The average of n numbers x_{1}, x_{2}, ldots, | vedclass

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(B) The average $\bar{x}$ of $n$ numbers $x_{1}, x_{2}, \ldots, x_{n}$ is given by the formula:$\bar{x} = \frac{\sum_{i=1}^{n} x_{i}}{n}$This implies

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(B) The average $\bar{x}$ of $n$ numbers $x_{1}, x_{2}, \ldots, x_{n}$ is given by the formula:$\bar{x} = \frac{\sum_{i=1}^{n} x_{i}}{n}$This implies that $\sum_{i=1}^{n} x_{i} = n\bar{x}$ .....$(i)$Now,we need to find the value of $\sum_{i=1}^{n}(x_{i}-\bar{x})$:$\sum_{i=1}^{n}(x_{i}-\bar{x}) = \sum_{i=1}^{n} x_{i} - \sum_{i=1}^{n} \bar{x}$Since $\bar{x}$ is a constant,$\sum_{i=1}^{n} \bar{x} = n\bar{x}$.Substituting the value from $(i)$:$\sum_{i=1}^{n}(x_{i}-\bar{x}) = n\bar{x} - n\bar{x} = 0$Therefore,the sum of deviations of observations from their mean is always $0$.

The average of $n$ numbers $x_{1}, x_{2}, \ldots, x_{n}$ is $\bar{x}$. Then,the value of $\sum_{i=1}^{n}(x_{i}-\bar{x})$ is equal to

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