RS Aggarwal Solutions Chapter 9 Mean, Mode and Median Exercise 9B Class 10 Maths

Chapter Name

RS Aggarwal Chapter 9 Mean, Mode and Median

Book Name

RS Aggarwal Mathematics for Class 10

Other Exercises

  • Exercise 9A
  • Exercise 9C
  • Exercise 9D
  • Exercise 9E
  • Exercise 9F

Related Study

NCERT Solutions for Class 10 Maths

Exercise 9B Solutions

1. In a hospital, the ages of diabetic patient were recorded as follows. Find the median age.

Age (in years)

0 - 15

15 - 30

30 - 45

45 - 60

60 - 75

No. of patients

5

20

40

50

25

Solution 

We prepare the cumulative frequency table, as shown below:

Age (in years)

Number of patients (fi)

Cumulative Frequency (cf)

0 - 15

5

5

15 - 30

20

25

30 - 45

40

65

45 - 60

50

115

60 – 75

25

140

Total

N = Σ f= 140

 

 Now, N = 140

⇒ N/2 = 70

The cumulative frequency just greater than 70 is 115 and the corresponding class is 45 – 60.

Thus, the median class is 45 – 60.

∴ l = 45, h = 15, f = 50, N = 140 and cf = 65.

Now,

Median = l + ((N/2 – cf)/f) × h

= 45 + (140/2 – 65)/50 × 15

= 45 + (70 – 65)/50 × 15

= 45 + 1.5

= 46.5

Hence, the median age is 46.5 years.


2. Compute mean from the following data:

Marks

0 - 7

7 - 14

14 - 21

21 - 28

28 - 35

35 - 42

42 - 49

Number of students

3

4

7

11

0

16

9

Solution

Class

Frequency (f)

Cumulative Frequency (cf)

0 - 7

3

3

7 - 14

4

7

14 - 21

7

14

21 - 28

11

25

28 – 35

0

25

35 – 42

16

41

42 - 49

9

50

 

N = Σf = 50

 

Now, N = 50

⇒ N/2 = 25

The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 - 42

Thus, the median class is 35 - 42.

∴ l = 35, h = 7, f = 16, cf = c.f. of preceding class=25 and N/2 = 25.

Now,

Median = l + ((N/2 – cf)/f) × h

= 35 + 7 × (25 – 25)/16

= 35 + 0

= 35.

Hence, the median age is 46.5 years


3. The following table shows the daily wages of workers in a factory:

Daily wages in (₹)

0 - 100

100 - 200

200 - 300

300 - 400

400 - 500

Number of workers

40

32

48

22

8

Find the median daily wage income of the workers.

Solution 

Class

Frequency (f)

Cumulative Frequency (cf)

0 – 100

40

40

100 - 200

32

72

200 - 300

48

120

300 - 400

22

142

400 - 500

8

150

 

N = Σf = 150

 

Now, N = 150

⇒ N/2 = 75

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

Thus, the median class is 200 – 300.

∴ l = 200, h = 100, f = 48, cf = c.f. of preceding class=72 and N/2 = 75.

Now,

Median, M = l + {h × ((N/2 – cf)/f)} × h

 = 200 + {100 × (75 – 72)/48}

= 200 + 6.25

= 206.25

Hence, the median daily wage income of the workers is RS 206.25.


4. Calculate the median from the following frequency distribution table:

Class

5 - 10

10 - 15

15 - 20

20 - 25

25 - 30

30 - 35

35 - 40

40 - 45

Frequency

5

6

15

10

5

4

2

2

Solution 

Class

Frequency (f)

Cumulative Frequency (cf)

5 - 10

5

5

10 - 15

6

11

15 - 20

15

26

20 - 25

10

36

25 – 30

5

41

30 – 35

4

45

35 – 40

2

47

40 - 45

2

49

 

N = Σf = 49

 

Now, N = 49

⇒ N/2 = 24.5

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

Thus, the median class is 15 – 20.

