Q16 of 37 Page 14

Find the mean of the following data, using step - deviation method: [CBSE 2013]


Class

5 - 15

15 - 25

25 - 35

35 - 45

45 - 55

55 - 65

65 - 75

Frequency

6

10

16

15

24

8

7

We will find the mean of the data using step - deviation method, where A = Assumed mean and h = length of class interval.

Here, let A = 40 and h = 10



CLASS

MID - POINT(xi)

DEVIATION(di)


di = xi – 40

FREQUENCY(fi)

ui = di/h

fiui

5 - 15

10

- 30

6

- 3

- 18

15 - 25

20

- 20

10

- 2

- 20

25 - 35

30

- 10

16

- 1

- 16

35 - 45

40 = A

0

15

0

0

45 - 55

50

10

24

1

24

55 - 65

60

20

8

2

16

65 - 75

70

30

7

3

21

TOTAL

    

86

  

7

We have got


A = 40, h = 10, Σfi = 86 & Σfiui = 7


∵ mean is given by



⇒


⇒


Thus, mean is 40.81

More from this chapter

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14

Compute the mean of the following data: [CBSE 2013]


Class

1 - 3

3 - 5

5 - 7

7 - 9

Frequency

12

22

27

19
15

If the mean of the following frequency distribution is 24, find the value of p. [CBSE 2013]


Class

0 - 10

10 - 20

20 - 30

30 - 40

40 - 50

Frequency

3

4

p

3

2
17

The weights of tea in 70 packets are shown in the following table: [CBSE 2013]


Weight (in grams)

200 - 201

201 - 202

202 - 203

203 - 204

204 - 205

205 - 206

Number of packets

13

27

18

10

1

1

Find the mean weight of packets using step - deviation method.

 18

Find the mean of the following frequency distribution using a suitable method: 

[CBSE 2013]


Class

20 - 30

30 - 40

40 - 50

50 - 60

60 - 70

Frequency

25

40

42

33

10