Question

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Answer 1

Answer:

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Answer 2

Given that:

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$\bar {X}$ is the mean of n observations ${x_1} , {x_2} ,....., {x_n}$

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To prove:

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$\displaystyle\sum\limits_{i = 1}^{n} (x_i –\bar {X}$) = 0

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Solution:

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Here as per information provided in the question,

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$\bar {X}$ = 1/n$\displaystyle\sum\limits_{i = 1}^{n}(x_i)$ = n$\bar {X}$ = $\displaystyle\sum\limits_{i = 1}^{n} {x_i}$ ........(i)

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Now,

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$\displaystyle\sum\limits_{i = 1}^{n} ({x_i} – \bar {X})$ = $({x_i} –\bar {X}$) + $({x_2 }– \bar {X}$ ) + .... +

(${x_n} – \bar {X})$

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$\implies$$\displaystyle\sum\limits_{i = 1}^{n} ({x_i} – \bar {X})$ = ${x_1} + {x_2} +..... {x_n}$ – n $\bar {X}$

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$\implies$$\displaystyle\sum\limits_{i = 1}^{n} ({x_i} – \bar {X})$ = $\displaystyle\sum\limits_{i = 1}^{n} ({x_i} – n\bar {X})$

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Now using equation 1,

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$\displaystyle\sum\limits_{i = 1}^{n} ({x_i} –\bar {X})$ = n$\bar {X}$ – n$\bar {X}$

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$\implies$$\displaystyle\sum\limits_{i = 1}^{n} (x_i –\bar {X}$) = 0

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Thus, proved.