Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths

RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

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 | RD Sharma Solutions for Class 9 Mathematics

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 | RD Sharma Solutions for Class 9 Mathematics

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 | RD Sharma Solutions for Class 9 Mathematics

⇒ 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 | RD Sharma Solutions for Class 9 Mathematics

⇒ 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 | RD Sharma Solutions for Class 9 Mathematics

⇒ 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 | RD Sharma Solutions for Class 9 Mathematics

⇒ 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 | RD Sharma Solutions for Class 9 Mathematics

⇒ 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 | RD Sharma Solutions for Class 9 Mathematics

⇒ 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 | RD Sharma Solutions for Class 9 Mathematics

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 | RD Sharma Solutions for Class 9 Mathematics

⇒ 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 | RD Sharma Solutions for Class 9 Mathematics

⇒ 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

The document RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What are measures of central tendency in statistics?
Ans. Measures of central tendency in statistics are statistical measures that represent the center or average of a data set. They provide information about the typical or most common value in a data set. The three main measures of central tendency are the mean, median, and mode.
2. How is the mean calculated?
Ans. The mean is calculated by adding up all the values in a data set and then dividing the sum by the total number of values. It is also known as the average. For example, if we have a data set of {2, 4, 6, 8, 10}, the mean can be calculated as (2 + 4 + 6 + 8 + 10) / 5 = 6.
3. What is the median and how is it calculated?
Ans. The median is the middle value in a data set when the data is arranged in ascending or descending order. To calculate the median, first arrange the data set in order. If the number of values is odd, the median is the middle value. If the number of values is even, the median is the average of the two middle values.
4. How is the mode determined?
Ans. The mode is the value that appears most frequently in a data set. It can be determined by simply identifying the value that occurs with the highest frequency. In some cases, there may be multiple modes if multiple values occur with the same highest frequency.
5. How are measures of central tendency useful in data analysis?
Ans. Measures of central tendency provide valuable information about the typical or average value in a data set. They help in summarizing and understanding the data by providing a single representative value. They are widely used in various fields such as economics, social sciences, and market research to analyze and interpret data.
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