Let in a series of 2n observations,half of th | vedclass

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Question

(A) Let the observations be $x_{i}$ for $1 \leq i \leq 2n$. The mean of the original observations is $\bar{x} = \frac{\sum x_{i}}{2n} = \frac{n(a) + n

Vedclass · Accepted Answer

$425$

Vedclass · Answer

(A) Let the observations be $x_{i}$ for $1 \leq i \leq 2n$. The mean of the original observations is $\bar{x} = \frac{\sum x_{i}}{2n} = \frac{n(a) + n(-a)}{2n} = 0$. The variance of the original observations is $\sigma_{x}^{2} = \frac{\sum x_{i}^{2}}{2n} - (\bar{x})^{2} = \frac{n(a^{2}) + n(-a)^{2}}{2n} - 0 = \frac{2na^{2}}{2n} = a^{2}$. Thus,the standard deviation is $\sigma_{x} = \sqrt{a^{2}} = |a|$. When a constant $b$ is added to each observation,the new mean $\bar{y} = \bar{x} + b = 0 + b = 5$,so $b = 5$. The standard deviation remains unchanged by adding a constant,so $\sigma_{y} = \sigma_{x} = |a| = 20$. Therefore,$a^{2} = 20^{2} = 400$ and $b^{2} = 5^{2} = 25$. Finally,$a^{2} + b^{2} = 400 + 25 = 425$.

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