Mean of given number is $\frac{2+3+11+x}{4}$ i.e. $\frac{16+x}{4}$
$$
\begin{aligned}
& \text { Variance }=\frac{1}{\mathrm{n}} \sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2 \\
& \therefore \frac{49}{4}=\frac{1}{4}\left[\left(\frac{16+\mathrm{x}}{4}-2\right)^2+\left(\frac{16+\mathrm{x}}{4}-3\right)^2+\left(\frac{16+\mathrm{x}}{4}-11\right)^2+\left(\frac{16+\mathrm{x}}{4}-\mathrm{x}\right)^2\right] \\
& \therefore 49=\frac{(8+\mathrm{x})^2}{16}+\frac{(4+\mathrm{x})^2}{16}+\frac{(\mathrm{x}-28)^2}{16}+\frac{(16-3 \mathrm{x})^2}{16} \\
& \therefore(49)(16)=\left(64+\mathrm{x}^2+16 \mathrm{x}\right)+\left(16+\mathrm{x}^2+8 \mathrm{x}\right)
\end{aligned}
$$

$$
\begin{aligned}
& +\left(\mathrm{x}^2-56 \mathrm{x}+784\right)+\left(256+9 \mathrm{x}^2-96 \mathrm{x}\right) \\
& \therefore 784=12 \mathrm{x}^2-128 \mathrm{x}+80+784+256 \\
& \therefore 12 \mathrm{x}^2-128 \mathrm{x}+336=0 \\
& \Rightarrow 3 \mathrm{x}^2-32 \mathrm{x}+84=0 \\
& \therefore \mathrm{x}=\frac{32 \pm \sqrt{(32)^2-(4)(3)(84)}}{2(3)}=\frac{32 \pm \sqrt{1024-1008}}{6}=\frac{32 \pm 4}{6} \\
& \therefore \mathrm{x}=\frac{36}{6} \text { or } \mathrm{x}=\frac{28}{6} \\
& \Rightarrow \mathrm{x}=6, \frac{14}{3}
\end{aligned}
$$