My lecture notes states the Chebyshev inequality as follows:
If k is the number of entries of x that satisfy $|x_i − avg(x)| \ge a$, then $k/n \le {(std(x)/a)}^2$.
x is a 100-vector with rms(x) = 1. Let $q_{90}$ be the 90th percentile of the absolute values of the entries of the vector x. How big and how small can $q_{90}$ be?
First, I obtain possible range of values of avg(x) from the equality $${rms(x)}^2 = {avg(x)}^2 + {std(x)}^2$$ I get ${avg(x)}^2 = {rms(x)}^2 - {std(x)}^2 \le {rms(x)}^2 = 1$ $\Rightarrow |avg(x)| \le 1$. Similarly, ${std(x)}^2 = {rms(x)}^2 - {avg(x)}^2 \le {rms(x)}^2 = 1$.
Since $q_{90}$ is the the 90th percentile, there are k=10 values that with $|x_i − avg(x)|\ge a$. Therefore, $1/10 \le {(std(x)/a)}^2 \le 1/a$ $\Rightarrow a \le {\sqrt 10}$.
I feel that possible $q_{90}$ should fall in the range between $min\{avg(x)\}-max\{a\} = -1-{\sqrt 10}$ and $max\{avg(x)\}+max\{a\} = 1+ {\sqrt 10}$, but I am not sure how to justify it.
Is this even correct? Can anyone point me in the right direction? I would appreciate any help!