Central Tendency & DispersionMTP Mar 21Question 3106 of 473
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The equation of a line is 5x+2y=17\displaystyle 5x + 2y = 17. Mean deviation of y\displaystyle y about mean is 5\displaystyle 5. Calculate mean deviation of x\displaystyle x about mean.

Options

A−2\displaystyle -2
B2\displaystyle 2
C−4\displaystyle -4
DNone
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Correct Answer

✅ Option b — 2\displaystyle 2

All Options:

  • A−2\displaystyle -2
  • B2\displaystyle 2
  • C−4\displaystyle -4
  • DNone

Detailed Solution & Explanation

We are given the equation of a line: 5x+2y=17  ⟹  5x=−2y+17  ⟹  x=−0.4y+3.45x + 2y = 17 \implies 5x = -2y + 17 \implies x = -0.4y + 3.4 Since mean deviation is independent of change of origin but affected by change of scale, the mean deviation of x\displaystyle x (M.D.x\displaystyle \text{M.D.}_x) is related to the mean deviation of y\displaystyle y (M.D.y\displaystyle \text{M.D.}_y) by: M.D.x=∣a∣×M.D.y\text{M.D.}_x = |a| \times \text{M.D.}_y Substitute the given values (a=−0.4\displaystyle a = -0.4 and M.D.y=5\displaystyle \text{M.D.}_y = 5): M.D.x=∣−0.4∣×5=0.4×5=2\text{M.D.}_x = |-0.4| \times 5 = 0.4 \times 5 = 2 Hence, **Option B** is the correct answer.

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