Q15 of 37 Page 14

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

For equal class intervals, we will solve by finding mid points of these classes using direct method.


CLASS

MID - POINT(xi)

FREQUENCY(fi)

fixi

0 - 10

5

3

15

10 - 20

15

4

60

20 - 30

25

p

25p

30 - 40

35

3

105

40 - 50

45

2

90

TOTAL

  

12 + p

270 + 25p

We have got


Σfi = 12 + p & Σfixi = 270 + 25p


∵ mean is given by



⇒


⇒ 288 + 24p = 270 + 25p


⇒ 25p – 24p = 288 – 270


⇒ p = 18


Thus, p is 18

More from this chapter

All 37 →
13

If the mean of the following frequency distribution is 54, find the value of p. [CBSE 2006]


Class

0 - 20

20 - 40

40 - 60

60 - 80

80 - 100

Frequency

7

p

10

9

13
14

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


Class

1 - 3

3 - 5

5 - 7

7 - 9

Frequency

12

22

27

19
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
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.