The raw data $x_1, x_2, \ldots \ldots, x_{\mathrm{n}}$ is an A.P. with common difference $\mathrm{d}$ and first term $0 . \bar{x}$ and $\sigma^2$ are mean and variance of $x_{\mathrm{i}}, \mathrm{i}=1,2, \ldots \ldots \mathrm{n}$, then $\sigma^2$ is

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Aicharya · Accepted Answer

$\begin{aligned} \bar{x}= & \frac{x_1+x_2+\ldots+x_{\mathrm{n}}}{\mathrm{n}} \ & =\frac{\frac{\mathrm{n}}{2}\left[2 x_1+(\mathrm{n}-1) \mathrm{d}ight]}{\mathrm{n}} \ & =\frac{(\mathrm{n}-1) \mathrm{d}}{2} \ \sum x_{\mathrm{i}}^2 & =x_1^2+x_2^2+\ldots+x_{\mathrm{n}}^2 \ & =0+\mathrm{d}^2+(2 \mathrm{~d})^2+\ldots+[(\mathrm{n}-1) \mathrm{d}]^2 \ & =\mathrm{d}^2\left[1+2^2+3^2+\ldots+(\mathrm{n}-1)^2ight] \ & =\mathrm{d}^2\left[\frac{\mathrm{n}(\mathrm{n}-1)(2 \mathrm{n}-1)}{6}ight] \ \sigma^2 & =\frac{1}{\mathrm{n}} \sum x_{\mathrm{i}}^2-(\bar{x})^2 \ & =\frac{\mathrm{d}^2(\mathrm{n}-1)(2 \mathrm{n}-1)}{6}-\left[\frac{(\mathrm{n}-1) \mathrm{d}}{2}ight]^2 \ & =\frac{\mathrm{d}^2(\mathrm{n}-1)}{2}\left(\frac{2 \mathrm{n}-1}{3}-\frac{\mathrm{n}-1}{2}ight) \ & =\frac{\mathrm{d}^2(\mathrm{n}-1)}{2}\left(\frac{\mathrm{n}+1}{6}ight)=\frac{\left(\mathrm{n}^2-1ight) \mathrm{d}^2}{12}\end{aligned}$

The raw data $x_1, x_2, \ldots \ldots, x_{\mathrm{n}}$ is an A.P. with common difference $\mathrm{d}$ and…

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