<p>If \(\mu\) is the mean of distribution \((y_i, f_i)\), then \(\Sigma f_i(y_i - \mu) =\)</p>
Step-by-Step Solution
Key Concept: By definition of mean, the sum of all deviations from the mean weighted by their frequencies equals zero, since mean is the balance point of the distribution.
<p><strong>Step 1:</strong> By definition, the mean μ of distribution (y_i, f_i) is: μ = Σf_i·y_i / Σf_i</p><p><strong>Step 2:</strong> Expand Σf_i(y_i - μ): Σf_i(y_i - μ) = Σf_i·y_i - μ·Σf_i</p><p><strong>Step 3:</strong> Substitute μ = Σf_i·y_i / Σf_i: Σf_i·y_i - (Σf_i·y_i / Σf_i)·Σf_i = Σf_i·y_i - Σf_i·y_i = 0</p><p><strong>Step 4:</strong> This is the <strong>fundamental property of mean</strong>: algebraic sum of deviations from mean is always zero.</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: C