∴ l = 15, h = 5, f = 15, cf = c.f. of preceding class=11 and N/2 = 24.5.

Now,

Median, M = l + {h × ((N/2 – cf))/f)}

= 15 + {5 × (24.5 – 11)/15)}

= 15 + 4.5

= 19.5

Hence, the median = 19.5.


5. Given below is the number of units of electricity consumed in a week in a certain locality:

Class

65 - 85

85 - 105

105 - 125

125 - 145

145 - 165

165 - 185

185 - 200

Frequency

4

5

13

20

14

7

4

Calculate the median.

Solution 

Class

Frequency (f)

Cumulative Frequency (cf)

65 - 85

4

4

85 - 105

5

9

105 - 125

13

22

125 - 145

20

42

145 - 165

14

56

165 – 185

7

63

185 - 205

4

67

 

N = Σf = 67

 

Now, N = 67

⇒ N/2 = 33.5

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

Thus, the median class is 125 - 145.

∴ l = 125, h = 20, f = 20, 

cf = c.f. of preceding class =22 and N/2 = 33.5. 

Now, 

Median, M = l + {h 

Median, M = l + {h × ((N/2 – cf))/f)}

= 125 + {20 × (33.5 – 22)/20)}

= 125 + 11.5

= 136.5

Hence, the median = 136.5.


6. Calculate the median from the following data:

Height (in cm)

135 - 140

140 - 145

145 - 150

150 - 160

155 - 160

160 - 165

165 - 170

170 - 175

Frequency

6

10

18

22

20

15

6

3

Solution

Class

Frequency (f)

Cumulative Frequency (cf)

135 - 140

6

6

140 - 145

10

16

145 - 150

18

34

150 - 155

22

56

155 - 160

20

76

160 – 165

15

91

165 – 170

6

97

170 - 175

3

100

 

N = Σf = 100

 

Now, N = 100

⇒ N/2 = 50.

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, cf = c.f. of preceding class=34 = N/2 = 50.

Now,

Median, M = l + {h × ((N/2 – cf)/f)}

= 150 + {5 × ((50 – 34)/22)}

= 150 + 3.64

= 153.64

Hence, the median = 153.64


7. Calculate the missing frequency from the following distribution, it being given that the median of distribution is 24.

Class

0 - 10

10 - 20

20 - 30

30 - 40

40 - 50

Frequency

5

25

?

18

7

Solution 

Class

Frequency (fi)

Cumulative Frequency (cf)

0 - 10

5

5

10 - 20

25

30

20 - 30

x

x + 30

30 - 40

18

x + 48

40 - 50

7

x + 55

Median is 24 which lies in 20 – 30

∴ Median class=20 – 30

Let the unknown frequency be x.

Here, l = 20, n/2 = (x + 55)/2, c.f. of the preceding class=c.f = 30, f = x, h = 10

Now,

Median, M = {l + (n/2 – cf)/f} × h

⇒ 24 = 20 + ((x + 55)/2 – 30)/x × 10

⇒ 24 = 20 + ((x + 55 – 60)/2)/x × 10

⇒ 24 = 20 + (x – 5)/2x × 10

⇒ 24 = 20 + (5x – 25)/x

⇒ 24 = (20x + 5x – 25)/x

⇒ 24x = 25x – 25

⇒ -x = - 25

⇒ x = 25

Hence, the unknown frequency is 25.


8. The median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.

Class

0 - 5

5 - 10

10 - 15

15 - 20

20 - 25

25 - 30

30 - 35

35 - 40

Frequency

12

a

12

15

b

6

6

4

Solution

Class

Frequency (f)

Cumulative Frequency (cf)

0 - 5

12

12

5 - 10

a

12 + a

10 - 15

12

24 + a

15 - 20

15

39 + a

20 – 25

b

39 + a + b

25 – 30

6

45 + a + b

30 – 35

6

51 + a + b

35 – 40

4

55 + a + b

Total

N = Σfi = 70

 

Let a and b be the missing frequencies of class intervals 5 – 10 and 20 – 25 respectively.

