An inequality regarding mean and standard deviation : $\frac{1}{n}\#x_i \in (\bar{x}-ks,\bar{x}+ks) \geq 1-\frac{1}{k^2}$

Question

Let $x_1,\cdots,x_n \in \mathbb{R}$. Define $$\bar{x}=\frac{1}{n}\sum_{i=1}^n x_i, \quad s=\sqrt{\frac{1}{n-1}\sum_{i=1}^n (x_i-\bar{x})^2}$$ To show that the proportion of $x_i$'s in the interval $(\bar{x}-ks,\bar{x}+ks)$ exceeds $1-\frac{1}{k^2}$.

In other words, we have to show that $$\frac{1}{n}\#\{x_i : \frac{x_i-\bar{x}}{s} \in (-k,k)\} \geq 1-\frac{1}{k^2}$$

Any help would be much appreciated. Thank you.

Answer 1

Let $y_i = \frac{x_i - \bar{x}}{s}$, the problem becomes

Given $y_1, \ldots, y_n \in \mathbb{R}$ such that $\sum_{i=1}^n y_i = 0$ and $\sum_{i=1}^n y_i^2 = n -1$. How to show $$\# \{ y_i : |y_i| < k \} > n\left(1-\frac{1}{k^2}\right) $$

Notice

$$n > n-1 = \sum_{i=1}^n y_i^2 \ge \sum_{i=1,|y_i|\ge k}^n y_i^2 \ge k^2 \sum_{i=1,|y_i|\ge k}^n 1 = k^2 \# \{ y_i : |y_i| \ge k \}$$

This leads to $$\#\{ y_i : |y_i| \ge k \} < \frac{n}{k^2} \quad\implies\quad \# \{ y_i : | y_i | < k \} > n - \frac{n}{k^2} = n\left(1- \frac{1}{k^2}\right)$$