Answer
Range = $4$
Sample variance = $s^{2}=2.3$
Sample standard deviation = $s = 1.52$
Work Step by Step
Recall the shortcut formula for calculating the variance ($s^{2}$)$:$
$$s^{2}=\frac{\sum x_{i}^{2}-\frac{(\sum x_{i})^{2}}{n}}{n-1}$$
Let's create a table in which we compute $\displaystyle \sum x_{i}$ and $\displaystyle \sum x_{i}^{2}$:
$$ \begin{array}{lll}
& x & x^{2}\\
\hline & 4 & 16\\
& 2 & 4\\
& 1 & 1\\
& 0 & 0\\
& 1 & 1\\
\hline\rm Sum & 8 & 22
\end{array}$$
Plug in the given values (there are $n=5$ data items):
$$\begin{align*}
s^{2}&= \displaystyle \frac{22-\frac{(8)^{2}}{5}}{5-1} & & \\
& = 2.3 \\\\
s&= \sqrt{2.3} =1.52 \end{align*}$$
The range is the difference between the largest and smallest data item:
$4-0=4$
