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 $q_2=1-q_1.$

Let's call your original CI a 'probability-symmetric' confidence interval. For a symmetrical distribution, such an interval may be the narrowest one.
However, the probability-symmetric 95% CI for normal $\sigma^2,$ based on pivoting $$\frac{(n-1)S^2}{\sigma^2}\sim\mathsf{Chisq}(\nu = n-1)$$
is not the shortest because chi-squared distributions are
not symmetrical. For convenience, the probability-symmetric 95% CI is often used. (Also, 'minimum width' may not be the most important criterion, so the narrowest CI may not be the
most useful.)
Example: Suppose a random normal sample of size $n=50$ has $S^2 = 13.52.$
Then the probability-symmetric 95% CI $(9.43,20.99)$ has width $11.56,$ while the 95% CI $(9.28,20.61)$
has width $11.33.$  [Using R below.]
CI=49*(13.52)/qchisq(c(.975, .025), 49);  CI
[1]  9.434025 20.994510
diff(CI)
[1] 11.56048

CI = 49*(13.52)/qchisq(c(.98, .03), 49);  CI
[1]  9.277642 20.611959
diff(CI)
[1] 11.33432

CI = 49*(13.52)/qchisq(c(1, .05), 49);  CI
[1]  0.00000 19.52473
diff(CI)
[1] 19.52473    # one-sided

In case width is especially important, one could search for the narrowest 95% CI.
