Confidence interval for variance Let's say we have X, a random vector that follows $\mathcal N(\mu,\sigma^2)$. My book asks me to find a confidence interval for $σ^2$ , the variance, if $μ$ is known.
It begins by taking that $$S^2=\sum_{i=1}^n\frac{(X_{i}-μ)^2}{n}$$
Next step, it considers the following function $$Y=\frac{n*S^2}{σ^2}$$
which follows $\chi_n^2$ .
Next step, we need to find $c_{1},c_{2}$ so that $P=[c_{1}<Y<c_{2}]=γ$
where $γ=1-α$ is our confidence level.
At other examples, where our distribution was symmetrical, it defined $c_{1},c_{2}$ as $-z_{α/2},z_{α/2}$ respectively, where $α/2$ is the probability of our distribution to be above $c_{2}$ or below $c_{1}$.
However, in this example our distribution isn't $\mathcal N(\mu,\sigma^2)$, but instead $\chi_n^2$, which is not symmetrical. For our convenience, it says, we will still consider it symmetrical and consider $c_{1},c_{2}$ as $x_{n,1-α/2}$ and $x_{n,α/2}$ respectively.
And this is what I can't understand. Can you please explain me why at the first $x$ the index has $1-a/2$ but in the second it has $α/2$, even thought it said it considers it symmetrical?
I don't know if I'm making any sense guys because different teachers might teach the same stuff in different ways. If you can, please help me.
Thank you
 A: If you have a random sample of $n$ observations $X_i$ from $\mathsf{N}(\mu, \sigma^2),$
with $\mu$ known and $\sigma^2$ unknown, then $\frac{nV}{\sigma^2} \sim
\mathsf{Chisq}(df = n),$ where $V = \frac{1}{n}\sum_{i=1}^n(X_i = \mu)^2.$
[I prefer to reserve the notation $S^2$ for the case where $\mu$ is
unknown and estimated by $\bar X.$]
Then, to make a $1 - \alpha = .95$ confidence interval (CI) for $\sigma^2,$ we can find values $L$ and $U$ that cut 2.5% from the lower and
upper tails, respectively of $\mathsf{Chisq}(n).$ That is, $L$ is quantile
$\alpha/2 =.025$ and $U$ is quantile $1-\alpha/2 =.975.$
Thus $P\left(L \le \frac{nV}{\sigma^2} \le U \right) = .95.$ After some
manipulation of inequalities we obtain 
$P\left(\frac{nV}{U} \le \sigma^2 \le \frac{nV}{L}\right) = .95),$ so that
a 95% CI for $\sigma^2$ is of the form $\left(\frac{nV}{U},\; \frac{nV}{L}\right).$
This is sometimes called a 'probability-symmetrical' CI; while $V$ is
contained in the CI, it is not at the center of the CI. (So the CI
is not symmetrical about $V$.)
As an example, suppose $n = 20$ so that $L = 3.247$ and $U = 20.483$ from R
statistical software as below or from printed tables of the chi-squared distribution. Then if $V = 20.5,$ the 95% CI is $(10.01,\, 63.14).$
qchisq(c(.025,.975), 10)
##  3.246973 20.483177
10*20.5/qchisq(c(.975,.025),10)
## 10.00821 63.13573

For simplicity, this is the CI most commonly used in practice. However,
two variations are possible: 
(a) One variation is to use software to search for values
$L_c$ and $U_c$ so that $V$ turns out to be at the center of the CI.
People accustomed to symmetrical CIs for $\mu$ of the form
$\bar X \pm M,$ where the margin of error $M$ comes from a (symmetrical)
normal or t distribution, may prefer to see CIs where the point
estimate is at the center of the interval estimate. But in practice,
many types of CIs are not symmetrical in this way, and there seems to
be no practical advantage to symmetrical CIs--particularly when it is
extra work to find them.
(b) Perhaps a more useful variation is to search for $L_m$ and $U_m$ so that
the length $\left(\frac{nV}{L_m} - \frac{nV}{U_m}\right)$ of the CI is as small as possible, while maintaining $P\left(\frac{nV}{U_m} \le \sigma^2 \le \frac{nV}{L_m}\right) = .95,$ Roughly speaking larger sample sizes $n$ result in shorter CIs, and
so there is some focus on the length of CIs. There may be practical reasons
for taking the extra trouble to find the shortest CI for the population
variance.
Note: Response to comment. Here is a figure showing the PDF of $\mathsf{Chisq}(10).$ Quantiles .025 (3,247) and .975 (20.483) are shown as vertical red lines. Areas in each tail beyond the quantile are 2.5%, leaving 95% area between the red lines.

