Let the mean and variance of 5 observations ${\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4,{\mathrm{x}}_5$ be$\frac{24}{5}\text{ and }\frac{194}{25}$respectively .If the mean and variance of the first 4 observations are$\frac{7}{2}$and ‘a’ respectively then $4\mathrm{a}+{\mathrm{x}}_5=$

$$\begin{gathered} \begin{array}{l}\overline{\mathrm{x}}=\frac{24}{5}\Rightarrow \frac{\sum {\mathrm{x}}_{\mathrm{i}}}{5}=\frac{24}{5}\Rightarrow \sum {\mathrm{x}}_{\mathrm{i}}=24, \\ {\mathrm{\sigma }}^2=\frac{194}{25} \\ \Rightarrow \frac{\sum {\mathrm{x}}_{\mathrm{i}}^2}{5}-{\left(\frac{24}{5}\right)}^2=\frac{194}{25} \\ \Rightarrow \frac{\sum {\mathrm{x}}_{\mathrm{i}}^2}{5}-\frac{576}{25}=\frac{194}{25}\\ \Rightarrow 5\sum {\mathrm{x}}_{\mathrm{i}}^2-576=194 \\ \Rightarrow 5\sum {\mathrm{x}}_{\mathrm{i}}^2=194+576=770\Rightarrow \sum {\mathrm{x}}_{\mathrm{i}}^2=154\\ \frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4}{4}=\frac{7}{2}, \\ \Rightarrow {\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4=14 \\ \Rightarrow {\mathrm{x}}_5=10 \\ \frac{{\mathrm{x}}_1^2+{\mathrm{x}}_2^2+{\mathrm{x}}_3^2+{\mathrm{x}}_4^2}{4}-{\left(\frac{7}{2}\right)}^2=\mathrm{a}\\ \Rightarrow \frac{{\mathrm{x}}_1^2+{\mathrm{x}}_2^2+{\mathrm{x}}_3^2+{\mathrm{x}}_4^2}{4}-\frac{49}{4}=\mathrm{a} \\ \Rightarrow {\mathrm{x}}_1^2+{\mathrm{x}}_2^2+{\mathrm{x}}_3^2+{\mathrm{x}}_4^2=4\mathrm{a}+49\\ {\mathrm{x}}_5^2=154-4\mathrm{a}-49 \\ \Rightarrow 100=105-4\mathrm{a} \\ \Rightarrow 4\mathrm{a}=105-100=5\Rightarrow \mathrm{a}=\frac{5}{4}\\ 4\mathrm{a}+{\mathrm{x}}_5=4\left(\frac{5}{4}\right)+10=15\end{array} \end{gathered}$$