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Question-1

The following table shows marks secured by 140 students in an examination:
 

Marks

0-10

10-20

20-30

30-40

40-50

    Number of

    students

20

24

40

36

20

Calculate mean marks by using all the three methods, i.e., direct method, assumed mean method and step-deviation method.

Solution:
 Direct Method

Marks

Number of Students (fi)

Mid-point (xi)

fixi

0 - 10

20

5

100

10 - 20

24

15

360

20 - 30

40

25

1000

30 - 40

36

35

1260

40 - 50

20

45

900

Total

140

 

3620

Hence mean = = = 25.86 marks
Assumed Mean Method
 

Marks

Number of Students(fi)

Mid-point (xi)

Deviation
di= xi - 25

fidi

0 - 10

20

5

-20

-400

10 - 20

24

15

-10

-240

20 - 30

40

a = 25

0

0

30 - 40

36

35

10

360

40 - 50

20

45

20

400

Total

140

   

120

 

Hence mean = a + = 25 + = 25.86 marks

Step-Deviation Method
 

Marks

Number of Students (fi)

Mid-point (xi)

Deviation
ui= (xi - 25)/10

fiui

0-10

20

5

-2

-40

10 - 20

24

15

-1

-24

20 - 30

40

a=25

0

0

30 - 40

36

35

1

36

40 - 50

20

45

2

40

Total

140

   

12


Hence mean = a + h × [where h = 10]

 

                        = 25 + 10 × = 25.86 marks
 

Question-2

Find the mean age in years from the frequency distribution given below:
 

Class Interval of Age in years

Frequency
ƒi

25 - 29

30 - 34

35 - 39

40 - 44

45 - 49

50 - 54

55 - 59

4

14

22

16

6

5

3

Total

70



Solution

Class Interval of Age in years

Frequency
(fi)

Mid-point (xi)

Deviation
ui= (xi - 42)/5

fiui

25 - 29

4

27

-3

-12

30 - 34

14

32

-2

-28

35 - 39

22

37

-1

-22

40 - 44

16

a = 42

0

0

45 - 49

6

47

1

6

50 - 54

5

52

2

10

55 - 59

3

57

3

9

Total

70

   

-37


Hence mean = a + h ×  [where h = 5]

                     = 42 + 5 ×  = 39.36 years
 

Question-3

New Page 2 Find the mean marks from the following data:

Marks

Number of Students


Below 10

Below 20

Below 30

Below 40

Below 50

Below 60

Below 70

Below 80

Below 90

Below 100


5

9

17

29

45

60

70

78

83

85


Solution:
 

Marks

Number of Students (fi)

Mid-point (xi)

Deviation
ui =(xi - 55)/10

fiui

0 - 10

5

5

-5

-25

10 - 20

9 - 5 = 4

15

-4

-16

20 - 30

17 - 9 = 8

25

-3

-24

30 - 40

29 - 17 = 12

35

-2

-24

40 - 50

45 - 29 = 16

45

-1

-16

50 - 60

60 - 45 = 15

a = 55

0

0

60 - 70

70 - 60 = 10

65

1

10

70 - 80

78 - 70 = 8

75

2

16

80 - 90

83 - 78 = 5

85

3

15

90 - 100

85 - 83 = 2

95

4

8

Total

85

   

-56


 Hence mean = a + h ×  [where h = 10]

                           = 55 + 10 ×  = 48.41 marks
 

Question-4

 New Page 2 Find the mean age of 100 residents of a colony from the following data:

Age in years (Greater than or equal to)

Number of Persons

0

10

20

30

40

50

60

70

100

90

75

50

25

15

5

0



Solution:
 

Age

Number of Persons (fi)

Mid-point (xi)

Deviation
ui= (xi - 35)/10

fiui

0 - 10

100 - 90 = 10

5

-3

-30

10 - 20

90 - 75 = 15

15

-2

-30

20 - 30

75 - 50 = 25

25

-1

-25

30 - 40

50 - 25 = 25

a = 35

0

0

40 - 50

25 - 15 = 10

45

1

10

50 - 60

15 - 5 = 10

55

2

20

60 - 70

5 - 0 = 5

65

3

15

Total

100

   

-40

Hence mean = a + h ×    [where h = 10]

                      = 35 + 10 × = 31 years


Question-5

If the mean of the following data is 21.5, find the value of k:
            

xi

5

15

25

35

45

fi

6

4

3

k

2


Solution:

xi

fi

xifi

5 

6 

30 

15 

4 

60 

25 

3 

75 

35 

k 

35k 

45

2

90

Total

15 + k

255 + 35k


Mean =
= 21.5
21.5 (15 + k) = 255 + 35k
21.5
×15 + 21.5k = 255 + 35k
35k - 21.5k = 322.5 - 255
13.5k = 67.5

 k = = 5

Question-6

Find the values of f1 and f2 of the frequency, if the mean of the following frequency distribution is 21.4 and the total frequency is 40.

Class-interval

0 - 8

8 - 16

16 - 24

24 - 32

32 - 40

Frequency

6

f1

10

f2

9


Solution:
Take a = 10 and h = 8.

Class interval

Mid-point xi

fi

fi xi

0 - 8

4

6

24

8 - 16

12

f1

12 f1

16 - 24

20

10

200

24 - 32

28

f2

28 f2

32 - 40

36

9

324

25 + f1+ f2 = 40

548 + 12 f1 + 28 f2


 f1 + f2 = 15 ……………………..(i)
=
21.4 =
856 = 548 + 12f1 + 28f2
12f1 + 28f2 = 308
3f1 + 7f2 = 77 …………………..(ii)

(ii) - 3 × (i) 

3f1 + 7f2 = 77
3f1 + 3f2 = 45
------------------
        4f2 = 32
------------------

f2 = 8 f1 = 15 - 8 = 7

Question-7

The median of the observations 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find the value of x.

Solution:
The number of observations = n = 10. Since n is even ,
Median =
 24 =   = =
x = 21.

Question-8

Find the median of the following data: 19, 25, 59, 48, 35, 31, 30, 32, 51. If 25 is replaced by 52 and 19 by 29, what will be the new median?

Solution:
Arranging the data in ascending order: 19, 25, 30, 31, 32, 35, 48, 51, 59.
The number of observations n = 9 (odd)
Median = observation = 5th observation = 32.
Hence Median = 32.
If 25 is replaced by 52 and 19 by 29, the new observations in ascending order are :
29, 30, 31, 32, 35, 48, 51, 52, 59.
New Median = 5th observation = 35.

Question-9

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:
 

xi

fi

xi fi

15

2

30

17

3

51

19

4

76

20 + p

5p

100p + 5p2

23

6

138

Total

15 + 5p

5p2+100p + 295


Mean = =
 
20 =
5p2+100p + 295 = 300 + 100p

             5p2 - 5 = 0

           5(p2 - 1) = 0

    (p + 1)(p - 1) = 0

        p = 1 or p = -1




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