The meaning of upper 100 alpha(th) percentile Recently, I learned about the chi-square distribution. In my class, I was told about the upper $100\alpha^{th}$ percentile $\chi^{2}_{\alpha}(k)$ and given the following definition:
$$P(X<\chi^{2}_{\alpha}(k))=1-\alpha$$
$$P(X>\chi^{2}_{\alpha}(k))=\alpha$$
I have tried my best while looking online for additional information using a given name but got almost nothing( I do admit I might have looked for the wrong thing). Can someone tell me the formal name of this concept? Additionally, why would we even have something like this if a general concept of percentile should be sufficient? Also, do we use it anywhere else outside of the chi-square distribution?
 A: This is just standard definition to make notation less ugly. One area you will often see this kind of notation is in building confidence intervals. Basically, $\chi^2_\alpha(k)$ is the value such that $P(X > \chi_\alpha^2(k)) = a$. This is common when building confidence intervals: you have some theory that tells you that the test statistic follows some distribution that you know how to explicitly work with, such as chi square distribution or the normal distribution (or asymptotically does).
A common situation is when you want to build a confidence interval for the mean given data $x_1,\dots,x_n$. Let $\hat{\theta}$ be the sample mean. Assuming the CLT holds, we know that asymptotically,
$$Z={\frac {{\hat{\theta}}-\theta }{\sigma /{\sqrt {n}}}} \sim N(0,1),$$
and so if I want to build a 95% CI for $\theta$, then I claim it's enough to just know values $z^*_l,z^*_u$ such that
$$P( z^*_l \leq Z \leq z^*_u) = .95$$
Why? Well I can just massage that equation given that I know $Z$ to build the CI as follows:
\begin{align*}
 .95 = P( z^*_l \leq Z \leq z^*_u) & = P\bigg( z^*_l \leq {\frac {{\hat{\theta}}-\theta }{\sigma /{\sqrt {n}}}} \leq z^*_u \bigg) \\
& = P\bigg( \hat{\theta} - z^*_l\frac{\sigma}{\sqrt{n}} \leq \theta \leq \hat{\theta} + z^*_u\frac{\sigma}{\sqrt{n}} \bigg).
\end{align*}
So great, I just need to know the values $z^*_l,z^*_u$, but I know those: they are precisely the values such that $P( z^*_l \leq Z \leq z^*_u) = .95$ holds when $Z \sim N(0,1)$, and I know exactly how to work with the CDF of the standard normal distribution, so I'm all set after I decide to choose values that are symmetric around $0$. Since I know those values, when denoting them, I may decide to just shorthand $z^*_\alpha$ as the value such that
$$P(-z^*_\alpha\leq Z\leq z^*_\alpha)=1-\alpha.$$
