Asymmetric confidence intervals Suppose we have iid data $X_i$ with known variance $\sigma^2$, and wish to write an asymptotic $1-\alpha $ coverage CI for the population mean $\mu$. CLT implies that if $z_q$ represents the $q$ quantile of a standard normal,
$$z_{\alpha/2}=-z_{1-\alpha/2}\leq \frac{\bar X-\mu}{\sigma/\sqrt  n}\leq z_{1-\alpha/2}$$
occurs (asymptotically) with probability $1-\alpha$ and thus implies a CI for $\mu$ of $\bar X\pm z_{1-\alpha/2}\frac{\sigma}{\sqrt n}.$
Any particular reason we take symmetric bounds, or is this just a matter of simplicity? For instance, it seems to me we could have also used
$$ z_{q_1}\leq \frac{\bar X-\mu}{\sigma/\sqrt  n}\leq z_{q_2}$$
for any $q_2-q_1=1-\alpha.$

Update: By "symmetric," I mean using $q_2=1-q_1.$

Building on BruceET's point, I thought it would be interesting to include an addendum to my post, namely the idea that we may choose a confidence region to minimize its volume subject to meeting its coverage constraint.
For simplicity, I will work with the one dimensional case, letting $T_n$ denote some absolutely continuous pivotal statistic with known invertible CDF $F$, differentiable density $f$, and $q$ quantile given by $t_q\equiv F^{-1}(q).$
Then we wish to choose $q_1$ to minimize the length of the $1-\alpha$ CI:
$$t_{1-\alpha+q_1}-t_{q_1}=F^{-1}(1-\alpha+q_1)-F^{-1}(q_1),$$
giving first order condition
$$ (f(t_{1-\alpha+q_1}))^{-1}-(f(t_{q_1}))^{-1}=0\\
\implies f(t_{1-\alpha+q_1})=f(t_{q_1}),$$
and it suffices that $f'(t_{1-\alpha+q_1})<0<f'(t_{q_1})$ for the second order condition to be met.
For symmetric density, such as in the normal case, this implies $q_1=\alpha/2.$
