JEE Main 2024 — 1 February, Shift 2 — Question 18

$\mathrm{x}_{1}, \mathrm{x}_{2} \ldots \ldots . \mathrm{x}_{10}$

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

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

$\therefore \quad \Sigma \mathrm{x}_{\mathrm{i}}=12$

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

Now $\sum_{i=1}^{10}\left(x_{i}-\beta\right)^{2}=40$

Let $y_{i}=x_{i}-\beta$

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

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

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

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

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

or $\beta=2$

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