Central Tendency & DispersionMTP June 24 Series IIQuestion 3114 of 473
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If two variables x\displaystyle x and y\displaystyle y are related by 2x\displaystyle 2x and 3y−7=0\displaystyle 3y - 7 = 0 and the mean and mean deviation about mean of x\displaystyle x are 1\displaystyle 1 and 0.3\displaystyle 0.3 respectively, then the coefficient of mean deviation of y\displaystyle y about mean is:

Options

A−.5\displaystyle -.5
B.4\displaystyle .4
C.12\displaystyle .12
D.50\displaystyle .50
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Correct Answer

✅ Option c — .12\displaystyle .12

All Options:

  • A−.5\displaystyle -.5
  • B.4\displaystyle .4
  • C.12\displaystyle .12
  • D.50\displaystyle .50

Detailed Solution & Explanation

We are given: 2x+3y−7=0  ⟹  3y=−2x+7  ⟹  y=−23x+73\displaystyle 2x + 3y - 7 = 0 \implies 3y = -2x + 7 \implies y = -\frac{2}{3}x + \frac{7}{3}. - Mean of x\displaystyle x (xˉ\displaystyle \bar{x}) = 1\displaystyle 1 - Mean deviation of x\displaystyle x (M.D.x\displaystyle \text{M.D.}_x) = 0.3\displaystyle 0.3 1. Find Mean of y\displaystyle y (yˉ\displaystyle \bar{y}): yˉ=−23xˉ+73=−23(1)+73=53≈1.67\bar{y} = -\frac{2}{3}\bar{x} + \frac{7}{3} = -\frac{2}{3}(1) + \frac{7}{3} = \frac{5}{3} \approx 1.67 2. Find Mean deviation of y\displaystyle y (M.D.y\displaystyle \text{M.D.}_y): M.D.y=∣−23∣×M.D.x=23×0.3=0.2\text{M.D.}_y = \left|-\frac{2}{3}\right| \times \text{M.D.}_x = \frac{2}{3} \times 0.3 = 0.2 3. Calculate coefficient of mean deviation of y\displaystyle y: Coefficient=M.D.y∣yˉ∣=0.25/3=0.12\text{Coefficient} = \frac{\text{M.D.}_y}{|\bar{y}|} = \frac{0.2}{5/3} = 0.12 Hence, **Option C** is the correct answer.

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