Let \(\{x_1,x_2,\ldots,x_n\}\) be the data set. The variance of the data is
\(s^2=\displaystyle\sum_{i=1}^n\frac{(x_i-\bar x)^2}{n-1}\)
where \(\bar x\) is the mean of the data. You have
\(\bar x=\displaystyle\sum_{i=1}^n\frac{x_i}n\)
so if each data point was multiplied by 2.4, you have a "new" mean of
\({\bar x}^*=\displaystyle\sum_{i=1}^n\frac{2.4x_i}n=2.4\sum_{i=1}^n\frac{x_i}n=2.4\bar x\)
So in the variance formula, you have
\({s^2}^*=\displaystyle\sum_{i=1}^n\frac{(2.4x_i-2.4\bar x)^2}{n-1}=2.4^2\sum_{i=1}^n\frac{(x_i-\bar x)^2}{n-1}=5.76s^2\)
so the standard deviation of the new data set would be
\(s^*=\sqrt{{s^2}^*}=\sqrt{5.76s^2}=2.4s\)
The original standard deviation was \(s=5\) so the new one would be \(s^*=2.4\times5=12\).
