For $\operatorname{Exp}$ with mean $\theta$ we have $\operatorname{Var}(X_1)=\theta^{2}$ , so the CLT gives

$\frac{\sqrt{n}\bigl(\overline{X}_n-\theta\bigr)}{\theta} \overset{d}{\longrightarrow} N(0,1).$

Note the standard deviation is $\theta$ itself — the parameter appears in the scaling, so we must invert the pivot rather than just add and subtract a margin.

Start from

$\mathbb{P}\left(z_{\alpha/2}\le \frac{\sqrt{n}\bigl(\overline{X}_n-\theta\bigr)}{\theta} \le z_{1-\alpha/2}\right)\approx1-\alpha ,$

where $z_{\alpha/2}<0<z_{1-\alpha/2}$ . Take the right-hand inequality:

$\sqrt{n}\,\overline{X}_n-\sqrt{n}\,\theta\le z_{1-\alpha/2}\,\theta \Longrightarrow\ \sqrt{n}\,\overline{X}_n\le\theta\bigl(z_{1-\alpha/2}+\sqrt{n}\bigr) \Longrightarrow\ \theta \ge \frac{\sqrt{n}\,\overline{X}_n}{z_{1-\alpha/2}+\sqrt{n}} .$

Similarly the left-hand inequality gives

$\theta \le \frac{\sqrt{n}\,\overline{X}_n}{z_{\alpha/2}+\sqrt{n}} .$

So the interval is

$\left(\frac{\sqrt{n}\,\overline{X}_n}{z_{1-\alpha/2}+\sqrt{n}},\ \frac{\sqrt{n}\,\overline{X}_n}{z_{\alpha/2}+\sqrt{n}}\right).$

The upper quantile lands in the lower limit — inverting a decreasing relation swaps the endpoints, which is exactly what distinguishes (B) from the look-alike (A).