उत्तर (1)

समाधान

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

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

माध्य $\mu=\frac{6}{5}=\frac{\sum x _i}{10}$

$\therefore \quad \sum x _i=12$

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

अब $\sum _{i=1}^{10}\left(x _i-\beta\right)^{2}=40 \quad$ मान लीजिए $y _i=x _i-\beta$

$\therefore \sigma _y^{2}=\frac{1}{10} \sum y _i^{2}-(\overline{y})^{2}$

$\sigma _x^{2}=\frac{1}{10} \sum\left(x _i-\beta\right)^{2}-\left(\frac{\sum _{i=1}^{10}\left(x _i-\beta\right)}{10}\right)^{2}$

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

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

$6-5 \beta= \pm 4 \Rightarrow \beta=\frac{2}{5}$ (संभव नहीं) या $\beta=2$

अतः $\frac{\beta}{\alpha}=2$