Consider 10 observation mathrm{x}_1, mathrm{x | vedclass

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Question

$ \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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$ \mathrm{x}_1, \mathrm{x}_2 \ldots \ldots . \mathrm{x}_{10} $

$ \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\alphaight)=2 \Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}-10 \alpha=2 $

$ 	ext { Mean } \mu=\frac{6}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}}{10} $

$ 	herefore \quad \sum_{\mathrm{i}}=12 $

$ \quad 10 \alpha+2=12 \quad 	herefore \alpha=1 $

$ 	ext { Now } \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\betaight)^2=40 	ext { Let } \mathrm{y}_{\mathrm{i}}=\mathrm{x}_{\mathrm{i}}-\beta $

$ 	herefore \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ight)^2-\left(\frac{\sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\betaight)}{10}ight)^2 $

$ \frac{84}{25}=4-\left(\frac{12-10 \beta}{10}ight)^2 $

$ 	herefore\left(\frac{6-5 \beta}{5}ight)^2=4-\frac{84}{25}=\frac{16}{25} $

$ 6-5 \beta= \pm 4 \Rightarrow \beta=\frac{2}{5} 	ext { (not possible) or } \beta=2$

Hence $\frac{\beta}{\alpha}=2$

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