RD Sharma Solutions Class 9 Measures Of Central Tendency Exercise 24.2

RD Sharma Solutions Class 9 Chapter 24 Exercise 24.2

RD Sharma Class 9 Solutions Chapter 24 Ex 24.2 Free Download

Q 1 . Calculate the mean for the following distribution :

x :           5              6              7              8              9

f :            4              8              14           11           3

SOLUTION :

x f fx
5

6

7

8

9

4

8

14

11

3

20

48

98

88

27

N=40 \(\sum fx=281\)

\(∴ Mean\;\bar{x}=\frac{\sum fx }{N}\)

=\(\frac{281}{40}\) = 7.025.

Q 2 . Find the mean of the following data :

x :           19           21           23           25           27           29           31

f :            13           15           16           18           16           15           13

SOLUTION :

x f fx
19

21

23

25

27

29

31

13

15

16

18

16

15

13

247

315

368

450

432

435

403

  N=106 \(\sum fx=2650\)

\(∴ Mean\;\bar{x}=\frac{\sum fx }{N}\)

=\(\frac{2650}{106}\) = 25.

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

x :           10           15           p             25           35

f :            3              10           25           7              5

SOLUTION :

x f fx
10

15

P

25

35

3

10

25

7

5

30

150

25p

175

175

  N = 50 \(\sum fx=25p+530\)

It is given that ,

Mean = 20.6

\(\Rightarrow \frac{\sum fx}{N}=20.6\)

\(\Rightarrow \frac{25p+530}{50}=20.6\)

\(\Rightarrow 25p+530=20.6\times 50\)

\(\Rightarrow 25p=1030-530\)

\(\Rightarrow 25p=500\)

\(\Rightarrow p=\frac{500}{25}\) =20

\(\Rightarrow p=20\)

\(∴ p=20\).

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

x :           5              10           15           20           25

f :            6              p             6              10           5

SOLUTION :

x f fx
5

10

15

20

25

6

P

6

10

5

30

10p

90

200

125

  N=p+27 \(\sum fx=10p+445\)

It is given that ,

Mean = 15

\(\Rightarrow \frac{\sum fx}{N}=15\)

\(\Rightarrow \frac{10p+445}{p+27}=15\)

\(\Rightarrow 10p+445=15\times \left ( p+27 \right )\)

\(\Rightarrow 10p +445=15p+405\)

\(\Rightarrow 15p-10p=445-405\)

\(\Rightarrow 5p=40\)

\(\Rightarrow p=\frac{40}{5}\) =8

\(\Rightarrow p=8\)

\(∴ p=8\).

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

x :           8              12           15           p             20           25           30

f :            12           16           20           24           16           8              4

SOLUTION :

x f fx
8

12

15

P

20

25

30

12

16

20

24

16

8

4

96

192

300

24p

320

200

120

  N=100 \(\sum fx=24p+1228\)

It is given that ,

Mean = 16.6

\(\Rightarrow \frac{\sum fx}{N}=16.6\)

\(\Rightarrow \frac{24p+1228}{100}=16.6\)

\(\Rightarrow 24p+1228=1660\)

\(\Rightarrow 24p =1660-1228\)

\(\Rightarrow 24p=432\)

\(\Rightarrow p=\frac{432}{24}\) =18

\(\Rightarrow p=18\)

\(∴ p=18\).

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

x :           5              8              10           12           p             20           25

f :            2              5              8              22           7              4              2

SOLUTION :

x f fx
5

8

10

12

P

20

25

2

5

8

22

7

4

2

10

40

80

264

7p

80

50

  N = 50 \(\sum fx=7p+524\)

It is given that ,

Mean = 12.58

\(\Rightarrow \frac{\sum fx}{N}=12.58\)

\(\Rightarrow \frac{7p+524}{50}=12.58\)

\(\Rightarrow 7p+524=629\)

\(\Rightarrow 7p =629-524\)

\(\Rightarrow 7p=105\)

\(\Rightarrow p=\frac{105}{7}\) =15

\(\Rightarrow p=15\)

\(∴ p=18\).

Q 7 . Find the missing frequency (p) for the following distribution whose mean is 7.68 .

x :           3              5              7              9              11           13

f :            6              8              15           p             8              4

SOLUTION :

x f fx
3

5

7

9

11

13

6

8

15

P

8

4

18

40

105

9p

88

52

  N=p+41 \(\sum fx=9p+303\)

It is given that ,

Mean = 7.68

\(\Rightarrow \frac{\sum fx}{N}=7.68\)

\(\Rightarrow \frac{9p+303}{ p+41}=7.68 \)

\(\Rightarrow 9p+303=7.68p +314.88\)

\(\Rightarrow 9p-7.68p =314.88-303\)

\(\Rightarrow 1.32p=11.88\)

\(\Rightarrow p=\frac{11.88}{1.32}\) =9

\(\Rightarrow p=9\)

\(∴ p=9\).

