Question

Let $r$ be the range and $S^2=\frac{1}{n-1}{\sum }_{i=1}^n{{(x}_i-\overline{x})}^2$ be the S.D. of a set of observations $x_1,x_2,.....x_n$, then

• A. $S\le r\sqrt{\frac{n}{n-1}}$
• B. $S=r\sqrt{\frac{n}{n-1}}$
• C. $S\ge r\sqrt{\frac{n}{n-1}}$
• D. none of these

Answer 1

The correct option is A $S\le r\sqrt{\frac{n}{n-1}}$

Here range $=r=$ largest value $-$ smallest value $=Max\left|\right(x_i-x_j\left)\right|(i\ne j)$ and
$S=\frac{1}{n-1}\sum _{i=1}^n{(x_i-\overline{x})}^2$
now ${(x_i-\overline{x})}^2={[x_i-\frac{x_1+x_2+.....+x_n}{n}]}^2$
$=\frac{1}{n^2}{[\left(x_{i-}{x}_1\right)+\left(x_{i-}{x}_2\right)+...+(x_i-x_n)]}^2$ $=\frac{1}{n^2}{[(x_i-x_1)+(x_i-x_2)+...+\left(x_{i-}{x}_{i-1}\right)+(x_i-x_{i+1})+...+(x_i-x_n)]}^2$
$\Rightarrow .{(x_i-\overline{x})}^2\le \frac{1}{n^2}{\left[(n-1)r\right]}^2(\because \left|x_i{x}_j\right|\le r)$
$\Rightarrow \frac{{\sum }_{i=1}^n{(x_i-\overline{x})}^2}{n-1}\le \frac{1}{n^2(n-1)}\sum {\left[(n-1)r\right]}^2$ (summing up and dividing (n-1) both sides)
$\Rightarrow \frac{1}{n^2}\frac{1}{n-1}n{(n-1)}^2{r}^2$ $=\frac{n-1}{n}{r}^2$ $<\frac{n}{n-1}.r^2$ $(\frac{\because \forall n>1}{n>\frac{1}{n}})$
$\therefore S^2<\frac{n}{n-1}.r^2\text{or}S<r\sqrt{\frac{n}{n-1}}$