A known-scale normal mean has a nonrandom interval length

Let $n\ge2$ and let $X_1,\ldots,X_n$ be independent $N(\mu,\sigma^2)$ observations, where $\mu\in\mathbb R$ is unknown and the scale $\sigma>0$ is known. Starting from the joint sampling law, derive the exact distribution of $$ Z=\frac{\sqrt n\,(\bar X-\mu)}{\sigma}. $$ For a fixed $\alpha\in(0,1)$, let $z_{\alpha/2}$ denote the upper $\alpha/2$ quantile of $N(0,1)$, so $P(Z>z_{\alpha/2})=\alpha/2$. Invert the event $$ -z_{\alpha/2}\le Z\le z_{\alpha/2} $$ to obtain a confidence interval $C(X)$ for $\mu$. Prove that the coverage equals $1-\alpha$ for every $\mu\in\mathbb R$. Compute the length of $C(X)$ and prove that this length is a nonrandom function of $(n,\sigma,\alpha)$ only. Explain why substituting a realized sample mean $\bar x$ is legitimate when reporting the observed interval, but illegitimate when rederiving the coverage probability by treating the unknown $\mu$ as random. Do not replace $\sigma$ by a sample scale, and do not invoke a posterior for $\mu$.

Standardize $\bar X$ using its exact normal law. The scale $\sigma$ is a known constant, so it may appear in the pivot.