RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions for Class 9 Mathematics

Created by: Abhishek Kapoor

Class 9 : RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev

The document RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
All you need of Class 9 at this link: Class 9

Q 1 . Calculate the mean for the following distribution :

x : 56789
f : 4814113


SOLUTION :

xffx
5

6

7

8

9

4

8

14

11

3

20

48

98

88

27

 N=40∑fx = 281

RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev

Q 2 . Find the mean of the following data :

x : 192123 25272931
f : 1315161816 1513

 

SOLUTION :

xffx
19

21

23

25

27

29

31

13

15

16

18

16

15

13

247

315

368

450

432

435

403

 N=106∑fx=2650

 

RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev

Q 3 . The mean of the following data is 20.6 .Find the value of p.

x :1015p 2535
f : 102575


SOLUTION :

xffx
10

15

P

25

35

3

10

25

7

5

30

150

25p

175

175

 N = 50∑fx=25p+530


It is given that ,

Mean = 20.6

RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev

⇒ 25p+530 = 20.6×50

⇒ 25p = 1030−530

⇒ 25p = 500

⇒ p = 50025 = 20

⇒ p = 20

∴ p = 20.


Q 4 . If the mean of the following data is 15 , find p.

x :  510152025
f :   6p6105


SOLUTION :

xffx
5

10

15

20

25

6

P

6

10

5

30

10p

90

200

125

 N = p+27∑fx = 10p+445


It is given that ,

Mean = 15

RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev

⇒ 10p+445 = 15×(p+27)

⇒10p+445 = 15p+405

⇒15p−10p = 445−405

⇒ 5p = 40

⇒ p = 40/5 =8

⇒ p = 8

∴ p = 8.


Q 5. Find the value of p for the following distribution whose mean is 16.6.

x : 8 1215p202530
f : 121620241684

 

SOLUTION :

xffx
8

12

15

P

20

25

30

12

16

20

24

16

8

4

96

192

300

24p

320

200

120

 N = 100∑fx = 24p+1228

 

It is given that ,

Mean = 16.6

RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev

⇒ 24p+1228 = 1660

⇒ 24p = 1660−1228

⇒ 24p = 432

⇒ p = 432/24 = 18

⇒ p = 18

∴ p = 18.


Q 6 . Find the missing value of p for the following distribution whose mean is 12.58 .

x : 581012p2025
f :25822742

 

SOLUTION :

xffx
5

8

10

12

P

20

25

2

5

8

22

7

4

2

10

40

80

264

7p

80

50

 N = 50∑fx=7p+524


It is given that ,

Mean = 12.58

RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev

⇒ 7p+524 = 629

⇒ 7p = 629−524

⇒ 7p = 105

⇒ p = 1057 = 15

⇒ p = 15

∴ p = 18.


Q7 . Find the missing frequency (p) for the following distribution whose mean is 7.68 .

x :  35791113
f :6815p84

SOLUTION :

xffx
3

5

7

9

11

13

6

8

15

P

8

4

18

40

105

9p

88

52

 N=p+41∑fx=9p+303


It is given that ,

Mean = 7.68

RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev

⇒ 9p+303 = 7.68p+314.88

⇒ 9p−7.68p = 314.88−303

⇒ 1.32p = 11.88

⇒ p = 11.881.32 = 9

⇒ p = 9

∴ p = 9.


Q 8. Find the value of p, if the mean of the following distribution is 20 .

x :15171920+p23
f: 2345p6

SOLUTION :

            xffx
15

17

19

20+p

23

2

3

4

5p

6

30

51

76

100p+ 5p2

138

 N=5p+15fx= 5p2+100p+295

It is given that ,

Mean = 20

RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev

⇒ 5p2+100p+295 = 20(5p+15)

⇒ 5p2+100p+295 = 100p+300

⇒ 5p2 =300−295

⇒ 5p2 = 5

⇒ p2 = 1

⇒ p = ±1

Frequency can’t be negative.

Hence, value of p is 1.

Q 9 . Find the mean of the following distribution :

x : 1012202530
f :3101575

SOLUTION :

  xffx
10

12

20

25

35

3

10

15

7

5

30

120

300

175

175

 N = 40∑fx = 800

 

RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev

Q 10. Candidates of four schools appear in a mathematics test. The data were as follows :

SchoolsNo. Of CandidatesAverage Score
I

II

III

IV

60

48

Not Available

40

75

80

55

50

If the average score of the candidates of all four schools is 66 , Find the number of candidates that appeared from school III .

SOLUTION :

SchoolsNo. Of CandidatesAverage Score
I

II

III

IV

60

48

x

40

75

80

55

50


Given the average score of all schools = 66

RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev

⇒ 10340+55x = 66x+9768

⇒ 10340−9768 = 66x−55x

⇒ 11x = 572

⇒ x = 572/11 = 52

∴ No. of candidates appeared from school III = 52.

Q.11 . Five coins were simultaneously tossed 1000 times and at each, toss the number of heads was observed. The number of tosses during which 0 , 1 , 2 , 3 , 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

No . of heads per tossNo.of tosses
0

1

2

3

4

5

38

144

342

287

164

25

Total1000

SOLUTION :

No . of heads per toss(x)No.of tosses(f)fx
0

1

2

3

4

5

38

144

342

287

164

25

0

144

684

861

656

125

 N = 1000∑fx = 2470

∴ Mean number of heads per toss

= 2470/1000

= 2.47

Q 12 . Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

x : 10  30 507090
f :17f132f219

Total = 120

SOLUTION :

xffx
10

 

30

50

70

90

17

 

f1

32

f2

19

170

 

30f1

1600

70f2

1710

 N=120∑fx = 3480+30f1+70f2

It is given that

Mean = 50

RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev

⇒ 3480+30f1+70f= 50×120

⇒ 30f1+70f= 6000−3480

⇒10(3f1+7f2) = 10(252)

⇒3f1+7f= 252⋅⋅⋅⋅⋅(1)                         [∵Divideby10]

And N = 20

⇒17+f1+32+f2+19 = 120

⇒ 68+f1+f= 120

⇒ f1+f= 120−68

⇒ f1+f= 52

Multiply with 3 on both sides

⇒3f1+3f= 156⋅⋅⋅⋅⋅⋅(2)

Subtracting equation (2) from equation (1)

⇒3f1+7f2−3f1−3f= 252−156

⇒ 4f2=96

⇒ f= 96/4 = 24

Put the value of f2 in equation (1)

⇒ 3f1+7×24 = 252

⇒ 3f= 252−168

⇒ f= 84/3 = 28

⇒ f= 28

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