First slide

Introduction to statistics

Question

$Let r be the range and S^{2} = \frac{1}{n - 1} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} be the SD$ $of a set of observations x_{1}, x_{2}, \dots, x_{n}, then$

Very difficult

A

$S \leq r \sqrt{\frac{n}{n - 1}}$
B

$S = r \sqrt{\frac{n}{n - 1}}$
C

$S \geq r \sqrt{\frac{n}{n - 1}}$
D

None of these

Solution

$We have r = max |x_{i} - x_{j}|$
$\begin{array}{l} and S^{2} = \frac{1}{n - 1} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} \\ Now, {(x_{i} - \bar{x})}^{2} = {(x_{i} - \frac{x_{1} + x_{2} + \dots + x_{n}}{n})}^{2} \end{array}$
$= \frac{1}{n^{2}} [(x_{i} - x_{1}) + (x_{i} - x_{2}) + \dots + (x_{i} - x_{i} - 1) + (x_{i} - x_{i} + 1) + \dots + (x_{i} - x_{n})] \leq \frac{1}{n^{2}} [(n - 1) r]^{2}$ $[∵] x_{i} - x_{j} ∣\leq r]$
$\begin{array}{l} \Rightarrow {(x_{i} - \bar{x})}^{2} \leq r^{2} \Rightarrow \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} \leq n r^{2} \\ \Rightarrow \frac{1}{n - 1} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} \leq \frac{n r^{2}}{(n - 1)} \Rightarrow S^{2} \leq \frac{n r^{2}}{(n - 1)} \\ \Rightarrow S \leq r \sqrt{\frac{n}{n - 1}} \end{array}$

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