Let the mean and the variance of 5 observations ${\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4,{\mathrm{x}}_5$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first 4 observation are $\frac{7}{2}$ and a respectively, then $\left(4a+x_5\right)$ is equal to :

$\begin{array}{l}\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4+{\mathrm{x}}_5}{5}=\frac{24}{5}\\ \Rightarrow {\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4+{\mathrm{x}}_5=24......\left(1\right)\\ \text{ and }\frac{{\mathrm{x}}_1^2+{\mathrm{x}}_2^2+{\mathrm{x}}_3^2+{\mathrm{x}}_4^2+{\mathrm{x}}_5^2}{5}-{\left(\frac{24}{5}\right)}^2=\frac{194}{25}\end{array}$
$\begin{array}{l}\frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}{5}=\frac{194}{25}+\frac{576}{25}=\frac{770}{25}\\ x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=154......\left(2\right)\\ now \frac{x_1+x_2+x_3+x_4}{4}=\frac{7}{2}\\ \Rightarrow x_1+x_2+x_3+x_4=14.......\left(3\right)\\ \Rightarrow \frac{x_1^2+x_2^2+x_3^2+x_4^2}{4}-{\left(\frac{7}{2}\right)}^2=a\\ \frac{154-x_5^2}{4}-\frac{49}{4}=a\\ \frac{154}{4}-\frac{49}{4}-\frac{x_5^2}{4}=a\\ 4a+x_5^2=105\\ \text{ from equation }\left(1\right)\text{ and (3) }\\ x_5=10,4a=5\\ 4a+x_5=5+10=15\end{array}$