RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.4

RD Sharma Class 10 Solutions Chapter 7 Ex 7.4 PDF Free Download

The median is the middle of a distribution. In this exercise, students will practice to find the median of a discrete and grouped frequency distribution. For quick access to solutions, the RD Sharma Solutions Class 10 is the best place. It has well-structured solutions in simple language to match the level of all the students. For detailed studies regarding this exercise, download RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.4 PDF provided below.

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.4 Download PDF

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Access RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.4

1. Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median: 

715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719.

Solution:

Arranging the given data in ascending order, we have

694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745

As the number of terms is an old number i.e., N = 15

We use the following procedure to find the median.

Median = (N + 1)/2 th term

= (15 + 1)/2 th term

= 8th term

So, the 8th term in the arranged order of the given data should be the median.

Therefore, 716 is the median of the data.

2. The following is the distribution of height of students of a certain class in a certain city:

Height (in cm):

160 – 162

163 – 165

166 – 168

169 – 171

172 – 174

No of students:

15

118

142

127

18

Find the median height.

Solution:

Class interval (exclusive)

Class interval  (inclusive)

Class interval frequency

Cumulative frequency

160 – 162

159.5 – 162.5

15

15

163 – 165

162.5 – 165.5

118

133(F)

166 – 168

165.5 – 168.5

142(f)

275

169 – 171

168.5 – 171.5

127

402

172 – 174

171.5 – 174.5

18

420

N = 420

Here, we have N = 420,

So, N/2 = 420/ 2 = 210

The cumulative frequency just greater than N/2 is 275 then 165.5 – 168.5 is the median class such, that

L = 165.5, f = 142, F = 133 and h = (168.5 – 165.5) = 3

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 1

= 165.5 + 1.63

= 167.13

3. Following is the distribution of I.Q of 100 students. Find the median I.Q.

I.Q:

55 – 64

65 – 74

75 – 84

85 – 94

95 – 104

105 – 114

115 – 124

125 – 134

135 – 144

No of students:

1

2

9

22

33

22

8

2

1

Solution:

Class interval (exclusive)

Class interval  (inclusive)

Class interval frequency

Cumulative frequency

55 – 64

54.5 – 64-5

1

1

65 – 74

64.5 – 74.5

2

3

75 – 84

74.5 – 84.5

9

12

85 – 94

84.5 – 94.5

22

34(F)

95 – 104

94.5 – 104.5

33(f)

67

105 – 114

104.5 – 114.5

22

89

115 – 124

114.5 – 124.5

8

97

125 – 134

124.5 – 134.5

2

98

135 – 144

134.5 – 144.5

1

100

N = 100

Here, we have N = 100,

So, N/2 = 100/ 2 = 50

The cumulative frequency just greater than N/ 2 is 67 then the median class is (94.5 – 104.5) such that L = 94.5, F = 33, h = (104.5 – 94.5) = 10

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 2

= 94.5 + 4.85

= 99.35

4. Calculate the median from the following data:

Rent (in Rs):

15 – 25

25 – 35

35 – 45

45 – 55

55 – 65

65 – 75

75 – 85

85 – 95

No of houses:

8

10

15

25

40

20

15

7

Solution:

Class interval

Frequency

Cumulative frequency

15 – 25

8

8

25 – 35

10

18

35 – 45

15

33

45 – 55

25

58(F)

55 – 65

40(f)

98

65 – 75

20

118

75 – 85

15

133

85 – 95

7

140

N = 140

Here, we have N = 140,

So, N/2 = 140/ 2 = 70

The cumulative frequency just greater than N/ 2 is 98 then median class is 55 – 65 such that L = 55, f = 40, F = 58, h = 65 – 55 = 10

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 3

= 55 + 3 = 58

 

5. Calculate the median from the following data:

Marks below:

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

60 – 70

70 – 80

85 – 95

No of students:

15

35

60

84

96

127

198

250

Solution:

Marks below

No. of students

Class interval

Frequency

Cumulative frequency

10

15

0 – 10

15

15

20

35

10 – 20

20

35

30

60

20 – 30

25

60

40

84

30 – 40

24

84

50

96

40 – 50

12

96(F)

60

127

50 – 60

31(f)

127

70

198

60 – 70

71

198

80

250

70 – 80

52

250

N = 250

Here, we have N = 250,

So, N/2 = 250/ 2 = 125

The cumulative frequency just greater than N/ 2 is 127 then median class is 50 – 60 such that L = 50, f = 31, F = 96, h = 60 -50 = 10


R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 4

= 50 + 9.35

= 59.35

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

Age in years:

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

No of persons:

5

25

?

