Consider 10 observation x _ 1 , x _ 2 , . .. x _ 10 , such that ∑ _...

Given observations are ${x}_{1}, {x}_{2}, . . . , {x}_{10}$

$⇒ \mathop{∑}_{i = 1}^{10}\left({x}_{i} - \alpha\right) = 2$

$⇒ \mathop{∑}_{i = 1}^{10}{x}_{i} - 10\alpha = 2$

Mean, $\mu = \dfrac{6}{5} = \dfrac{∑ {x}_{i}}{10}$

$⇒ ∑ {x}_{i} = 12$

$⇒ 10\alpha + 2 = 12$

$⇒ \alpha = 1$

Now, $\mathop{∑}_{i = 1}^{10}{\left({x}_{i} - \beta\right)}^{2} = 40$. Let ${y}_{i} = {x}_{i} - \beta$

$⇒ {\sigma}_{y}^{2} = \dfrac{1}{10} ∑ {y}_{i}^{2} - {\left(\overset{¯}{y}\right)}^{2}$

$⇒ \dfrac{84}{25} = \dfrac{1}{10} ∑ {\left({x}_{i} - \beta\right)}^{2} - {\left(\dfrac{\mathop{∑}_{i = 1}^{10}\left({x}_{i} - \beta\right)}{10}\right)}^{2}$

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

$⇒ {\left(\dfrac{6 - 5\beta}{5}\right)}^{2} = 4 - \dfrac{84}{25}$

$⇒ {\left(\dfrac{6 - 5\beta}{5}\right)}^{2} = \dfrac{16}{25}$

$⇒ 6 - 5\beta = \pm 4$

$⇒ \beta = \dfrac{2}{5}$ (not possible) or $\beta = 2$

Hence, $\dfrac{\beta}{\alpha} = 2$