Show that the two formulae for the standard deviation of ungrouped data: `sigma=sqrt(1/nsum(x_i- barX )^2)` `a n d\ sigma^(prime)=sqrt(1/nsumx_i^2- barX ^2)` are equivalent where `barX =1/nsumx_i`.

Answer

Answer:



`bar{X}=frac{1}{n} sum x_{i}`

Explanation

\begin{equation}
\begin{aligned}
&\sigma=\sqrt{\frac{1}{n} \sum\left(x_{i}-\bar{X}\right)^{2}} \\
&=\sqrt{\frac{1}{n} \sum\left(x_{i}^{2}-2 x_{i} \bar{X}+\bar{X}\right)^{2}}=\sqrt{\frac{1}{n} \sum x_{i}^{2}-\frac{1}{n} \sum 2 x_{i} \bar{X}+\frac{1}{n} \sum \bar{X}^{2}} \\
&=\sqrt{\frac{1}{n} \sum x_{i}^{2}-\frac{1}{n} \times 2 \bar{X} \sum x_{i}+\frac{1}{n} \times \bar{X}^{2} \sum 1} \\
&=\sqrt{\frac{1}{n} \sum x_{i}^{2}-\frac{1}{n} \times 2 \bar{X} \times n \bar{X}+\frac{1}{n} \times \bar{X}^{2} \times n}\left(\bar{X}=\frac{1}{n} \sum x_{i}\right) \\
&=\sqrt{\frac{1}{n} \sum x_{i}^{2}-2 \bar{X}+\bar{X}^{2}}=\sqrt{\frac{1}{n} \sum x_{i}^{2}-\bar{X}^{2}}=\sigma^{1} \\
&\text { Hence, the formula } \\
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