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(C) Let the total number of observations be $n$.Given that $\frac{n}{4}$ observations are $a$,$\frac{n}{4}$ are $-a$,$\frac{n}{4}$ are $b$,and $\frac{

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$a = b$

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(C) Let the total number of observations be $n$.Given that $\frac{n}{4}$ observations are $a$,$\frac{n}{4}$ are $-a$,$\frac{n}{4}$ are $b$,and $\frac{n}{4}$ are $-b$.The mean $\bar{x} = \frac{\frac{n}{4}(a - a + b - b)}{n} = 0$.The variance is given by $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.Substituting the values: $ab = \frac{\frac{n}{4}(a^2 + (-a)^2 + b^2 + (-b)^2)}{n} - 0$.$ab = \frac{a^2 + a^2 + b^2 + b^2}{4} = \frac{2a^2 + 2b^2}{4} = \frac{a^2 + b^2}{2}$.$2ab = a^2 + b^2 \Rightarrow a^2 + b^2 - 2ab = 0$.$(a - b)^2 = 0 \Rightarrow a = b$.

In a discrete data,$\frac{1}{4}$ of the observations are equal to $a$,another $\frac{1}{4}$ of the observations are equal to $-a$. Out of the remaining,half of them are equal to $b$ and the rest are equal to $-b$. If the variance of all the observations is $ab$,then:

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