Central Tendency & DispersionMTP June 24 Series IQuestion 3208 of 473
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The sum of the squares of deviations of a set of observations has the smallest value, when the deviations are taken from their

Options

AA. M.
BH. M.
CG. M.
DNone of these
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Correct Answer

✅ Option a — A. M.

All Options:

  • AA. M.
  • BH. M.
  • CG. M.
  • DNone of these

Detailed Solution & Explanation

**Theorem:** The sum of squared deviations ∑(xi−c)2\displaystyle \sum(x_i - c)^2 is minimized when c=xˉ\displaystyle c = \bar{x} (the Arithmetic Mean). **Proof:** Let f(c)=∑i=1n(xi−c)2\displaystyle f(c) = \sum_{i=1}^{n}(x_i - c)^2. f′(c)=−2∑i=1n(xi−c)=0f'(c) = -2\sum_{i=1}^{n}(x_i - c) = 0 ∑xi=nc\sum x_i = nc c=∑xin=xˉc = \frac{\sum x_i}{n} = \bar{x} f′′(c)=2n>0\displaystyle f''(c) = 2n > 0, confirming this is a minimum. Therefore, the sum of squared deviations is smallest when deviations are taken from the **Arithmetic Mean (A.M.)**. Hence, **Option A** is the correct answer.

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