Question

The algebraic sum of the deviations of a set of 'n' values from their mean is :

• A. 0
• B. n-1
• C. n
• D. n+1

Hint:

This is a fundamental property of the arithmetic mean. The mean is the balancing point of the data; thus, the sum of positive deviations from the mean exactly balances the sum of negative deviations, resulting in a net sum of zero.

Answer 1

The Correct Option is A

Step 1: Define the terms.
Let a set of 'n' values be $x_1, x_2, \ldots, x_n$. The mean of these values, denoted by $\bar{x}$, is given by: $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$ The deviation of a value $x_i$ from the mean $\bar{x}$ is $(x_i - \bar{x})$. Step 2: Calculate the algebraic sum of the deviations.
The algebraic sum of the deviations is $\sum_{i=1}^{n} (x_i - \bar{x})$. Expand the sum: $\sum_{i=1}^{n} (x_i - \bar{x}) = (x_1 - \bar{x}) + (x_2 - \bar{x}) + \ldots + (x_n - \bar{x})$ Rearrange the terms: $= (x_1 + x_2 + \ldots + x_n) - (\bar{x} + \bar{x} + \ldots + \bar{x} \text{ (n times)})$ $= \sum_{i=1}^{n} x_i - n\bar{x}$ Step 3: Substitute the definition of the mean into the expression.
We know that $\sum_{i=1}^{n} x_i = n\bar{x}$ from the definition of the mean. Substitute this into the sum of deviations: $\sum_{i=1}^{n} x_i - n\bar{x} = n\bar{x} - n\bar{x} = 0$ Step 4: Conclude the result.
The algebraic sum of the deviations of a set of 'n' values from their mean is always 0. Step 5: Compare with the given options.
The result is 0, which matches option (1). (1) 0