Question
Let $$r$$ be the range and $${S^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} $$ be the S.D. of a set of observations $${x_1},\,{x_2}......,\,{x_n},$$ then :
A.
$$S \leqslant r\sqrt {\frac{n}{{n - 1}}} $$
B.
$$S = r\sqrt {\frac{n}{{n - 1}}} $$
C.
$$S \geqslant r\sqrt {\frac{n}{{n - 1}}} $$
D.
None of these
Answer :
$$S \leqslant r\sqrt {\frac{n}{{n - 1}}} $$
Solution :
$$\eqalign{
& {\text{We have, }}r = \mathop {\max }\limits_{i \ne j} \left| {{x_i} - {x_j}} \right|{\text{ and}} \cr
& {S^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} \cr
& {\text{Now, consider}} \cr
& {\left( {{x_i} - \overline x } \right)^2} = {\left( {{x_i} - \frac{{{x_1} + {x_2} + ...... + {x_n}}}{n}} \right)^2} \cr
& = \frac{1}{{{n^2}}}\left[ {\left( {{x_i} - {x_1}} \right) + \left( {{x_i} - {x_2}} \right) + ...... + \left( {{x_i} - {x_i} - 1} \right)} \right] + \left[ {\left( {{x_i} - {x_i} + 1} \right) + ...... + \left( {{x_i} - {x_n}} \right)} \right] \leqslant \frac{1}{{{n^2}}}{\left[ {\left( {n - 1} \right)r} \right]^2} \cr
& \Rightarrow {\left( {{x_i} - \overline x } \right)^2} \leqslant {r^2} \cr
& \Rightarrow \sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} \leqslant n{r^2} \cr
& \Rightarrow \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} \leqslant \frac{{n{r^2}}}{{\left( {n - 1} \right)}} \cr
& \Rightarrow S \leqslant r\sqrt {\frac{n}{{n - 1}}} \cr} $$