If `x_(i)'s` are the mid-points of the class intervals of grouped data, `f_(i)'s` are the corresponding frequencies and `bar(x)` is the mean, then `sum(f_(i)x_(i)-bar(x))` equal to

Options

A. 0
B. `-1`
C. 1
D. 2

Explanation

To solve the problem, we need to analyze the expression given in the question:

\[ \sum (f_i x_i - \bar{x}) \]

Where:
- \(x_i\) are the mid-points of the class intervals,
- \(f_i\) are the corresponding frequencies,
- \(\bar{x}\) is the mean of the data.

### Step-by-Step Solution:

1. **Understanding the Mean**:
The mean \(\bar{x}\) of grouped data can be calculated using the formula:
\[ \bar{x} = \frac{\sum (f_i x_i)}{n} \]
where \(n = \sum f_i\) (the total frequency).

2. **Rearranging the Expression**:
We can rewrite the expression \(\sum (f_i x_i - \bar{x})\) as:
\[ \sum (f_i x_i - \bar{x}) = \sum f_i x_i - \sum \bar{x} \]
Since \(\bar{x}\) is a constant, we can express \(\sum \bar{x}\) as:
\[ \sum \bar{x} = n \bar{x} \]
Thus, we have:
\[ \sum (f_i x_i - \bar{x}) = \sum f_i x_i - n \bar{x} \]

3. **Substituting the Mean**:
From the mean formula, we know that:
\[ n \bar{x} = \sum f_i x_i \]
Therefore, we can substitute this into our expression:
\[ \sum (f_i x_i - \bar{x}) = \sum f_i x_i - \sum f_i x_i \]

4. **Simplifying the Expression**:
This simplifies to:
\[ \sum (f_i x_i - \bar{x}) = 0 \]

### Final Result:
Thus, the value of \(\sum (f_i x_i - \bar{x})\) is equal to **0**.