Consider 10 observations x_1, x_2, ldots, x_{ | vedclass

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(A) Given $n=10$ observations $x_1, x_2, \ldots, x_{10}$.From $\sum_{i=1}^{10}(x_i-\alpha)=2$,we have $\sum x_i - 10\alpha = 2$.Given mean $\mu = \fra

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$2$

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(A) Given $n=10$ observations $x_1, x_2, \ldots, x_{10}$.From $\sum_{i=1}^{10}(x_i-\alpha)=2$,we have $\sum x_i - 10\alpha = 2$.Given mean $\mu = \frac{\sum x_i}{10} = \frac{6}{5}$,so $\sum x_i = 12$.Substituting $\sum x_i = 12$ into the first equation: $12 - 10\alpha = 2$ $\Rightarrow 10\alpha = 10$ $\Rightarrow \alpha = 1$.Variance $\sigma^2 = \frac{1}{n} \sum x_i^2 - \mu^2 = \frac{84}{25}$.We know $\sum (x_i - \beta)^2 = \sum x_i^2 - 2\beta \sum x_i + 10\beta^2 = 40$.From $\sigma^2 = \frac{\sum x_i^2}{10} - (\frac{6}{5})^2 = \frac{84}{25}$,we get $\frac{\sum x_i^2}{10} = \frac{84}{25} + \frac{36}{25} = \frac{120}{25} = \frac{24}{5}$,so $\sum x_i^2 = 48$.Substituting $\sum x_i^2 = 48$ and $\sum x_i = 12$ into the equation $48 - 2\beta(12) + 10\beta^2 = 40$:$10\beta^2 - 24\beta + 8 = 0 \Rightarrow 5\beta^2 - 12\beta + 4 = 0$.Solving the quadratic: $(5\beta - 2)(\beta - 2) = 0$.Since $\beta$ is a positive integer,$\beta = 2$.Thus,$\frac{\beta}{\alpha} = \frac{2}{1} = 2$.

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