Q75 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Mathematics

If 10 is the mean deviation of ' $n$ ' observations $x_1, x_2$, $x_3, \ldots, x_n$ then the mean deviation of the observations $\frac{2 x_1+5}{3}, \frac{2 x_2+5}{3}, \frac{2 x_3+5}{3}, \ldots ., \frac{2 x_n+5}{3}$ is

  • A
    $\frac{25}{3}$
  • B
    $\frac{40}{9}$
  • Correct answer
    C
    $\frac{20}{3}$
  • D
    15

Solution

Given observation are $x_1, x_2, x_3 \ldots \ldots x_n$. $$ \bar{x}=\frac{x_1+x_2+x_3+\ldots . .+x_n}{n} $$ New $\bar{x}=\frac{1}{n}\left[\frac{2 x_1+5}{3}+\frac{2 x_2+5}{3}+\ldots . .+\frac{2 x_n+5}{3}\right]$ New $\overline{\mathrm{x}}=\frac{2 \overline{\mathrm{x}}_1+5}{3}$ Mean deviation of the observations $$ \begin{aligned} & \frac{2 \mathrm{x}_1+5}{3}, \frac{2 \mathrm{x}_2+5}{3}, \ldots \ldots \frac{2 \mathrm{x}_{\mathrm{n}}+5}{3} \\ & \text { New M.D. }=\frac{1}{\mathrm{n}} \sum_{\mathrm{x}=1}^{\mathrm{n}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right| \\ & =\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}}\left|\frac{2 \mathrm{x}_{\mathrm{i}}+5}{3}-\frac{2 \overline{\mathrm{x}}+5}{3}\right| \\ & =\frac{1}{\mathrm{n}}\left[\left|\frac{2 \mathrm{x}_1+5-2 \overline{\mathrm{x}}-5}{3}\right|+\left|\frac{2 \mathrm{x}_2+5-2 \overline{\mathrm{x}}-5}{3}\right|+\right. \\ & \left.\ldots+\left|\frac{2 \mathrm{x}_{\mathrm{n}}+5-2 \overline{\mathrm{x}}-5}{3}\right|\right] \\ & =\left[\frac{2}{3} \frac{1}{\mathrm{n}}\left|\mathrm{x}_1-\overline{\mathrm{x}}\right|+\left|\mathrm{x}_2-\overline{\mathrm{x}}\right|+\left|\mathrm{x}_3-\overline{\mathrm{x}}\right|+\ldots .+\left|\overline{\mathrm{x}}_{\mathrm{n}}-\overline{\mathrm{x}}\right|\right] \\ & =\frac{2}{3} \times 10=\frac{20}{3} \end{aligned} $$ So, option (c) is correct

Hence, the correct answer is option: 'C'.

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