If bar{x} is the mean of x_{1}, x_{2}, ldots, | vedclass

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(B) We know that if $\bar{x}$ is the mean of $x_{1}, x_{2}, \ldots, x_{n},$ then $\bar{x} = \frac{\sum_{i=1}^{n} x_{i}}{n},$ which implies $\sum_{i=1}

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$\left(a+\frac{1}{a}\right) \frac{\bar{x}}{2}$

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(B) We know that if $\bar{x}$ is the mean of $x_{1}, x_{2}, \ldots, x_{n},$ then $\bar{x} = \frac{\sum_{i=1}^{n} x_{i}}{n},$ which implies $\sum_{i=1}^{n} x_{i} = n \bar{x}.$The new set of observations consists of $2n$ terms: $a x_{1}, a x_{2}, \ldots, a x_{n}$ and $\frac{x_{1}}{a}, \frac{x_{2}}{a}, \ldots, \frac{x_{n}}{a}.$The mean of these $2n$ observations is given by:Mean $= \frac{(a x_{1} + a x_{2} + \ldots + a x_{n}) + (\frac{x_{1}}{a} + \frac{x_{2}}{a} + \ldots + \frac{x_{n}}{a})}{2n}$$= \frac{a(x_{1} + x_{2} + \ldots + x_{n}) + \frac{1}{a}(x_{1} + x_{2} + \ldots + x_{n})}{2n}$$= \frac{a(n \bar{x}) + \frac{1}{a}(n \bar{x})}{2n}$$= \frac{n \bar{x} (a + \frac{1}{a})}{2n}$$= \frac{\bar{x}}{2} (a + \frac{1}{a}).$

If $\bar{x}$ is the mean of $x_{1}, x_{2}, \ldots, x_{n},$ then for $a \neq 0,$ the mean of $a x_{1}, a x_{2}, \ldots, a x_{n}, \frac{x_{1}}{a}, \frac{x_{2}}{a}, \ldots, \frac{x_{n}}{a}$ is

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