= Solution
<Independence> and normality give $\overline X\sim N(\mu,1/n)$. Consequently $Z=\sqrt n(\overline X-\mu)$ is standard normal, and
$$
\mathbb P\left(\overline X<\mu+\frac{z_\alpha}{\sqrt n}\right)=\Phi(z_\alpha)=1-\alpha.
$$
Rearranging the event gives $\mu>\overline X-z_\alpha/\sqrt n$. Thus the random <interval> $(\overline X-z_\alpha/\sqrt n,\infty)$ covers the fixed parameter with <probability> $1-\alpha$. After observing the sample, its realized <confidence interval> is
$$
\boxed{\left(\overline x-\frac{z_\alpha}{\sqrt n},\infty\right).}
$$
The coverage refers to repeated samples, not to a <posterior probability> for the fixed parameter.
Back to article page
