If 4 coins are tossed then the mean and variance of X where X is the number of heads 
 

Let x denote the number of heads in the four tosses of the coin then probability distribution of x is

Mean$\left(\overline{\mathrm{x}}\right)=\sum {\mathrm{x}}_{\mathrm{i}} \mathrm{P}(\mathrm{x}={\mathrm{x}}_{\mathrm{i}})$

                $$\begin{gathered} =0+1\left(\frac{1}{4}\right)+2\left(\frac{3}{8}\right)+3\left(\frac{1}{4}\right)+4\left(\frac{1}{16}\right) \\ =\frac{1}{4}+\frac{3}{4}+\frac{3}{4}+\frac{1}{4} \\ =\frac{8}{4}=2 \end{gathered}$$

Variance $\left({\mathrm{a}}^2\right)=\sum {\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^2 \mathrm{P}(\mathrm{x}={\mathrm{x}}_{\mathrm{i}})$

                         $$\begin{gathered} ={(0-2)}^2\left(\frac{1}{16}\right)+{\left(1-2\right)}^2\left(\frac{1}{4}\right)+{\left(2-2\right)}^2\left(\frac{3}{8}\right)+{\left(3-2\right)}^2\left(\frac{1}{4}\right)+{\left(4-2\right)}^2\left(\frac{1}{16}\right) \\ =\frac{1}{4}+\frac{1}{4}+0+\frac{1}{4}+\frac{1}{4} \\ =1 \end{gathered}$$