Answer
$3$
Work Step by Step
The mean of $n$ numbers is the sum of the numbers divided by $n$.
Hence the mean: $\frac{3\cdot6+1\cdot9+3\cdot12}{7}=9$
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: $\sqrt{\frac{3(6-9)^2+(9-9)^2+3(12-9)^2}{7-1}}=3$
