Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics III / 8 / a / ii / Solution

The sample mean is exactly $N(\mu ,{\sigma }^2/n)$, so $(\overline{X}-\mu )/(\sigma /\sqrt{n})$ has the standard normal distribution. The given quantile gives probability $0.95$ to the interval $[-1.96,1.96]$. Inverting this pivot yields the exact $95\%$ confidence interval