$\begin{array}{|c|c|c|c|}
\hline x & 0 & 1 & -1 \
\hline P(x) & \frac{10}{16} & \frac{3}{16} & \frac{3}{16} \
\hline
\end{array}\begin{aligned} & \operatorname{Var}(x)=E\left(x^2\right)-[E(x)]^2 \ & =\sum_{i=1}^3 x_i^2 P\left(x_i\right)-(\mu)^2 \ & =1 \times \frac{3}{16}+1 \times \frac{3}{16} \quad[\mu=0] \ & =\frac{6}{16}=\frac{3}{8}\end{aligned}$