Then, 55 + a + b = 70 ⇒ a + b = 15 ...(1)

Median is 16, which lies in 15 – 20. So, the median class is 15 – 20.

∴ l = 15, h = 5, N = 70, f = 15 and cf = 24 + a

Now,

Median, M = l + (N/2 – cf/f)) × h

⇒ 16 = 15 + (70/2 – (24 + a)/15) × 5

⇒ 16 = 15 + (35 – 24 – a)/3

⇒ 16 = 15 + (11 – a)/3

⇒ 16 – 15 = (11 – a)/3

⇒ 1 × 3 = 11 – a

⇒ a = 11 – 3

⇒ a = 8

∴ b = 15 – a [From (1)]

⇒ b = 15 – 8

⇒ b = 7

Hence, a = 8, and b = 7.


9. In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies x and y. 

Runs scored

2500 - 3500

3500 - 4500

4500 - 5500

5500 - 6500

6500 - 7500

7500 - 8500

Number of batsman

5

x

y

12

6

2

Solution 

We prepare the cumulative frequency table, as shown below:

Runs scored

Number of batsman (fi)

Cumulative Frequency (cf)

2500 - 3500

5

5

3500 – 4500

x

5 + x

4500 - 5500

y

5 + x + y

5500 - 6500

12

17 + x + y

6500 - 7500

6

23 + x + y

7500 - 8500

2

25 + x + y

Total

N = Σfi = 60

 

Let x and y be the missing frequencies of class intervals 3500 – 4500 respectively. Then,

25 + x + y = 60 ⇒ x + y = 35 ...(1)

Median is 5000, which lies in 4500 – 5500. So, the median class is 4500 – 5500.

∴ l = 4500, h = 1000, N = 60, f = y and cf = 5 + x

Now,

Median, M = l + ((N/2 – cf)/f) × h

⇒ 5000 = 4500 + (60/2 – (5 + x)/y) × 1000

⇒ 5000 – 4500 = (30 – 5 – x)/y × 1000

⇒ 500 = (25 – x)/y × 1000

⇒ y = 50 – 2x

⇒ 35 – x = 50 – 2x [From (1)]

⇒ 2x – x = 50 – 35

⇒ x = 15

∴ y = 35 – x [From (1)]

⇒ y = 35 – 15

⇒ y = 20

Hence, x = 15 and y = 20.


10. If the median of the following frequency distribution is 32.5, find the values of f1and f2.

Class

0 - 10

10 - 20

20 – 30

30 - 40

40 - 50

50 – 60

60 - 70

Total

Frequency

fi

5

9

12

f2

3

2

40

 Solution 

Class

Frequency (f)

Cumulative Frequency (cf)

0 – 10

f1

f1

10 - 20

5

f1 + 5

20 - 30

9

f1 + 14

30 – 40

12

f1 + 26

40 - 50

f2

f1 + f2 + 26

50 - 60

3

f1 + f2 + 29

60 - 70

2

f1 + f2 + 31

 

N = Σf = 40

 

Now, f1 + f2 + 31 = 40

⇒ f1 + f2 = 9

⇒ f2 = 9 – f1

The median is 32.5 which lies in 30 – 40.

Hence, Median class=30 - 40

Hence, l = 30, N/2 = 40/2 = 20, f = 12 and cf = 14 + f1

Now, Median = 32.5.

⇒ l = ((N/2 – cf)/f) × h = 32.5

⇒ 30 + {(20 – 14 + f1)/12) × 10 = 32.5

⇒ (6 – f1)/12 × 10 = 2.5

⇒ (60 – 10f1)/12 = 2.5

⇒ 60 – 10f1 = 30

⇒ 10f1 = 30

⇒ f1 = 3

From equation (i), we have:

f2 = 9 – 3

⇒ f2 = 6


11. Calculate the median for the following data:

Class

19 - 25

26 - 32

33 - 39

40 - 46

47 - 53

54 - 60

Frequency

35

96

68

102

35

4

Solution

First, we will convert the data into exclusive form.

