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

Moderate

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

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