(A) The mean $\bar{x}$ is calculated as: $\bar{x} = \frac{n(a) + n(-a)}{2n} = \frac{0}{2n} = 0$. The variance $\sigma^2$ is given by $\sigma^2 = \frac

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(A) The mean $\bar{x}$ is calculated as: $\bar{x} = \frac{n(a) + n(-a)}{2n} = \frac{0}{2n} = 0$. The variance $\sigma^2$ is given by $\sigma^2 = \frac{1}{N} \sum x_i^2 - \bar{x}^2$. Here,$N = 2n$,$\sum x_i^2 = n(a^2) + n(-a)^2 = 2na^2$,and $\bar{x} = 0$. So,$\sigma^2 = \frac{2na^2}{2n} - 0^2 = a^2$. Given the standard deviation $\sigma = 2$,we have $\sigma^2 = 4$. Therefore,$a^2 = 4$,which implies $|a| = 2$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

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$\frac{1}{n}$
D
$\frac{\sqrt{2}}{n}$

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