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

x :           15           17           19           20+p      23

f :            2              3              4              5p           6

SOLUTION :

            x f fx
15

17

19

20+p

23

2

3

4

5p

6

30

51

76

100p+ \(5p^{2}\)

138

  N=5p+15 fx= \(5p^{2}+100p+295\)

It is given that ,

Mean = 20

\(\Rightarrow \frac{\sum fx}{N}=20\)

\(\Rightarrow\frac{ 5p^{2}+100p+295}{5p+15}=20\)

\(\Rightarrow5p^{2}+100p+295=20\left ( 5p+15 \right )\)

\(\Rightarrow5p^{2}+100p+295=100p+300\)

\(\Rightarrow5p^{2}=300-295\)

\(\Rightarrow5p^{2}=5\)

\(\Rightarrow p^{2}=1\)

\(\Rightarrow p=\pm 1\)

Frequency can’t be negative.

Hence, value of p is 1.

Q 9 . Find the mean of the following distribution :

x :           10           12           20           25           35

f :            3              10           15           7              5

SOLUTION :

            x f fx
10

12

20

25

35

3

10

15

7

5

30

120

300

175

175

  N=40 \(\sum fx=800\)

\(∴ Mean\;\bar{x}=\frac{\sum fx }{N}\)

=\(\frac{800}{40}\) = 20.

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

Schools No. Of Candidates Average 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 :

Schools No. Of Candidates Average Score
I

II

III

IV

60

48

x

40

75

80

55

50

Given the average score of all schools =66

\(\Rightarrow \frac{N_{1}\bar{x}_{1}+N_{2}\bar{x}_{2}+N_{3}\bar{x}_{3}+N_{4}\bar{x}_{4}}{N_{1}+N_{2}+N_{3}+N_{4}}=66\)

\(\Rightarrow \frac{60\times 75+48\times 80+x\times 55+40\times 50}{60+48+x+40}=66\)

\(\Rightarrow \frac{4500+3840+55x+2000}{148+x}=66\)

\(\Rightarrow \frac{10340+55x}{148+x}=66\)

\(\Rightarrow 10340+55x=66x+9768\)

\(\Rightarrow 10340-9768=66x-55x\)

\(\Rightarrow 11x=572\)

\(\Rightarrow x=\frac{572}{11}=52\)

\(∴\) 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 toss No.of tosses
0

1

2

3

4

5

38

144

342

287

164

25

Total 1000

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 \(\sum fx=2470\)

\(∴ Mean\) number of heads per toss = \(\frac{\sum fx}{N}\)

= \(\frac{2470}{1000}\)

= 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                                           50           70                                           90

f :            17           \(f_{1}\)        32           \(f_{2}\)        19          

Total=120

SOLUTION :

x f fx
10

30

50

70

90

17

\(f_{1}\)

32

\(f_{2}\)

19

170

30\(f_{1}\)

1600

70\(f_{2}\)

1710

  N=120 \(\sum fx=3480+30f_{1}+70f_{2}\)

It is given that

Mean = 50

\(\Rightarrow \frac{\sum fx}{N}=50\)

\(\Rightarrow \frac{3480+30f_{1}+70f_{2}}{N}\)=50

\(\Rightarrow 3480+30f_{1}+70f_{2}=50\times 120\)

\(\Rightarrow 30f_{1}+70f_{2}=6000-3480\)

\(\Rightarrow 10\left (3f_{1}+7f_{2} \right )=10\left (252 \right )\)

\(\Rightarrow 3f_{1}+7f_{2} =252 \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )\)                         \(\left [∵ Divide \;by \;10 \right ]\)

And N = 20

\(\Rightarrow 17+f_{1}+32+f_{2}+19=120\)

\(\Rightarrow 68+f_{1}+f_{2}=120\)

\(\Rightarrow f_{1}+f_{2}=120-68\)

\(\Rightarrow f_{1}+f_{2}=52\)

Multiply with 3 on both sides

\(\Rightarrow 3f_{1}+3f_{2}=156\cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )\)

Subtracting equation (2) from equation (1)

\(\Rightarrow 3f_{1}+7f_{2}-3f_{1}-3f_{2}=252-156\)

\(\Rightarrow 4f_{2}=96\)

\(\Rightarrow f_{2}=\frac{96}{4}\) = 24

Put the value of \(f_{2}\) in equation (1)

\(\Rightarrow 3f_{1}+7\times 24=252\)

\(\Rightarrow 3f_{1}=252-168\)

\(\Rightarrow f_{1}=\frac{84}{3}=28\)

\(\Rightarrow f_{1}=28\)