If \(x_i\) are the class marks and \(f_i\) the corresponding frequencies with mean \(\bar{x}\), then \(\sum f_i(x_i-\bar{x})\) equals
0
-1
1
2
Step 1: Recall the formula for the mean of grouped data.
The mean is given by:
\(\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}\)
Step 2: Rearrange this formula.
Multiply both sides by \(\sum f_i\):
\(\bar{x} \cdot \sum f_i = \sum f_i x_i\)
Step 3: Now look at the given expression:
\(\sum f_i (x_i - \bar{x})\)
Step 4: Expand this expression.
\(\sum f_i (x_i - \bar{x}) = \sum f_i x_i - \sum f_i \bar{x}\)
Step 5: Notice that \(\bar{x}\) is the same for every term, so:
\(\sum f_i \bar{x} = \bar{x} \cdot \sum f_i\)
Step 6: From Step 2, we know that:
\(\sum f_i x_i = \bar{x} \cdot \sum f_i\)
Step 7: Substitute this back.
\(\sum f_i (x_i - \bar{x}) = (\bar{x} \cdot \sum f_i) - (\bar{x} \cdot \sum f_i)\)
Step 8: Simplify.
This becomes: \(0\).
Final Answer: \(\sum f_i (x_i - \bar{x}) = 0\). So the correct option is (A).