Can a percentile simultaneously refer to the % below or equal to a score and the % above and equal to a score? I realise there are several definitions of a percentile. Recently I was given the following definition
Definition 1- A percentile is the percentage of scores lower than or equal to it
I was then asked whether one could infer that the percentile could refer to the percentage of scores above and equal to the percentile.
e.g. 75th percentile means that 25% of observations were greater than or equal to the rank
It seems to me that if you use the definition I was given, you would be doubling up and it would be more accurate to say that 24% of observations where above the 75th percentile. Is this right?
 A: By convention, the cumulative distribution function is defined by
$$F_X(x)=\mathbb{P}(X\le x)$$
rather than
$$F_X(x)=\mathbb{P}(X< x)$$
Therefore, it is also conventional to define an $\alpha$-quantile $q_\alpha$ by
$$F_X(q_\alpha)=\mathbb{P}(X\le q_\alpha)=\alpha$$
and to define its empirical version as
$$\frac{1}{n}\sum_{i=1}^n \mathbb{I}(X_i\le \hat{q}_\alpha)\le\alpha$$
i.e., $$\hat{q}_\alpha=X_{(\lfloor n\alpha \rfloor)}$$
the $\lfloor n\alpha \rfloor$-th order statistic for the sample $X_1,\ldots,X_n$.
In terms of your question, an $\alpha$-quantile $q_\alpha$ also satisfies
$$1-F_X(q_\alpha)=\mathbb{P}(X> q_\alpha)=1-\alpha$$
which means that the proportion of realisations of $X\sim F_X$ strictly larger than $q_\alpha$ is $1-\alpha$. In your terms, this means the percentage of scores strictly above to the $n$-th percentile is $(100-n)\%$. Not above or equal. In absolutely continuous settings, it does not make a difference to use above or above or equal, because $\mathbb{P}(X=x)=0$ but in discrete cases it obviously may. So, if $q_{.75}$ is the $75\%$ percentile, we cannnot deduce which percentage of the sample is above or equal to $q_{.75}$. It is at least $25\%$.

To state that $24\%$ of the sample is strictly above the $75\%$ percentile is
  thus incorrect since the right percentage is $25\%$.

A: $X$ is a random variable, so it has a set of possible outcomes, where all the possible $x$'s are elements of this set. This means that if 
$$ \Pr(X \leq x) = p $$
then
$$ \Pr(X > x) = 1-p $$
so knowing percentage below, you also know the percentage above. Saying it differently, if you take subset of all the possible $x$'s such that $X \leq x$, than subset $X > x$ is a complement of the previous subset. This is directly related to Kolmogorov's axioms, since second axiom states that probability of all the possible outcomes is one (i.e. all the individual probabilities sum up to $1$), so if we subtract from the total probability probabilities of some outcomes, then we will be left with sum of probabilities for all the other outcomes.
