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(D) Let $y_{i} = x_{i} - 5$. Then $\sum_{i=1}^{10} y_{i} = 10$ and $\sum_{i=1}^{10} y_{i}^{2} = 40$. Mean of $y_{i}$ is $\bar{y} = \frac{1}{10} \sum y

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$(3, 3)$

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(D) Let $y_{i} = x_{i} - 5$. Then $\sum_{i=1}^{10} y_{i} = 10$ and $\sum_{i=1}^{10} y_{i}^{2} = 40$. Mean of $y_{i}$ is $\bar{y} = \frac{1}{10} \sum y_{i} = \frac{10}{10} = 1$. Variance of $y_{i}$ is $\sigma_{y}^{2} = \frac{1}{10} \sum y_{i}^{2} - (\bar{y})^{2} = \frac{40}{10} - (1)^{2} = 4 - 1 = 3$. Now,let $z_{i} = x_{i} - 3$. We can write $z_{i} = (x_{i} - 5) + 2 = y_{i} + 2$. The mean $\mu$ of $z_{i}$ is $\bar{z} = \bar{y} + 2 = 1 + 2 = 3$. The variance $\lambda$ of $z_{i}$ is $	ext{Var}(y_{i} + 2) = 	ext{Var}(y_{i}) = 3$. Thus,the ordered pair $(\mu, \lambda) = (3, 3)$.

Let the observations $x_{i} (1 \leq i \leq 10)$ satisfy the equations $\sum_{i=1}^{10}(x_{i}-5)=10$ and $\sum_{i=1}^{10}(x_{i}-5)^{2}=40$. If $\mu$ and $\lambda$ are the mean and the variance of the observations $x_{1}-3, x_{2}-3, \dots, x_{10}-3$,then the ordered pair $(\mu, \lambda)$ is equal to:

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