Probability about confidence interval 
Let $X_1,...X_n$ be iid $N(\theta,1)$. A 95% confidence interval for
  $\theta$ is $\overline{X}\pm\frac{1.96}{\sqrt{n}}$.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 .95?Prove
  your answer.
Casella & Berger, Chapter 9 - Interval Estimation

First $\overline{X}\approx N(\theta,\frac{1}{n})\Rightarrow Y=\sqrt{n}\overline{X}$~$N(\theta,1)$
I know that
$$P(\overline{X}-\frac{1.96}{\sqrt{n}}\leq\theta\leq \overline{X}+\frac{1.96}{\sqrt{n}})=0.95$$
$$P(\theta-\frac{1.96}{\sqrt{n}}\leq\overline{X}\leq\theta+\frac{1.96}{\sqrt{n}})=0.95$$
$$P(\theta-1.96\leq\overline{X}-\theta\leq\theta+1.96)=0.95$$
$$P(\theta-1.96\leq Y\leq\theta+1.96)=0.95$$
That's why I think it's the same, I tried other ways but got nowhere.
Someone give me a hint?
 A: Perhaps a little more than a hint but here goes...
The question appears to be asking if 
$$\begin{align}
p = P\left\{\bar X - 1.96\sqrt\frac{1}{n} \le X_{n+1} \le \bar X + 1.96\sqrt\frac{1}{n} \right\} \\ 
 = P\left\{-1.96\sqrt\frac{1}{n} \le X_{n+1} - \bar X \le 1.96\sqrt\frac{1}{n} \right\} \\
\end{align}$$
is less than, equal to, or greater than $0.95$.
Now, it turns out $X_{n+1} - \bar X$ (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 $\bar X$). 
Then we have that
$P\left\{ -1.96\sqrt{1+\frac{1}{n}} \le X_{n+1} - \bar 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
$$ 
\left[ -1.96\sqrt{1+\frac{1}{n}}, \ \ 1.96\sqrt{1+\frac{1}{n}} \ \right] 
= \left[ -1.96\sqrt{1+\frac{1}{n}}, \ \ -1.96\sqrt{\frac{1}{n}} \ \right) \\
\bigcup \color{red}{ \left[ -1.96\sqrt{\frac{1}{n}}, \ \ 1.96\sqrt{\frac{1}{n}} \ \right]} \\
\bigcup \left( 1.96\sqrt{\frac{1}{n}}, \ \ 1.96\sqrt{1+\frac{1}{n}} \ \right]
$$
Can you deduce where $p$ stands relative to 0.95?
