If `mu` the mean of distribution `(y_(i),f_(i))`, then `sum f_(i)(y_(i)-mu)`=

Options

A. M.D.
B. S.D.
C. 0
D. Relative frequency

Explanation

To solve the problem, we need to find the value of \( \sum f_i (y_i - \mu) \), where \( \mu \) is the mean of the distribution given by the pairs \( (y_i, f_i) \).

### Step-by-Step Solution:

1. **Understand the Mean**: The mean \( \mu \) of the distribution is defined as:
\[ \mu = \frac{\sum f_i y_i}{\sum f_i} \]
This means that the mean is the total weighted value divided by the total weight.

2. **Rearranging the Mean Formula**: From the mean formula, we can express \( \sum f_i y_i \) in terms of \( \mu \):
\[ \sum f_i y_i = \mu \sum f_i \]

3. **Substituting into the Summation**: We need to evaluate \( \sum f_i (y_i - \mu) \):
\[ \sum f_i (y_i - \mu) = \sum f_i y_i - \sum f_i \mu \]

4. **Breaking Down the Summation**: We can separate the terms:
\[ \sum f_i (y_i - \mu) = \sum f_i y_i - \mu \sum f_i \]

5. **Using the Rearranged Mean**: Now, substitute \( \sum f_i y_i \) from step 2 into the equation:
\[ \sum f_i (y_i - \mu) = \mu \sum f_i - \mu \sum f_i \]

6. **Simplifying the Expression**: The two terms on the right-hand side cancel each other out:
\[ \sum f_i (y_i - \mu) = 0 \]

### Final Answer:
Thus, we conclude that:
\[ \sum f_i (y_i - \mu) = 0 \]