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4.Average
hard
The average of $n$ numbers $x_{1}, x_{2}, \ldots, x_{n}$ is $x$. Then, the value of $\sum \limits_{i=1}^{n}\left(x_{i}-\bar{x}\right)$ is equal to
A
$n$
B
$0$
C
$n \bar{x}$
D
$\bar{x}$
Solution
By the formula
$\frac{\left(x_{1}+x_{2}+x_{3} \cdots+x_{n}\right)}{n}=\bar{x}$ …..$(i)$
$\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)$
Putting value of $i$ from $(i)$ to $n$.
$\left(x_{1}-\bar{x}\right)+\left(x_{2}-\bar{x}\right)+\left(x_{3}-\bar{x}\right)+\ldots .+\left(x_{n}-\bar{x}\right)$ from $(i)$
$=n \frac{\left(x_{1}+x_{2}+\ldots+x_{n}\right)}{n}-n \bar{x}$
$\left[\frac{\left(x_{1}+x_{2}+\ldots .+x_{n}\right)}{n}=\bar{x}\right]=n \bar{x}-n \bar{x}=0$
Standard 13
Quantitative Aptitude
