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  • Mean deviation for n observations x1, x2, ..., xn from their mean x bar is given by A. i from 1 to n ximinus xbar

Mean deviation for n observations x1, x2, ..., xn from their mean \bar{X} is given by

\\A. \sum _{i=1}^{n}\left(x_{1}-\bar{x}\right)\\\\ B.\frac{1}{n} \sum _{i=1}^{n} \left |\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}} \right |\\\\C. \sum _{i=1}^{n} {\left(x_{i}-\bar{x}\right)^{2}}\\\\D \frac{1}{n}\sum _{i=1}^{n}(x-\bar{x})^{2}

Answers (1)

\\ \text{We know for n observations }x_{1},~x_{2}, \ldots \ldots ., x_{n}~having~\overline{X}\text{ is given by } \\ \\ M.D=\frac{ \Sigma d_{i}}{n}~ \\ \\ \text{ But we know that }d_{i}= \vert x_{i}-\overline{x} \vert ~~ \\ \\ \text{So,~mean~deviation becomes M.D}=\frac{ \Sigma \vert x_{i}-\overline{x} \vert }{n}~~ \\ \\ Or\: \: M.D=\frac{1}{n} \sum _{i=0}^{n} \vert x_{i}-\overline{x~} \vert \\

 

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infoexpert21

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