The average of n numbers x1, x2…..xn is $\overline{\boldsymbol{x}} .$ Then the value of $\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)$ s equal to.
A. n B. 0 C. n$\overline{\boldsymbol{x}}$ D. $\overline{\boldsymbol{x}}$ Answer: Option BShow Answer
Solution(By Apex Team)
According to the question,
Average of ‘n’ number’s x1, x2…..xn is $\overline{\boldsymbol{x}}$
Sum of n numbers = n$\overline{\boldsymbol{x}}$
$\begin{array}{l}\therefore\large\sum_{i=1}^n\left(x_1-\overline{x}\right)\\
\text{ Put }\mathrm{i}=1,2,3\ldots..\text{ n then }\\
\left\{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\ldots..(\mathrm{xn}-\mathrm{n}\overline{x})\right\}\\
\text{ As we know that }\\
x_1+x_2+x_3+\ldots..+x_nx=n\overline{x}\\
=(n\overline{x}-n\overline{x})\\
=0\end{array}$
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