Correct answer is (b)0
We know that \(\bar{x}\) = \(\cfrac{\sum f_ix_i}{\sum f_i}\)
⇒ \(\bar{x}\) Σ fi = Σ fixi … (i)
Now, Σ fi (xi − \(\bar{x}\)) = Σ fixi − \(\bar{x}\)Σ fi
⇒ Σ fi (x − \(\bar{x}\)) = Σ fixi − Σ fixi [Using (i)]
⇒ Σ fi (xi − \(\bar{x}\)) = 0