Class

Frequency (f)

Cumulative frequency (cf)

18.5 – 25.5

35

35

25.5 – 32.5

96

131

32.5 – 39.5

68

199

39.5 – 46.5

102

301

46.5 – 53.5

35

336

53.5 – 60.5

435

340

 

N = Σf = 340

 

Now, N = 340

⇒ N/2 = 70

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

Thus, the median class is 32.5 – 39.5

∴ l = 32.5, h = 7, f = 68, cf = c.f. of preceding class=131 and N/2 = 170.

∴ Median, M = l + {h × (N/2 – cf/f))}

= 32.5 + {7 × ((170 – 131)/68)}

= 32.5 + 4.01

= 36.51

Hence, the median = 36.51


12. Find the median wages for the following frequency distribution:

Wages per day (in ₹)

61 - 70

71 - 80

81 - 90

91 - 100

101 - 110

111 – 120

No. of women workers

5

15

20

30

20

8

Solution

Class

Frequency (f)

Cumulative Frequency (cf)

60.5 – 70.5

5

5

70.5 – 80.5

15

20

80.5 – 90.5

20

40

90.5 – 100.5

30

70

100.5 – 110.5

20

90

110.5 – 120.5

8

98

 

N = Σf = 98

 

Now, N = 98

⇒ N/2 = 49

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

Thus, the median class is 90.5 – 100.5.

Now, l = 90.5, h = 10, f = 30, cf = c.f of preceding class=40 and N/2 = 49.

∴ Median, M = l + {h × (N/2 – cf/f))}

= 90.5 + {10 × ((49 – 40)/30)}

= 90.5 + 3

= 93.5

Hence, median wages = Rs 93.50.


13. Find the median from the following data: 

Class

1 - 5

6 - 10

11 - 15

16 - 20

21 - 25

26 - 30

31 - 35

35 - 40

40 - 45

Frequency

7

10

16

32

24

16

11

5

2

Solution 

Converting into exclusive form we get:

Class

Frequency (f)

Cumulative Frequency (cf)

0.5 – 5.5

7

7

5.5 – 10.5

10

17

10.5 – 15.5

16

33

15.5 – 20.5

32

65

20.5 – 25.5

24

89

25.5 – 30.5

16

105

30.5 – 33.5

11

116

35.5 – 40.5

5

121

40.5 – 45.5

2

123

 

N = Σf = 123

 

 Now, N = 123

⇒ N/2 = 61.5

The cumulative frequency just greater than 61.5 is 65 and the corresponding class is 15.5 – 20.5.

Thus, the median class is 15.5 – 20.5

∴ l = 15.5, h = 5, f = 32, cf = c.f. of preceding class=33 and N/2 = 61.5.

∴ Median, M = l + {h × (N/2 – cf/f))}

= 15.5 + {5 × ((61.5 – 33)/32)}

= 15.5 + 4.45

= 19.95

Hence, median = 19.95.


14. Find the median from the following data:

Marks

No. of students

Below 10

12

Below 20

32

Below 30

57

Below 40

80

Below 50

92

Below 60

116

Below 70

164

Below 80

200

Solution

Class

Cumulative frequency (cf)

Frequency (f)

0 - 10

12

12

10 - 20

32

20

20 - 30

57

25

30 - 40

80

23

40 - 50

92

12

50 - 60

116

24

60 – 70

164

48

70 - 80

200

36

   

N = Σf = 200

Now, N = 200

⇒ N/2 = 100

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,

cf = c.f. pf preceding class=92 and N/2 = 100.

∴ Median, M = l + {h × (N/2 – cf/f))}

= 50 + {10 × (100 – 92)/24)}

= 50 + 3.33

= 53.33

Hence, median = 53.33.

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