Question

The algebraic sum of the deviations of a set of \(n\) values from their arithmetic mean is:

• A. \(n\)
• B. \(0\)
• C. \(2n\)
• D. None of these

Hint:

Remember the important identity: \[ \sum (x-\bar{x})=0. \] This property is one of the most important properties of the arithmetic mean.

Answer 1

The Correct Option is B

Concept: Let the observations be \[ x_1,x_2,\ldots,x_n \] with arithmetic mean \[ \bar{x} = \frac{\sum x_i}{n}. \] A fundamental property of the arithmetic mean is \[ \sum (x_i-\bar{x})=0. \]

Step 1: Write the sum of deviations. \[ \sum_{i=1}^{n}(x_i-\bar{x}). \]

Step 2: Separate the summation. \[ \sum_{i=1}^{n}x_i - \sum_{i=1}^{n}\bar{x}. \] Since \(\bar{x}\) is constant, \[ \sum_{i=1}^{n}\bar{x} = n\bar{x}. \] Therefore, \[ \sum_{i=1}^{n}(x_i-\bar{x}) = \sum_{i=1}^{n}x_i-n\bar{x}. \]

Step 3: Use the definition of mean. Since \[ \bar{x} = \frac{\sum x_i}{n}, \] we have \[ n\bar{x} = \sum x_i. \] Hence, \[ \sum_{i=1}^{n}(x_i-\bar{x}) = \sum x_i-\sum x_i = 0. \] Conclusion: \[ \boxed{0} \] Hence, the algebraic sum of deviations from the arithmetic mean is always zero.