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 evaluate the expression \( \sum (f_i x_i - \bar{x}) \), where \( x_i \) are the mid-points of the class intervals, \( f_i \) are the corresponding frequencies, and \( \bar{x} \) is the mean of the grouped data.

### Step-by-Step Solution:

1. **Understand the Mean Formula**:
The mean \( \bar{x} \) of the grouped data is given by:
\[ \bar{x} = \frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i} \]

2. **Rearranging the Expression**:
We need to evaluate:
\[ \sum_{i=1}^{n} (f_i x_i - \bar{x}) \]
This can be rewritten as:
\[ \sum_{i=1}^{n} f_i x_i - \sum_{i=1}^{n} \bar{x} \]

3. **Substituting the Mean**:
Since \( \bar{x} \) is a constant, we can express the second term as:
\[ \sum_{i=1}^{n} \bar{x} = \bar{x} \sum_{i=1}^{n} 1 = n \bar{x} \]
where \( n \) is the number of class intervals.

4. **Substituting Back into the Expression**:
Now, substituting back, we have:
\[ \sum_{i=1}^{n} f_i x_i - n \bar{x} \]

5. **Using the Mean Definition**:
From the mean formula, we know:
\[ \sum_{i=1}^{n} f_i x_i = \bar{x} \sum_{i=1}^{n} f_i \]
Therefore, we can substitute this into our expression:
\[ \bar{x} \sum_{i=1}^{n} f_i - n \bar{x} \]

6. **Factoring Out \( \bar{x} \)**:
Factoring \( \bar{x} \) out gives:
\[ \bar{x} \left( \sum_{i=1}^{n} f_i - n \right) \]

7. **Understanding the Result**:
The term \( \sum_{i=1}^{n} f_i \) represents the total frequency, which is equal to \( n \) if there are \( n \) class intervals with a frequency of 1 each. Thus, if \( \sum_{i=1}^{n} f_i = n \), we have:
\[ \bar{x} (n - n) = \bar{x} \cdot 0 = 0 \]

### Final Result:
Thus, we conclude that:
\[ \sum_{i=1}^{n} (f_i x_i - \bar{x}) = 0 \]