RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.2

RD Sharma Class 10 Solutions Chapter 7 Ex 7.2 PDF Free Download

In order to overcome minor errors in finding AM, the step deviation method is considered more appropriate for calculating the AM. Students wanting to build confidence in solving Class 10 Mathematics can access the RD Sharma Solutions Class 10. For any doubt clearing and explanations regarding this exercise, download RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.2, PDF provided below.

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.2 Download PDF

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.2 05
RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.2 06
RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.2 07

Access RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.2

1. The number of telephone calls received at an exchange per interval for 250 successive one- minute intervals are given in the following frequency table:

No. of calls (x):

0

1

2

3

4

5

6

No. of intervals (f):

15

24

29

46

54

43

39

Compute the mean number of calls per interval.

Solution:

Let the assumed mean(A) = 3

No. of calls xi

No. of intervals fi

ui = xi – A = xi – 3

fi ui

0

15

-3

-45

1

24

-2

-48

2

29

-1

-29

3

46

0

0

4

54

1

54

5

43

2

86

6

39

3

117

N = 250

Σ fixi = 135

Mean number of calls = A + Σ fixi / N

= 3 + 135/250

= (750 + 135)/ 250 = 885/ 250

= 3.54

2. 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 (x):

0

1

2

3

4

5

No. of tosses (f):

38

144

342

287

164

25

Solution:

Let the assumed mean(A) = 2

No. of heads per toss xi

No of intervals fi

ui = xi – A = xi – 2

fi ui

0

38

-2

-76

1

144

-1

-144

2

342

0

0

3

287

1

287

4

164

2

328

5

25

3

75

N = 1000

Σ fixi = 470

Mean number of heads per toss = A + Σ fixi / N

= 2 + 470/1000

= 2 + 0.470

= 2.470

3. The following table gives the number of branches and number of plants in the garden of a school.

No of branches (x):

2

3

4

5

6

No of plants (f):

49

43

57

38

13

Calculate the average number of branches per plant.

Solution:

Let the assumed mean (A) = 4

No of branches xi

No of plants fi

u= x− A = x− 4

fi ui

2

49

-2

-98

3

43

-1

-43

4

57

0

0

5

38

1

38

6

13

2

26

N = 200

Σ fixi = -77

Average number of branches per plant = A + Σ fixi / N = 4 + (-77/200) 

= 4 -77/200 

= (800 -77)/200

= 3.615

4. The following table gives the number of children of 150 families in a village

No of children (x):

0

1

2

3

4

5

No of families (f):

10

21

55

42

15

7

Find the average number of children per family.

Solution:

Let the assumed mean (A) = 2

No of children xi

No of families fi

u= xi − A = xi − 2

fi ui

0

10

-2

-20

1

21

-1

-21

2

55

0

0

3

42

1

42

4

15

2

30

5

7

3

21

N = 150

Σ fixi = 52

Average number of children for family = A + Σ fixi / N = 2 + 52/150

= (300 +52)/150 

= 352/150 

= 2.35 (corrected to neat decimal)

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

Exercise 7.1 Solutions

Exercise 7.3 Solutions

Exercise 7.4 Solutions

Exercise 7.5 Solutions

Exercise 7.6 Solutions

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