Solution 9

 

Mode is in 30-35 because it has the maximum frequency.

Solution 10

Arranging the given data in ascending order:

7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19

(i) Mode = 16 as it occurs maximum number of times.

(ii)

(iii)Total marks = 7+10+12+12+14+15+16+16+16+17+19 =

154

(iv)

Solution 11

Score

x

No. of shots

f

fx

0

0

0

1

3

3

2

6

12

3

4

12

4

7

28

5

5

25

Total

25

80

(i) Modal score = 4 as it has maximum frequency 7.

(ii)

(iii) Total score = 80

(iv)

Solution 1(a)

Correct option: (ii) 18

Arranging data in ascending order,

14, 15, 18, 21, 29

Number of observations, n = 5 (odd)

Then, median =

Solution 1(b)

Correct option: (iii) 5

Arranging data in ascending order,

0, 2, 3, 4, 6, 8, 11, 16

Number of observations, n = 8 (even)

Then, median =

Solution 1(c)

Correct option: (iv) 23

Data arranged in ascending order,

5, 15, 20, x, 28, 30, 35

Number of observations, n = 7 (odd)

Then, median =

Therefore, x = 23

Solution 1(d)

Correct option: (iii) 60 – 20

Data: 10, 20, 30, 40, 50, 60, 70 and 80

Number of observations, n = 8

Inter-quartile range = Q3 – Q1 = 60 – 20

Solution 1(e)

Correct option: (iii) 50-60

Modal class is the class with the highest frequency.

Therefore, modal class is 50-60.

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