Given the marks of 25 students in the class as \( \left\{\mathrm{m}_1, \mathrm{~m}_2, \ldots \ldots \ldots ., \mathrm{m}_{25}\right\} \) . Marks lie in the range of \( [1-100] \) and \( \overline{\mathrm{m}} \) is the mean. Which of the following quantity has the value zero?
Step-by-step Solution:
We are given the marks of 25 students as: \[ \{ m_1, m_2, \dots, m_{25} \} \] where the mean of these marks is denoted by \( \bar{m} \). Step 1: Understanding Mean The mean is defined as: \[ \bar{m} = \frac{1}{25} \sum_{i=1}^{25} m_i \] Rearranging: \[ \sum_{i=1}^{25} m_i = 25 \bar{m} \] Step 2: Evaluating the Given Options We analyze each given summation: 1. Option A: \[ \sum_{i=1}^{25} | m_i - \bar{m} | \] This represents the sum of absolute deviations from the mean. Since absolute values are always non-negative, this sum is not necessarily zero. 2. Option B: \[ \sum_{i=1}^{25} (m_i - \bar{m}) \] Expanding this: \[ \sum_{i=1}^{25} m_i - \sum_{i=1}^{25} \bar{m} \] \[ 25 \bar{m} - 25 \bar{m} = 0 \] Since this always evaluates to zero, this is the correct answer. 3. Option C: \[ \sum_{i=1}^{25} (m_i - \bar{m})^2 \] This is the sum of squared deviations, also known as the variance sum. Since squaring always results in non-negative values, this sum is not necessarily zero. 4. Option D: \[ \sum_{i=1}^{25} \frac{m_i}{\bar{m}} \] This does not have a general property that makes it zero. Final Answer: \[ \sum_{i=1}^{25} (m_i - \bar{m}) = 0 \] Thus, the correct option is B.