Question

The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50 . Compute the missing frequencies f1 & f2 class = frequency :0-20 = 5 ; 20-40 =f1; 40-60 = 10 ; 60-80 = f2; 80-100=7 ; 100-120 = 8 ; TOTAL =50

Answer 1

Hi ,

________________________
Class | f | class mark ( x ) | fx
________________________
0 - 20 | 5 | (0 + 20 ) /2 =10| 50
________________________
20- 40 | f1| 30 | 30f1
________________________
40-60 | 10| 50 | 500
________________________
60- 80| f2| 70 | 70f2
________________________
80-100| 7 | 90 | 630
________________________
100-120| 8| 110 | 880
_________________________

Sum of the frequencies =50

sigma f = 50

30 + f1 + f2 = 50

f1 + f2 = 50 - 30

f1 + f2 = 20 -----( 1 )

Sum of fixi = 2060 + 30f1 + 70f2

Mean = ( sum of fixi ) / ( sum of fi )

62.5 = ( 2060 + 30f1 + 70f2 ) /50

62.5 × 50 = 2060 + 30f1 + 70f2

3140 = 2060 + 30f1 + 70f2

3140 - 2060 = 30f1 + 70f2

1080 = 30f1 + 70f2

108 = 3f1 + 7f2 ------( 2 )

Sutract 3 × ( 1 ) from equation ( 2 ) , we get

4f2 = 48

f2 = 12

Put f2 = 12 in equation ( 1 ) , we get

f1 = 8

Therefore ,

f1 = 8 and f2 = 12

I hope this helps you.

:)