about the percentile of $f(X)$ Is the, say $90\%$, percentile of $X$ the same as the $90\%$ percentile of $aX+b$, where $a, b$ are constant?
I mean, to calculate the $90\%$ percentile of $X$, can I use the central limit law to calculate the $90\%$ percentile of $Y=\frac{X-\mu}{\sigma}$ instead of X?
Is the $90\%$ percentile of $f(X)$ always the same as $X$? (or iff $f(\cdot)$ is a linear function?)
Thank you so much!
 A: Roughly speaking (exactly for random variables $W$ with continuous distribution) a $90$-th percentile of $W$ is a number $k$ (usually unique) such that $P(W \le k)=0.9$.  
Let $\sigma$ be any positive constant, let $\mu$ be a constant. Let $Y=\frac{X-\mu}{\sigma}$.  Let $k$ be "the" $90$-th percentile of $Y$.
Then 
$$0.9=P(Y \le k)=P\left(\frac{X-\mu}{\sigma} \le k\right)=P(X \le \sigma k+\mu).$$
So $\sigma k+\mu$ is the $90$-th percentile of $X$.  Conversely, if $d$ is the $90$-th percentile of $X$, similar reasoning shows that $\frac{d-\mu}{\sigma}$ is the $90$-th percentile of $Y$.
Comment: The idea generalizes.  If $f$ is a strictly  increasing function, we can go back and forth between the $p$-th percentile of $X$ and the $p$-th percentile of $f(X)$ by doing what comes naturally. For such an $f$, the number $k$ is a $p$-th percentile of $X$ if and only if $f(k)$ is a $p$-th percentile of $f(X)$. So your intuition was right.  The actual expression you used was not quite right, the percentiles are not the same, but they are related in the "natural" way.  Of course things can, and usually  do break down for functions $f$ that are not everywhere increasing.
A: Um, no. The 90% percentile of $aX+b$ is $a($90% of $X)+b$, assuming that $a>0$.
So if you find the 90% percentile of $aX+b$ you have to subtract $b$ and divide by $a$ in order to find the 90% percentile of $X$.
