Let X_{1}, X_{2}, ldots, X_{18} be eighteen o | vedclass

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(A) Given $\sum_{i=1}^{18}(X_{i}-\alpha)=36 \implies \sum X_{i} - 18\alpha = 36 \implies \sum X_{i} = 18(\alpha+2)$.Given $\sum_{i=1}^{18}(X_{i}-\beta

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(A) Given $\sum_{i=1}^{18}(X_{i}-\alpha)=36 \implies \sum X_{i} - 18\alpha = 36 \implies \sum X_{i} = 18(\alpha+2)$.Given $\sum_{i=1}^{18}(X_{i}-\beta)^{2}=90 \implies \sum X_{i}^{2} - 2\beta \sum X_{i} + 18\beta^{2} = 90$.Substituting $\sum X_{i} = 18(\alpha+2)$,we get $\sum X_{i}^{2} = 90 - 18\beta^{2} + 36\beta(\alpha+2)$.The variance is given by $\sigma^{2} = \frac{\sum X_{i}^{2}}{18} - (\frac{\sum X_{i}}{18})^{2} = 1^{2} = 1$.Substituting the values: $\frac{90 - 18\beta^{2} + 36\beta(\alpha+2)}{18} - (\alpha+2)^{2} = 1$.$5 - \beta^{2} + 2\beta(\alpha+2) - (\alpha^{2} + 4\alpha + 4) = 1$.$5 - \beta^{2} + 2\alpha\beta + 4\beta - \alpha^{2} - 4\alpha - 4 = 1$.$-(\alpha^{2} - 2\alpha\beta + \beta^{2}) + 4(\beta - \alpha) = 0$.$-(\alpha-\beta)^{2} - 4(\alpha-\beta) = 0$.$(\alpha-\beta)^{2} + 4(\alpha-\beta) = 0$.$(\alpha-\beta)(\alpha-\beta+4) = 0$.Since $\alpha 
eq \beta$,we have $\alpha-\beta = -4$,so $|\alpha-\beta| = 4$.

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