19.Find the missing frequencies f1 , f2 and f3 in the following frequency distribution , when it is given taht f2 : f3 = 4 : 3 , and mean = 50
Class interval | Frequency |
0-20 20-40 40-60 60-80 80-100 |
17 f1 f2 f3 19 |
Total | 120 |
Dear Student,
Please find below the solution to the asked query:
Here we assume f1 = x ,
Given f2 : f3 = 4 : 3 , We assume ratio coefficient = y , So
f2 = 4 y and f3 = 3 y
We form table from given information , As :
Also given : Sum of frequency f = 120 , So
x + 7 y + 36 = 120 ,
x + 7 y = 84 --- ( 1 )
And mean of given data = 50
So,
Mean = = 50 , Substitute values we get :
30 x + 410 y + 1880 = 6000
30 x + 410 y = 4120
3 x + 41 y = 412 --- ( 2 )
Now we multiply by 3 in equation 1 and get :
3 x + 21 y = 252 --- ( 3 )
Now we subtract equation 3 from equation 2 and get :
20 y = 160 ,
y = 8 , Substitute that value in equation 1 we get :
x + 7 ( 8 ) = 84 ,
x + 56 = 84 ,
x = 28
Therefore,
f1 = 28 and f2 = 4 ( 8 ) = 32 and f3 = 3 ( 8 ) = 24 ( Ans )
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
Please find below the solution to the asked query:
Here we assume f1 = x ,
Given f2 : f3 = 4 : 3 , We assume ratio coefficient = y , So
f2 = 4 y and f3 = 3 y
We form table from given information , As :
Class | Mid point ( x ) | Frequency ( f ) | fx |
0 - 20 | 10 | 17 | 170 |
20 - 40 | 30 | x | 30 x |
40 - 60 | 50 | 4 y | 200 y |
60 - 80 | 70 | 3 y | 210 y |
80 - 100 | 70 | 19 | 1710 |
f = x + 7 y + 36 | fx = 30 x + 410 y + 1880 | ||
Also given : Sum of frequency f = 120 , So
x + 7 y + 36 = 120 ,
x + 7 y = 84 --- ( 1 )
And mean of given data = 50
So,
Mean = = 50 , Substitute values we get :
30 x + 410 y + 1880 = 6000
30 x + 410 y = 4120
3 x + 41 y = 412 --- ( 2 )
Now we multiply by 3 in equation 1 and get :
3 x + 21 y = 252 --- ( 3 )
Now we subtract equation 3 from equation 2 and get :
20 y = 160 ,
y = 8 , Substitute that value in equation 1 we get :
x + 7 ( 8 ) = 84 ,
x + 56 = 84 ,
x = 28
Therefore,
f1 = 28 and f2 = 4 ( 8 ) = 32 and f3 = 3 ( 8 ) = 24 ( Ans )
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards