= Solution
Let $z=\Phi^{-1}(1-\alpha/2)$, where $\Phi$ is the <standard normal distribution>'s <cumulative distribution function>. <Statistical consistency> and <Slutsky's theorem> allow replacement of $I_1(\theta)$ by $I_1(\widehat\theta)$:
$$
\sqrt{nI_1(\widehat\theta)}(\widehat\theta-\theta)\ \xrightarrow{d}\ N(0,1).
$$
Thus the probability that this standardized <estimation error> lies between $-z$ and $z$ tends to $1-\alpha$. Rearranging gives the <Wald confidence interval>
$$
\boxed{\left[\widehat\theta-\frac{z}{\sqrt{nI_1(\widehat\theta)}},\quad
\widehat\theta+\frac{z}{\sqrt{nI_1(\widehat\theta)}}\right].}
$$
The <coverage probability> is approximate, with limiting value $1-\alpha$; it is a repeated-sampling statement about the random <confidence interval>.
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