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Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.2
  • Difficulty Level : Easy
  • Last Updated : 21 Dec, 2020

Question1: The number of telephone calls received at an exchange per interval for 250 successive one-minute intervals are given in the following frequency table:

Number of calls (x)

0

1

2

3



4

5

6

Number of intervals (f)     

15

24

29

46



54

43

39

Compute the mean number of calls per interval.

Solution: 

Let the assumed mean(A) be =3 (Generally we choose the middle element to be the assumed mean, but it’s not mandatory),

hence, the table is,
 

Number of calls (x_i)

Number of intervals (f_i)

u_{i}=x_{i}-A=x_{i}-3

f_{i}*u_{i}

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

 

\displaystyle\sum_{}^{} f_{i}=250

 

\displaystyle\sum_{}^{} f_{i}*u_{i}=135

hence, mean of the calls = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                        = 3+\frac{135}{250}

                                        = 3.54

Therefore, mean number of calls per interval is 3.54

Question 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.

Number of heads per toss (x)      

0

1

2

3

4

5

Number of tosses (f)

38

144

342

287

164

25

Solution:

Let the assumed mean (A) be = 2

hence, the table is,

Number of heads per toss (x_i)

Number of tosses (f_i)

u_{i}=x_{i}-A=x_{i}-2

f_{i}*u_{i}

0

38

-2

-76

1

144

-1

-144

2

342

0

0

3

287

1

287

4

164

2

328

5

25

3

75

 

\displaystyle\sum_{}^{} f_{i}=1000

 

\displaystyle\sum_{}^{} f_{i}*u_{i}=470

Mean number of head per toss = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                                   = 2+\frac{470}{1000}

                                                   = 2.47

Therefore, mean number of head per toss is 2.47

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

Number of branches (x)

2

3

4

5

6

Number of plants (f)

49

43

57

38

13

Calculate the average number of branches per plant.

Solution:

Let the assumed mean(A) be = 4

hence, the table is,

Number of branches (x_i)

Number of plants (f_i)

u_{i}=x_{i}-A=x_{i}-4

f_{i}*u_{i}

2

49

-2

-98

3

43

-1

-43

4

57

0

0

5

38

1

38

6

13

2

26

 

\displaystyle\sum_{}^{} f_{i}=200

 

\displaystyle\sum_{}^{} f_{i}*u_{i}=-77

Average Number of branches per plant = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                                                = 2+(\frac{-77}{200})

                                                                = 3.615

Therefore, mean number of branches per plant is 3.615

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

Number of children (x)

0

1

2

3

4

5

Number of families (f)

10

21

55

42

15

7

Find the average number of children per family.

Solution:

Let the assumed mean(A) be = 2

Hence, the table is,

Number of children (x_i)

Number of families (f_i)

u_{i}=x_{i}-A=x_{i}-2

f_{i}*u_{i}

0

10

-2

-20

1

21

-1

-21

2

55

0

0

3

42

1

42

4

15

2

30

5

7

3

21

 

\displaystyle\sum_{}^{} f_{i}=150

 

\displaystyle\sum_{}^{} f_{i}*u_{i}=52

Average number of children per family = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                                               = 2+(\frac{52}{150})

                                                               = 2.35 (approximately)

Therefore, the average number of children per family is 2.35 (approximately)

Question 5: The marks obtained out of 50, by 102 students in a Physics test are given in the frequency table below:

Marks (x)

15

20

22

24

25

30

33

38

45

Frequency (f)

5

8

11

20

23

18

13

3

1

Find the average number of marks.

