Q17 of 37 Page 14

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.

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

Here, let A = 202.5 and h = 1



WEIGHT(g)

MID - POINT(xi)

DEVIATION(di)


di = xi – 202.5

NUMBER OF PACKETS(fi)

ui = di/h

fiui

200 - 201

200.5

- 2

13

- 2

- 26

201 - 202

201.5

- 1

27

- 1

- 27

202 - 203

202.5 = A

0

18

0

0

203 - 204

203.5

1

10

1

10

204 - 205

204.5

2

1

2

2

205 - 206

205.5

3

1

3

3

TOTAL

    

70

  

- 38

We have got


A = 202.5, h = 1, Σfi = 70 & Σfiui = - 38


∵ mean is given by



⇒


⇒


Thus, mean is 201.96 g.

More from this chapter

All 37 →
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
16

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
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
19

The median of the following data is 16. Find the missing frequencies a and b if the Total of frequencies is 70. [CBSE 2013]


Class

0 - 5

5 - 10

10 - 15

15 - 20

20 - 25

25 - 30

30 - 35

35 - 40

Frequency

12

A

12

15

b

6

6

4