$ \mathrm{x}_1, \mathrm{x}_2 \ldots \ldots . \mathrm{x}_{10} $

$ \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\alpha\right)=2 \Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}-10 \alpha=2 $

$ \text { Mean } \mu=\frac{6}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}}{10} $

$ \therefore \quad \sum_{\mathrm{i}}=12 $

$ \quad 10 \alpha+2=12 \quad \therefore \alpha=1 $

$ \text { Now } \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\beta\right)^2=40 \text { Let } \mathrm{y}_{\mathrm{i}}=\mathrm{x}_{\mathrm{i}}-\beta $

$ \therefore \sigma_{\mathrm{y}}^2=\frac{1}{10} \sum \mathrm{y}_{\mathrm{i}}^2-(\overline{\mathrm{y}})^2 $

$ \sigma_{\mathrm{x}}^2=\frac{1}{10} \sum\left(\mathrm{x}_{\mathrm{i}}-\beta\right)^2-\left(\frac{\sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\beta\right)}{10}\right)^2 $

$ \frac{84}{25}=4-\left(\frac{12-10 \beta}{10}\right)^2 $

$ \therefore\left(\frac{6-5 \beta}{5}\right)^2=4-\frac{84}{25}=\frac{16}{25} $

$ 6-5 \beta= \pm 4 \Rightarrow \beta=\frac{2}{5} \text { (not possible) or } \beta=2$

Hence $\frac{\beta}{\alpha}=2$