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If each of the observation x1,x2,...,xn is increased by 'a', where 'a' is a negative or position number, then show that the variance remains unchanged.

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Let \(\bar{x}\) be the mean of 

x1,x2,...,xn.

Then, the variance is given by

\(σ_1^2=\frac{1}{n}\displaystyle\sum_{i=1}^{n} (x_i-\bar{x})^2\)

If 'a' is added to each observation, then the new observation will be yi = xi+ a let the mean of the new observation be \(\bar{y}\)

Then :

\(\bar{y}=\frac{1}{n}\displaystyle\sum_{i=1}^{n} {y_i}\) 

\(=\frac{1}{n}\displaystyle\sum_{i=1}^{n} (x_i+a)\)

\(=\frac{1}{n}[\displaystyle\sum_{i=1}^{n}x_i+\displaystyle\sum_{i=1}^{n}a]\)

\(=\frac{1}{n}[\displaystyle\sum_{i=1}^{n}x_i+{n}a]\)⇒ \(\bar{y}=\bar{x}+a\)

New variance,

\(=\bar{x}+a\)

\(σ_1^2=\frac{1}{n}\displaystyle\sum_{i=1}^{n} (y_i-\bar{y})^2\)

\(=\frac{1}{n}\displaystyle\sum_{i=1}^{n} (x_i+a-\bar{x}-a)^2\)

\(=\frac{1}{n}\displaystyle\sum_{i=1}^{n} (x_i-\bar{x})^2\)

\(=σ_1^2\)

∴ Variance remains unchanged.

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