Let the mean and variance of 12 observations be $\frac{9}{2}$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is $\frac{m}{n}$, where $\mathrm{m}$ and $\mathrm{n}$ are coprime, then $\mathrm{m}+\mathrm{n}$ is equal to :

$$ \begin{aligned} & \text { Since, Mean }=\frac{9}{2} \\\\ & \Rightarrow \Sigma x=\frac{9}{2} \times 12=54 \end{aligned} $$ <br/><br/>Also, variance $=4$ <br/><br/>$$ \begin{aligned} & \Rightarrow \frac{\sum x^2}{12}=\left[\frac{\sum x_i}{12}\right]^2=4 \\\\ & \Rightarrow \frac{\sum x^2}{12}=4+\frac{81}{4}=\frac{97}{4} \\\\ & \Rightarrow \sum x^2=291 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \sum x^{\prime}=54-(9+10)+7+14 \\\\ & =54-19+21=56 \\\\ & \text { and } \sum x^2=291-(81+100)+49+196 \\\\ & =291-181+49+196=355 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { So, } \sigma_{\text {new }}^2=\frac{\sum x_{\text {new }}^2}{12}-\left(\frac{\sum x_{\text {new }}}{12}\right)^2 \\\\ & =\frac{355}{12}-\left(\frac{56}{12}\right)^2 \\\\ & =\frac{4260-3136}{144}=\frac{1124}{144}=\frac{281}{36} \\\\ & =\frac{m}{n} \\\\ & \Rightarrow m=281, n=36 \\\\ & \Rightarrow m+n=281+36=317 \end{aligned} $$