Consider 10 observations $x_1, x_2, \ldots, x_{10}$ such that $\sum\limits_{i=1}^{10}\left(x_i-\alpha\right)=2$ and $\sum\limits_{i=1}^{10}\left(x_i-\beta\right)^2=40$, where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. Then $\frac{\beta}{\alpha}$ is equal to :

We have given $\bar{x}($ mean $)=\frac{6}{5}$ <br/><br/>$$ \begin{aligned} & \text { Variance }=\frac{84}{25} \\\\ & \sum_{i=1}^{10}\left(x_i-\alpha\right)=2 \\\\ & \Rightarrow x_1+x_2+\ldots+x_{10}-10 \alpha=2 \\\\ & \Rightarrow \frac{x_1+x_2+\ldots+x_{10}}{10}-\alpha=\frac{2}{10} \\\\ & \Rightarrow \frac{6}{5}-\alpha=\frac{2}{10} \\\\ & \Rightarrow \alpha=1 \end{aligned} $$ <br/><br/>$\begin{aligned} & \text { and } \sum_{i=1}^{10}\left(x_i-\beta\right)^2=40 \\\\ & \left(x_1-\beta\right)^2+\left(x_2-\beta\right)^2+\ldots+\left(x_{10}-\beta\right)^2=40 \\\\ & x_1^2+x_2^2+\ldots+x_{10}^2+10 \beta^2-2 \beta\left(x_1+x_2+\ldots+x_{10}\right)=40 \\\\ & \Rightarrow \frac{x_1^2+x_2^2+\ldots+x_{10}^2}{10}+\beta^2-\frac{2 \beta\left(x_1+x_2+\ldots+x_{10}\right)}{10}=4 \\\\ & \Rightarrow \frac{x_1^2+x_2^2+\ldots+x_{10}^2}{10}-\frac{36}{25}+\frac{36}{25}+\beta^2-2 \beta \times \frac{6}{5}=4\end{aligned}$ <br/><br/>$\left[\right.$ Variance $\left.=\frac{\sum_{i=1}^n x_i^2}{n}-(\bar{x})^2\right]$ <br/><br/>$\begin{aligned} & \Rightarrow \frac{84}{25}+\frac{36}{25}+\beta^2-\frac{12 \beta}{5}-4=0 \\\\ & \Rightarrow \frac{120}{25}+\beta^2-\frac{12 \beta}{5}-4=0 \\\\ & \Rightarrow 25 \beta^2-60 \beta+20=0 \\\\ & \Rightarrow 5 \beta^2-12 \beta+4=0 \\\\ & \Rightarrow \beta=2, \frac{2}{5}\end{aligned}$ <br/><br/>Take $\beta=2$ <br/><br/>$\frac{\beta}{\alpha}=\frac{2}{1}=2$