18

7

Solution:

Let the unknown frequency be taken as x,

Class interval

Frequency

Cumulative frequency

0 – 10

5

5

10 – 20

25

30(F)

20 – 30

x (f)

30 + x

30 – 40

18

48 + x

40 – 50

7

55 + x

N = 170

It’s given that

Median = 24

Then, median class = 20 – 30; L = 20, h = 30 -20 = 10, f = x, F = 30

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 5

4x = 275 + 5x – 300

4x – 5x = – 25

– x = – 25

x = 25

Therefore, the Missing frequency = 25 

7. The following table gives the frequency distribution of married women by age at marriage.

Age (in years)

Frequency

Age (in years)

Frequency

15 – 19

53

40 – 44

9

20 – 24

140

45 – 49

5

25 – 29

98

45 – 49

3

30 – 34

32

55 – 59

3

35 – 39

12

60 and above

2

Calculate the median and interpret the results.

Solution:

Class interval (exclusive)

Class interval (inclusive)

Frequency

Cumulative frequency

15 – 19

14.5 – 19.5

53

53 (F)

20 – 24

19.5 – 24.5

140 (f)

193

25 – 29

24.5 – 29.5

98

291

30 – 34

29.5 – 34.5

32

323

35 – 39

34.5 – 39.5

12

335

40 – 44

39.5 – 44.5

9

344

45 – 49

44.5 – 49.5

5

349

50 – 54

49.5 – 54.5

3

352

55 – 54

54.5 – 59.5

3

355

60 and above

59.5 and above

2

357

N =357

Here, we have N = 357,

So, N/2 = 357/ 2 = 178.5

The cumulative frequency just greater than N/2 is 193, so then the median class is (19.5 – 24.5) such that l = 19.5, f = 140, F = 53, h = 25.5 – 19.5 = 5

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 6

Median = 23.98

Which means nearly half the women were married between the ages of 15 and 25 

8. 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.

Solution:

Life time

Number of lamps fi

Cumulative frequency (cf)

1500 – 2000

14

14

2000 – 2500

56

70

2500 – 3000

60

130(F)

3000 – 3500

86(f)

216

3500 – 4000

74

290

4000 – 4500

62

352

4500 – 5000

48

400

N = 400

It’s seen that, the cumulative frequency just greater than n/2 (400/2 = 200) is 216 and it belongs to the class interval 3000 – 3500 which becomes the Median class = 3000 – 3500

Lower limits (l) of median class = 3000 and,

Frequency (f) of median class = 86

Cumulative frequency (cf) of class preceding median class = 130

And, the Class size (h) = 500

Thus, calculating the median by the formula, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 7

= 3000 + (35000/86)

= 3406.98

Thus, the median life time of lamps is 3406.98 hours

9. The distribution below gives the weight of 30 students in a class. Find the median weight of students:

Weight (in kg):

40 – 45

45 – 50

50 – 55

55 – 60

60 – 65

65 – 70

70 – 75

No of students:

2

3

8

6

6

3

2

Solution:

Weight (in kg)

Number of students fi

Cumulative frequency (cf)

40 – 45

2

2

45 – 50

3

5

50 – 55

8

13

55 – 60

6

19

60 – 65

6

25

65 – 70

3

28

70 – 75

2

30

It’s seen that, the cumulative frequency just greater than n/ 2 (i.e. 30/ 2 = 15) is 19, belongs to class interval 55 – 60.

So, it’s chosen that

Median class = 55 – 60

Lower limit (l) of median class = 55

Frequency (f) of median class = 6

Cumulative frequency (cf) = 13

And, Class size (h) = 5

Thus, calculating the median by the formula, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 8

= 55 + 10/6 = 56.666

So, the median weight is 56.67 kg.

10. Find the missing frequencies and the median for the following distribution if the mean is 1.46

No. of accidents:

0

1

2

3

4

5

Total

Frequencies (no. of days):

46

?

?

25

10

5

200

Solution:

No. of accidents (x)

No. of days (f)

fx

0

46

0

1

x

x

2

y

2y

3

25

75

4

10

40

5

5

25

N = 200

Sum = x + 2y + 140

It’s given that, N = 200

⇒ 46 + x + y + 25 + 10 + 5 = 200

⇒ x + y = 200 – 46 – 25 – 10 – 5

⇒ x + y = 114 —- (i)

And also given, Mean = 1.46

⇒ Sum/ N = 1.46

⇒ (x + 2y + 140)/ 200 = 1.46

⇒ x + 2y = 292 – 140

⇒ x + 2y = 152 —- (ii)

Subtract equation (i) from equation (ii), we get

x + 2y – x – y = 152 – 114

⇒ y = 38

Now, on putting the value of y in equation (i), we find x = 114 – 38 = 76

Thus, the table become:

No. of accidents (x)

No. of days (f)

Cumulative frequency

0

46

46

1

76

122

2

38

160

3

25

185

4

10

195

5

5

200

N = 200

It’s seen that,

N = 200 N/2 = 200/2 = 100

So, the cumulative frequency just more than N/2 is 122

Therefore, the median is 1.

Also, access other exercise solutions of RD Sharma Class 10 Maths Chapter 7 Statistics

Exercise 7.1 Solutions

Exercise 7.2 Solutions

Exercise 7.3 Solutions

Exercise 7.5 Solutions

Exercise 7.6 Solutions

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