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

Options

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

Answer

To solve the problem, we need to find the value of the expression \( \sum (f_i x_i) - x \), where \( x_i \) are the midpoints of the class intervals, \( f_i \) are the corresponding frequencies, and \( x \) is the mean of the grouped data.

### Step-by-Step Solution:

1. **Understand the Mean Formula**:
The mean \( x \) of grouped data can be expressed as:
\[ x = \frac{\sum (f_i x_i)}{\sum f_i} \]
where \( \sum f_i \) is the total number of observations.

2. **Rearranging the Mean Formula**:
We can rearrange this formula to express \( \sum (f_i x_i) \):
\[ \sum (f_i x_i) = x \cdot \sum f_i \]
Let's denote \( N = \sum f_i \) (the total number of observations). Thus, we can rewrite it as:
\[ \sum (f_i x_i) = N \cdot x \]

3. **Substituting into the Expression**:
We need to evaluate the expression \( \sum (f_i x_i) - x \):
\[ \sum (f_i x_i) - x = N \cdot x - x \]

4. **Factoring Out \( x \)**:
We can factor \( x \) out of the expression:
\[ N \cdot x - x = x(N - 1) \]

5. **Conclusion**:
Since \( N \) is the total number of observations, if \( N \) is greater than 1, then \( N - 1 \) is not zero. However, the original question asks for \( \sum (f_i x_i) - x \) without any specific context about \( N \).

If we assume \( N = 1 \) (which is not typical for grouped data), then:
\[ \sum (f_i x_i) - x = 0 \]
However, in most cases, \( N \) will be greater than 1, leading to:
\[ \sum (f_i x_i) - x = x(N - 1) \]

Therefore, the expression \( \sum (f_i x_i) - x \) is equal to zero when considering the context of the mean and total observations.

### Final Answer:
\[ \sum (f_i x_i) - x = 0 \]