Solution:

Let the assume mean (A) be = 25

hence, the table is,

Marks (x_i)

Frequency (f_i)

u_{i}=x_{i}-A=x_{i}-25

f_{i}*u_{i}

15

5

-10

-50

20

8

-5

-40

22

11

-3

-33

24

20

-1

-20

25

23

0

0

30

18

5

90

33

13

8

104

38

3

13

39

45

1

20

20

 

\displaystyle\sum_{}^{} f_{i}=102

 

\displaystyle\sum_{}^{} f_{i}*u_{i}=110

Average number of marks =  A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                           = 25+(\frac{102}{110})

                                           = 26.08 (approximately)

Therefore, average number of marks is 26.08 (approximately)

Question 6: The number of students absent in a class was recorded every day for 120 days and the information is given in the following

Number of students absent (x)

0

1

2

3

4

5

6

7

Number of Days (f)

1

4

10

50

34

15

4

2

Find the mean number of students absent per day.

Solution:

Let mean assumed mean (A) be = 3

Number of students absent (x_i)

Number of Days (f)

u_{i}=x_{i}-A=x_{i}-3

f_{i}*u_{i}

0

1

-3

-3

1

4

-2

-8

2

10

-1

-10

3

50

0

0

4

34

1

34

5

15

2

30

6

4

3

12

7

2

4

8

 

\displaystyle\sum_{}^{} f_{i}=120

 

\displaystyle\sum_{}^{} f_{i}*u_{i}=63

Mean number of students absent per day = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                                                    = 3+(\frac{63}{120})

                                                                    = 3.525

Therefore, the mean number of students absent per day is 3.525

Question 7: In the first proof of reading of a book containing 300 pages the following distribution of misprints was obtained:

Number of misprints per page (x)

0

1

2

3

4

5

Number of page (f)

154

95

36

9

5

1

Find the average number of misprints per page.

Solution:

Let the assumed mean (A) be = 2

Number of misprints per page (x_i)

Number of page (f_i)

u_{i}=x_{i}-A=x_{i}-2

f_{i}*u_{i}

0

154

-2

-308

1

95

-1

-95

2

36

0

0

3

9

1

9

4

5

2

10

5

1

3

3

 

\displaystyle\sum_{}^{} f_{i}=300

 

\displaystyle\sum_{}^{} f_{i}*u_{i}=-381

Average number of misprints per day = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                                             = 2+(\frac{-381}{300})

                                                             = 0.73

Therefore, the average number of misprints per day is 0.73

Question 8: Find the mean from the following frequency distribution of marks at a test in statistics:

Number of accidents (x)

0

1

2

3

4

Number of workers (f)

70

52

34

3

1

Find the average number of misprints per page.

Solution:

Let the assumed mean (A) = 2

Number of accidents (x_i)

Number of workers (f_i)

u_{i}=x_{i}-A=x_{i}-2

f_{i}*u_{i}

0

70

-2

-140

1

52

-1

-52

2

34

0

0

3

3

1

3

4

1

2

2

 

\displaystyle\sum_{}^{} f_{i}=100

 

\displaystyle\sum_{}^{} f_{i}*u_{i}=-187

Average no of accidents per day workers = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

                                                                  = 2+(\frac{-187}{100})

                                                                  = 0.83

Therefore, average no of accidents per day workers 0.83

Question 9: Find the mean from the following frequency distribution of marks at a test in statistics:

Marks (x)

5

10

15

20

25

30

35

40

45

50

Number of students (f)

15

50

80

76

72

45

39

9

8

6

Solution:

Let the assumed mean (A) be = 25

Marks (x_i)

Number of students (f_i)

u_{i}=x_{i}-A=x_{i}-25

f_{i}*u_{i}

5

15

-20

-300

10

50

-15

-750

15

80

-10

-800

20

76

-5

-380

25

72

0

0

30

45

5

225

35

39

10

390

40

9

15

135

45

8

20

160

50

6

25

150

 

\displaystyle\sum_{}^{} f_{i}=400

 

\displaystyle\sum_{}^{} f_{i}*u_{i}=-1170

Mean = A+\frac{\displaystyle\sum_{}^{} f_{i}*u_{i}}{\displaystyle\sum_{}^{} f_{i}}

          = 25+(\frac{-1170}{400})

          = 22.075

Therefore, the mean is 22.075

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