Central Tendency & DispersionPYQ Dec. 21Question 2865 of 473
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If there are 3 observations 15, 20, 25 then sum of deviation of the observations from AM is

Options

A0\displaystyle 0
B5\displaystyle 5
C−5\displaystyle -5
D10\displaystyle 10
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Correct Answer

✅ Option a — 0\displaystyle 0

All Options:

  • A0\displaystyle 0
  • B5\displaystyle 5
  • C−5\displaystyle -5
  • D10\displaystyle 10

Detailed Solution & Explanation

**Step 1: Compute the arithmetic mean.** xˉ=15+20+253=603=20\bar{x} = \frac{15 + 20 + 25}{3} = \frac{60}{3} = 20 **Step 2: Compute deviations from AM.** | xi\displaystyle x_i | xi−xˉ\displaystyle x_i - \bar{x} | |--------|------------------| | 15 | 15−20=−5\displaystyle 15 - 20 = -5 | | 20 | 20−20=0\displaystyle 20 - 20 = 0 | | 25 | 25−20=5\displaystyle 25 - 20 = 5 | **Step 3: Sum of deviations.** ∑(xi−xˉ)=−5+0+5=0\sum(x_i - \bar{x}) = -5 + 0 + 5 = 0 This confirms the fundamental property: the algebraic sum of deviations from AM is always zero. Hence, **Option A** is the correct answer.

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