Answer
$\approx1.58$
Work Step by Step
The mean of $n$ numbers is the sum of the numbers divided by $n$.
Hence the mean: $\frac{1+2+3+4+5 }{5}=3$
The standard deviation of $x_1,x_2,...,x_n$ is (where $\overline{x}$ is the mean of the data values): $\sqrt{\frac{(x_1-\overline{x})^2+(x_2-\overline{x})^2+...+(x_n-\overline{x})^2}{n-1}}$.
Hence here the the standard deviation is: $\sqrt{\frac{(1-3)^2+(2-3)^2+(3-3)^2+(4-3)^2+(5-1)^2}{5-1}}\approx1.58$
