If the standard deviation of the numbers 2,3,2x and 11 is 3.5, then find the possible values of x.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ 2,3,2\mathrm{x} \mathrm{and} 11 \\ \mathrm{Mean}\left(\mathrm{m}\right)=\frac{2+3+2\mathrm{x}+11}{4}=\frac{16+2\mathrm{x}}{4}=4+\frac{\mathrm{x}}{2} \\ 4+\frac{\mathrm{x}}{2}-2=2+\frac{\mathrm{x}}{2} \\ 4+\frac{\mathrm{x}}{2}-3=1+\frac{\mathrm{x}}{2} \\ 4+\frac{\mathrm{x}}{2}-2\mathrm{x}=4-\frac{3\mathrm{x}}{2} \\ 4+\frac{\mathrm{x}}{2}-11=\frac{\mathrm{x}}{2}-7 \\ \mathrm{Standard} \mathrm{Deviation}=3.5 \\ \sqrt{\frac{{\left(2+\frac{\mathrm{x}}{2}\right)}^2+\left(1+\frac{\mathrm{x}}{2}\right)+{\left(4-\frac{3\mathrm{x}}{2}\right)}^2+{\left(\frac{\mathrm{x}}{2}-7\right)}^2}{4}}=3.5=\frac{35}{10}=\frac{7}{2} \\ \mathrm{Squarring} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get}: \\ \Rightarrow \frac{{\left(2+\frac{\mathrm{x}}{2}\right)}^2+\left(1+\frac{\mathrm{x}}{2}\right)+{\left(4-\frac{3\mathrm{x}}{2}\right)}^2+{\left(\frac{\mathrm{x}}{2}-7\right)}^2}{4}=\frac{49}{4} \\ \Rightarrow {\left(2+\frac{\mathrm{x}}{2}\right)}^2+\left(1+\frac{\mathrm{x}}{2}\right)+{\left(4-\frac{3\mathrm{x}}{2}\right)}^2+{\left(\frac{\mathrm{x}}{2}-7\right)}^2=49 \\ \Rightarrow \frac{{\left(4+\mathrm{x}\right)}^2}{4}+\frac{{\left(2+\mathrm{x}\right)}^2}{4}+\frac{{\left(8-3\mathrm{x}\right)}^2}{4}+\frac{{\left(\mathrm{x}-14\right)}^2}{4}=49 \\ \Rightarrow {\left(4+\mathrm{x}\right)}^2+{\left(2+\mathrm{x}\right)}^2+{\left(8-3\mathrm{x}\right)}^2+{\left(\mathrm{x}-14\right)}^2=196 \\ \Rightarrow 16+{\mathrm{x}}^2+8\mathrm{x}+4+{\mathrm{x}}^2+4\mathrm{x}+64+9{\mathrm{x}}^2-48\mathrm{x}+{\mathrm{x}}^2+196-28\mathrm{x}=196 \\ \Rightarrow 12{\mathrm{x}}^2-64\mathrm{x}+280-196=0 \\ \Rightarrow 12{\mathrm{x}}^2-64\mathrm{x}+84=0 \\ \Rightarrow 3{\mathrm{x}}^2-16\mathrm{x}+21=0 \\ \Rightarrow 3{\mathrm{x}}^2-9\mathrm{x}-7\mathrm{x}+21=0 \\ \Rightarrow 3\mathrm{x}\left(\mathrm{x}-3\right)-7\left(\mathrm{x}-3\right)=0 \\ \Rightarrow \left(\mathrm{x}-3\right)\left(3\mathrm{x}-7\right)=0 \\ \mathrm{Hence} \\ \mathbf{x}\mathbf{=}\mathbf{3}\mathbf{ }\mathbf{or}\mathbf{ }\mathbf{x}\mathbf{=}\frac{\mathbf{7}}{\mathbf{3}} \end{gathered}$$

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