Let x_1, x_2, ldots, x_{10} be ten observatio | vedclass

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(A) Given $\sum_{i=1}^{10}(x_i-2)=30 \implies \sum x_i - 20 = 30 \implies \sum x_i = 50$. The mean $\bar{x} = \frac{50}{10} = 5$. Variance $\sigma_x^2

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$100$

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(A) Given $\sum_{i=1}^{10}(x_i-2)=30 \implies \sum x_i - 20 = 30 \implies \sum x_i = 50$. The mean $\bar{x} = \frac{50}{10} = 5$. Variance $\sigma_x^2 = \frac{\sum x_i^2}{10} - (\bar{x})^2 = \frac{4}{5} \implies \frac{\sum x_i^2}{10} - 25 = 0.8 \implies \sum x_i^2 = 258$. Given $\sum_{i=1}^{10}(x_i-\beta)^2 = 98 \implies \sum x_i^2 - 2\beta \sum x_i + 10\beta^2 = 98$. Substituting values: $258 - 2\beta(50) + 10\beta^2 = 98 \implies 10\beta^2 - 100\beta + 160 = 0 \implies \beta^2 - 10\beta + 16 = 0$. $(\beta-8)(\beta-2) = 0$. Since $\beta > 2$,we have $\beta = 8$. Let $y_i = 2(x_i-1) + 4\beta = 2x_i - 2 + 32 = 2x_i + 30$. Mean $\mu = 2\bar{x} + 30 = 2(5) + 30 = 40$. Variance $\sigma^2 = 2^2 \cdot \sigma_x^2 = 4 \cdot \frac{4}{5} = \frac{16}{5}$. Then $\frac{\beta\mu}{\sigma^2} = \frac{8 \cdot 40}{16/5} = \frac{320 \cdot 5}{16} = 20 \cdot 5 = 100$.

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