We know,
Standard deviation \(\sigma = \sqrt{\frac{\Sigma(x_i-\bar X)^2}{n}}\)
And, mean \(\bar X\) = \(\frac{1}{n}\Sigma x_i
\)
Now, multiply by a in xi
New standard deviation, \(\sigma_{new} = \sqrt{\frac{\Sigma(ax_i-\bar X)^2}{n}}\)
\(\sigma_{new} = \sqrt{\frac{\Sigma(a^2x_i-\bar X)^2}{n}}\)
\(\sigma_{new} = |a| \sqrt{\frac{\Sigma(x_i-\bar X)^2}{n}}\)
\(\sigma_{new} = |a|\sigma_{old}\)
Hence, Proved