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.