The problem is this:
Let $k\geq 1$ . Show that, for any set of n measurements, the fraction included in the interval $\bar{y}-ks$ to $\bar{y}+ks$ is at least $(1-\frac{1}{k^2})$.
I tried using the $s^2=\frac{1}{n-1} \sum_{i=1}^{n}(y_i-\bar{y})^2$ and the fact that we have $|y_i-\bar{y}| \geq ks$ for some observations. I don't know how to continue. Help me, please.
Hint:
For any $k>0$, the probability of a random variable $Y$ with finite mean $\mu$ and finite variance $\sigma^2(>0)$ to lie within the interval $[\mu-k\sigma,\mu+k\sigma]$ is given by Chebyshev's inequality: $$\Pr(|Y-\mu|\le k\sigma)\ge1-\frac{1}{k^2}$$
Now this probability in the l.h.s is estimated by the proportion (or fraction) of the $y_i$'s satisfying $|y_i-\bar y|\le k s$, where $\bar y$ is the arithmetic mean and $s$ is the standard deviation of the set of observations $y_1,y_2,\cdots,y_n$. Think of $(y_1,\cdots,y_n)$ as a sample of size $n$. I think this provides some intuition for the desired result to follow. After all, it looks like an analogy of the probabilistic result.
By the way, here $s$ is the usual standard deviation with divisor $n$; there is no need to define $s$ with divisor $n-1$ as you say.
Just noticed that this is answered in detail in this thread.