standard deviation can not be $6$.

Question

For the frequency distribution : Variate x : $x_1 , x_2 , x_3 , x_ 4 ........, x_{15}$ Frequency f : $f_1 , f_2 , f_3 , f_ 4 ........, f_{15}$ where $0 < x_1 < x_2 < x_3 < x_ 4 ........< x_{15} = 10$ and $\sum f_i >0$.

Then the standard deviation can not be $6$.

Can anyone please help me by giving some hints? I have no idea.

Answer 1

There is a standard result that $$\sigma_x^2 \leq \frac{R_x^2}{4}$$ where $\sigma_x^2$ is the variance of $X$ and $R_x$ is the range of $X$

Note that here $R_x = 10 - 0 = 10$

Using this result, $$\sigma_x^2 \leq \frac{10^2}{4} = 25 $$ Or, $$\sigma_x \leq 5$$

Proof:

Let $\min_{1\leq i \leq n} x_i = a$ and $\max_{1\leq i \leq n} x_i = b$

Then $R_x = b - a$

$\sigma_x^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2$

Now $\sum_{i=1}^n (x_i - c)^2$ is least when $c = \bar{x}$

Hence $\sum_{i=1}^n (x_i - \bar{x})^2 \leq \sum_{i=1}^n (x_i - \frac{a+b}{2})^2 = \sum_1 (x_i - \frac{a+b}{2})^2 + \sum_2 (x_i - \frac{a+b}{2})^2$

where $\sum_1$ includes the values of $x$ less than or equal to $\frac{a+b}{2}$ and $\sum_2$ includes values of $x$ greater than $\frac{a+b}{2}$

Or, $\sum_{i=1}^n (x_i - \bar{x})^2 \leq \sum_1 (a - \frac{a+b}{2})^2 + \sum_2 (b - \frac{a+b}{2})^2 = \sum_1 \frac{R_x^2}{4} + \sum_2 \frac{R_x^2}{4} = n \frac{R_x^2}{4}$

Or, $\sigma_x^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 \leq \frac{R_x^2}{4}$