 # Let $$\displaystyle{X}_{{1}},\ldots{X}_{{n}}$$ be iid $$\displaystyle{N}{\left(\theta,\ {1}\right)}$$. Adolfo Hebert 2022-03-21 Answered
Let ${X}_{1},\dots {X}_{n}$ be iid . A $95\mathrm{%}$ confidence interval for $\theta$ is $\stackrel{―}{X}±\frac{1.96}{\sqrt{n}}$. Let p denote the probability that an additional independent observation, ${X}_{n+1}$, will fall in this interval. Is p greater than, less than, equal to $0.95$? Prove your answer.
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Step 1
Perhaps a little more than a hint but here goes
The question appears to be asking if
$p=P\left\{\stackrel{―}{X}-1.96\sqrt{\frac{1}{n}}\le {X}_{n+1}\le \stackrel{―}{X}+1.96\sqrt{\frac{1}{n}}\right\}$
$=P\left\{-1.96\sqrt{\frac{1}{n}}\le {X}_{n+1}-\stackrel{―}{X}\le 1.96\sqrt{\frac{1}{n}}\right\}$
is less than, equal to, or greater than 0.95.
Now, it turns out ${X}_{n+1}-\stackrel{―}{X}$ (itself a linear combination of normal random variables) is also normal, with mean 0 and variance $1+\frac{1}{n}$ (due to independence of ${X}_{n+1}$ from the other ${X}_{i}$ and hence from $\stackrel{―}{X}$).
Then we have that
$P\left\{-1.96\sqrt{1+\frac{1}{n}}\le {X}_{n+1}-\stackrel{―}{X}\le 1.96\sqrt{1+\frac{1}{n}}\right\}=0.95$ as well.
Now take a look at the interval above and notice that

Can you deduce where p stands relative to $0.95$?