= Solution
The sample mean $\overline X$ has <normal distribution> $N(\mu,1/n)$ because the observations are independent with known variance 1. The <pivotal quantity> $Z=\sqrt n(\overline X-\mu)$ has the <standard normal distribution>. Consequently
$$
\mathbb P_\mu\left\{-1.96\le\sqrt n(\overline X-\mu)\le1.96\right\}\simeq0.95,
$$
and inversion gives the <confidence interval>
$$
\boxed{\left[\overline X-\frac{1.96}{\sqrt n},\ \overline X+\frac{1.96}{\sqrt n}\right].}
$$
Replacing 1.96 by the exact $0.975$ normal quantile gives exact 95% coverage; the printed numerical value gives the usual rounded interval. No variance estimation or Student distribution is needed.
Back to article page
