Consider the following frequency distribution : If mean $=\frac{309}{22}$ and median $=14$, then th

Question

Consider the following frequency distribution :  If mean $=\frac{309}{22}$ and median $=14$, then the value $(a-b)^{2}$ is equal to\

Professor PC Thomas Classes · Answer

(4) : $66 a+198 b+11088$\ $=309 a+309 b+8034$ $$\begin{equation*}\Rightarrow \quad 81 a+37 b=1018 	ag{i} \end{equation*}$$ Clearly, median\ $=12+\frac{\frac{a+b+26}{2}-(a+b)}{12} 	imes 6=14$ (Given)\ $\Rightarrow \quad \frac{13}{2}-\left(\frac{a+b}{4}ight)=2$ $$\begin{equation*}\Rightarrow a+b=18 	ag{ii} \end{equation*}$$ By (i) and (ii), we get $a=8, b=10$\ $	herefore \quad(a-b)^{2}=(8-10)^{2}=4$\