If the variance of the numbers \(2,3,11\) and \(x\) is \(\frac{49}{4}\), then the values of \(x\) are

Mean of given number is \(\frac{2+3+11+x}{4}\) i.e. \(\frac{16+x}{4}\)
\( \text { Variance }=\frac{1}{\mathrm{n}} \sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2 \)
\( \therefore \frac{49}{4}=\frac{1}{4}[\left(\frac{16+\mathrm{x}}{4}-2\right)^2+\left(\frac{16+\mathrm{x}}{4}-3\right)^2+\) \((\frac{16+\mathrm{x}}{4}-11)^2+\left(\frac{16+\mathrm{x}}{4}-\mathrm{x}\right)^2] \)
\( \therefore 49=\frac{(8+\mathrm{x})^2}{16}+\frac{(4+\mathrm{x})^2}{16}+\frac{(\mathrm{x}-28)^2}{16}+\frac{(16-3 \mathrm{x})^2}{16} \)
\( \therefore(49)(16)=\left(64+\mathrm{x}^2+16 \mathrm{x}\right)+\left(16+\mathrm{x}^2+8 \mathrm{x}\right)\)
\( +\left(\mathrm{x}^2-56 \mathrm{x}+784\right)+\left(256+9 \mathrm{x}^2-96 \mathrm{x}\right) \)
\( \therefore 784=12 \mathrm{x}^2-128 \mathrm{x}+80+784+256 \)
\( \therefore 12 \mathrm{x}^2-128 \mathrm{x}+336=0 \)
\( \Rightarrow 3 \mathrm{x}^2-32 \mathrm{x}+84=0 \)
\( \therefore \mathrm{x}=\frac{32 \pm \sqrt{(32)^2-(4)(3)(84)}}{2(3)}=\frac{32 \pm \sqrt{1024-1008}}{6}=\) \(\frac{32 \pm 4}{6} \)
\( \therefore \mathrm{x}=\frac{36}{6} \text { or } \mathrm{x}=\frac{28}{6} \)
\( \Rightarrow \mathrm{x}=6, \frac{14}{3}\)