RD Sharma Solutions Class 10 Statistics Exercise 7.1

RD Sharma Class 10 Solutions Chapter 7 Ex 7.1 PDF Free Download

Exercise 7.1

Q.1: Calculate the mean for the following distribution:

x:

5 6 7 8 9

f:

4 8 14 11 3

Sol:

X

f

fx

5

4

20

6

8

48

7

14

98

8

11

88

9

3

27

N = 40

281

Using formula, Mean = fx/N

Mean = 281/40 =7.025

2. Find the mean of the following data:

x:

19 21 23 25 27 29 31

f:

13 15 16 18 16 15 13

Soln:

X

f

fx

19

13

247

21

15

315

23

16

368

25

18

450

27

16

432

29

15

435

31

13

403

N = 106

Sum = 2620

Mean = 2680/106 = 25

3. If 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

Soln:

X

f

fx

10

3

30

5

10

150

p

25

25p

25

7

175

35

5

175

N = 50

Sum = 530+25p

Mean = 20.6 (given)

Using formula, Mean = fx/N = (530+25p)/50

Now, (530+25p)/50 = 20.6

p = 20

The value of p is 20.

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

x:

5 10 15 20 25

f:

6 p 6 10 5

Soln:

X

f

fx

5

6

30

10

p

10p

15

6

90

20

10

200

25

5

125

N = p+27

Sum = 10p + 445

Mean =15 (Given)

(10p + 445)/(p+27) = 15

10p + 445 = 15p + 405

5p = 40

Or p = 8

The value of p is 8.

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

Soln:

X

f

fx

8

12

96

12

16

192

15

20

300

p

24

24p

20

16

320

25

8

200

30

4

120

N = 100

Sum = 24p + 1228

Mean = 16.6 (given)

(24p+1228)/100 = 16.6

24p + 1228 = 1660

24p = 1660 – 1228

p = 432/24 = 18

The value of p is 18.

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

Soln:

x

f

fx

5

2

10

8

5

40

10

8

80

12

22

264

p

7

7p

20

4

480

25

2

50

N = 50

Sum = 524 + 7p

Mean = 12.58 (given)

Sum/N = 12.58

(524 +7p)/50 = 12.58

524 + 7p = 629

7p = 105 = 15

The value of p is 15.

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

Soln:

X

f

fx

3

6

18

5

8

40

7

15

105

9

p

9p

11

8

88

13

4

52

N = p + 41

Sum= 9p + 303

Mean = 7.68 (given)

(9p+303)/p+41 = 7.68

9p + 303 = (7.68)(p+41)

9p – p(7.68) = 314.88 – 303

1.32p = 11.88

p = (11.88)/1.32 = 9

The value of p is 9

8.The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students.

Ages (in years): 15 16 17 18 19 20

No of students: 3 8 10 10 5 4

Soln:

x

f

fx

15

3

45

16

8

128

17

10

170

18

10

180

19

5

95

20

4

80

N = 40

Sum = 698

Mean age = fx/ N

= 698/ 40

= 17.45 years

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

Schools

No of candidates

Average score

I

60

75

II

48

80

III

P

55

IV

40

50

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

Soln:

Schools

No of candidates Ni

Average scores (xi)

I

60

75

II

48

80

III

P

55

IV

40

50

Average score or all schools = 66 (given)

\(\frac{N_{1}\overline{x_{1}}+N_{2}\overline{x_{2}}+N_{3}\overline{x_{3}}+ N_{4}\overline{x_{4}}}{N_{1}+ N_{2}+ N_{3}+ N_{4}}\)

\(\frac{4500 + 3340 + 55p + 2000 }{ 60 + 48 + p + 40}\)

10340 + 55p = 66p + 9768

10340 – 9768 = (66 – 55)p

p = 572/11 =52

The value of p is 52.

10.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

38

1

144

2

342

3

287

4

164

5

25

Total

1000

Soln:

No of heads per toss

No of tosses

0

38

1

144

2

342

3

287

4

164

5

25

No of heads per toss(x)

No of tosses(f)

fx

0

38

0

1

144

144

2

342

684

3

287

861

4

164

656

5

25

125

Mean number of heads per toss = 2470/1000 = 2.47

12.The arithmetic mean of the following data is 25. Find the value of k.

Xi:

5 15 25 35 45

fi:

3 k 3 6 2

Sol:

X

f

fx

5

3

15

15

k

15k

25

3

75

35

6

210

45

2

90

N = k + 120

Sum = 15k + 390

Given : Mean = 25

Sum/ N = 25

15k + 390 = 25k + 350

25k – 15k = 40

10k = 40

k = 4

13. If the mean of the following data is 18.75. Find the value of p.

Xi:

10 15 p 25 30

Fi:

5 10 7 8 2

Soln:

X

f

fx

10

5

50

15

10

150

p

7

7p

25

8

200

30

2

60

N = k + 120

Sum = 7p + 460

Given : Mean = 18.75

Sum/ N = 18.75

7p + 460 = 600

7p = 140

p = 20

14.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

Soln:

X

f

fx

15

2

30

17

3

51

19

4

76

20 + p

5p

100p+5p2

23

6

138

N = 5p +15

Sum = 295 +100p +5p2

Given : Mean= 20

Sum/ N = 20

(295 + 100p – 5p2) / (5p + 15) = 20

295 + 100p + 5p2 = 100p + 300

5p2 – 5 = 0

5(p2 – 1) = 0

p = 1 or p = -1

15.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 f1 32 f2 19

Sum of f is 120.

Soln:

x

f

fx

10

17

170

30

f1

30f1

50

32

1600

70

f2

70f2

90

19

1710

N = 120

Sum = 30f1+70f2+3480

Given : mean = 50

Sum/ N = 50

(30f1+70f2+3480)/ 120 = 50

30f1+70f2+3480 = 6000 —- (1)

Also, sum of f = 120

17 + f1 + 32 + f2 + 19 = 120

f1 + f2 = 52

f1 = 52 – f2

Substituting the value of f1 in (1)

30 (52 – f2) +70f2 + 3480 = 6000

f2 = 24

Hence f1 = 52 -24 = 28

f1 = 28 and f2 = 24

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