Find the missing frequency in the following distribution if N is 60 and median is 40.

Explanation

`:.N=60`
`:. 15+f_(1)+f_(2)=60 or f_(1)+f_(2)=45`
Median Size `=(N)/(2)=(60)/(2)=30`th term
(`:.` Median is 40 , so Median class is 30-60)
`M=l_(1)+(N/(2)-c.f.)/(f)xxi`
`rArr40=30+(30-(5+f_(1)))/(f_(2))xx30`
`rArr10=((25-f_(1)))/(f_(2))xx30`
`rArr10f_(2)=750-30f_(1)`
`30f_(1)+10f_(2)=750`
Thus, we get
`f_(1)+f_(2)=45` ... (i)
`30f_(1)+10f_(2)=750` ...(ii)
Dividing equation (ii) by 10, we get
`3f_(1)+f_(2)=75` ... (iii)
Now subtracting (i) from (iii), we get
`3f_(1)+f_(2)=75`
`(underset(-)(f_(1))underset(-)(+)f_(2)=underset(-)(45))/(2f_(1)=30)`
`rArr f_(1)=15`
Now substituting `f_(1)=15` in (i) , we get
`f_(2)=45-15=30`
Hence , `f_(1)=15,f_(2)=30`.