Let [tex]\mathbf x=\{x_i~:~1\le i\le n\}[/tex] be the sample, with [tex]n[/tex] the size of the sample.
The mean of the sample is
[tex]\displaystyle\bar x=\frac1n\sum_{i=1}^nx_i[/tex]
Let [tex]\mathbf y[/tex] be the same data set but with each value scaled up by 5. Then the mean here is
[tex]\bar y=\displaystyle\frac1n\sum_{i=1}^ny_i=\frac5n\sum_{i=1}^nx_i=5\bar x[/tex]
The standard deviation for [tex]\mathbf x[/tex] is given by
[tex]s_{\mathbf x}=\sqrt{\displaystyle\frac1{n-1}\sum_{i=1}^n(\bar x-x_i)^2}[/tex]
For [tex]\mathbf y[/tex], you would have
[tex]s_{\mathbf y}=\sqrt{\displaystyle\frac1{n-1}\sum_{i=1}^n(\bar y-y_i)^2}[/tex]
[tex]s_{\mathbf y}=\sqrt{\displaystyle\frac1{n-1}\sum_{i=1}^n(5\bar x-5x_i)^2}[/tex]
[tex]s_{\mathbf y}=\sqrt{\displaystyle\frac{25}{n-1}\sum_{i=1}^n(\bar x-x_i)^2}[/tex]
[tex]s_{\mathbf y}=5\sqrt{\displaystyle\frac1{n-1}\sum_{i=1}^n(\bar x-x_i)^2}[/tex]
[tex]s_{\mathbf y}=5s_{\mathbf x}[/tex]
