Let ${\mathrm{x}}_{1},{\mathrm{x}}_{2},\ldots \ldots {\mathrm{x}}_{10}$ be ten observations such that $\sum _{i=1}^{10}\left({x}_{i}-2\right)=30,\sum _{i=1}^{10}{\left({x}_{i}-\beta \right)}^{2}=98,\beta 2$ and their variance is $\frac{4}{5}$. If $\mu$ and ${\sigma }^{2}$ are respectively the mean and the variance of $2\left({x}_{1}-1\right)+4\beta ,2\left({x}_{2}-1\right)+4\beta ,$$\ldots ..,2\left({x}_{10}-1\right)+4\beta ,$ then $\frac{\beta \mu }{{\sigma }^{2}}$ is equal to : [JEE Main 2025, 29 Jan (Shift 1)]

Question

Let ${\mathrm{x}}_{1},{\mathrm{x}}_{2},\ldots \ldots {\mathrm{x}}_{10}$ be ten observations such that $\sum _{i=1}^{10}\left({x}_{i}-2\right)=30,\sum _{i=1}^{10}{\left({x}_{i}-\beta \right)}^{2}=98,\beta >2$ and their variance is $\frac{4}{5}$. If $\mu$ and ${\sigma }^{2}$ are respectively the mean and the variance of $2\left({x}_{1}-1\right)+4\beta ,2\left({x}_{2}-1\right)+4\beta ,$$\ldots ..,2\left({x}_{10}-1\right)+4\beta ,$ then $\frac{\beta \mu }{{\sigma }^{2}}$ is equal to : [JEE Main 2025, 29 Jan (Shift 1)]

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Let \({\mathrm{x}}_{1},{\mathrm{x}}_{2},\ldots \ldots {\mathrm{x}}_{10}\) be ten observations such that \(\sum _{i=1}^{1…

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