Q53 / 150 · MHT CET 2023 (14 May Shift 2) · Mathematics

If both mean and variance of 50 observations $x_1, x_2, \ldots, x_{50}$ are equal to 16 and 256 respectively, then mean of $\left(x_1-5\right)^2,\left(x_2-5\right)^2, \ldots \ldots,\left(x_{50}-5\right)^2$ is

  • A
    357
  • B
    367
  • Correct answer
    C
    377
  • D
    387

Solution

$\begin{aligned} & \text { Mean }=\frac{\sum x_i}{n} \\ & \Rightarrow 16=\frac{\sum x_{\mathrm{i}}}{50} \\ & \Rightarrow \sum x_{\mathrm{i}}=800 \\ & \text { Variance }=\frac{\sum x_{\mathrm{i}}^2}{\mathrm{n}}-(\bar{x})^2 \\ & \Rightarrow 256=\frac{\sum x_{\mathrm{i}}^2}{50}-(16)^2 \\ & \Rightarrow \frac{\sum x_{\mathrm{i}}^2}{50}=512 \\ & \Rightarrow \sum x_{\mathrm{i}}^2=25600 \\ & \text { New mean }=\frac{\sum\left(x_{\mathrm{i}}-5\right)^2}{\mathrm{n}} \\ & =\frac{\sum x_{\mathrm{i}}^2-10 \sum x_{\mathrm{i}}+\sum 25}{50} \\ & =\frac{25600-10(800)+25(50)}{50} \\ & =\frac{18850}{50} \\ & =377 \\ & \end{aligned}$

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

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