Compute the probability that $ | \bar{X} - \mu | > S$ Given $X_1, \ldots, X_n$ from $\mathcal{N} (\mu, \sigma^2)$.
I have to compute the probability:
$$P\left(|\bar{X} - \mu| > S\right)$$
where $\bar{X}$ is the sample mean and $S^2$ is the sample variance.
I tried to expand:
$$P\left(\bar{X}^2 + \mu^2 - \bar{X}\mu > \frac{1}{n}\sum {X_i}^2 + \frac{1}{n}\sum\bar{X} - 2\left(\frac{1}{n}\sum X_i\right) \bar{X} \right) $$
$$P\left( \mu^2 - \bar{X}\mu > \frac{1}{n}\sum {X_i}^2  - 2\bar{X}^2 \right) $$
but it does not seems to be helpful.
Can someone help me?
 A: The empirical mean and empirical variance of i.i.d. normal samples are independent and follow known distributions, which are respectively normal and chi-squared. This indicates that
$$
\mathrm P\left(|\bar X-\mu|\gt S\right)=\mathrm P\left((n-1)Z_1^2\gt n(Z_2^2+\cdots+Z_n^2)\right),
$$
where $(Z_k)_{1\leqslant k\leqslant n}$ is i.i.d. and standard normal. More simply, this is $\mathrm P(|T_{n-1}|\gt 1)$, where the distribution of $T_{n-1}$ is the Student's $t$-distribution with $n-1$ degrees of freedom. Hence,
$$
\mathrm P\left(|\bar X-\mu|\gt S\right)=\mathrm P\left(|T_{n-1}|\gt1\right)=I_{\frac{n-1}n}\left(\frac{n-1}2,\frac12\right),
$$
where $I$ denotes the regularized incomplete beta function. The cases $n=2$, $3$, $4$ and $\infty$ are somewhat explicit.
Edit: Recall that the empirical mean $\bar X$ and the empirical variance $S^2$ of the sample $(X_k)_{1\leqslant k\leqslant n}$ are defined as
$$
\bar X=\frac1n\sum\limits_{k=1}^nX_k,\qquad\qquad S^2=\frac1{n-1}\sum\limits_{k=1}^n(X_k-\bar X)^2.
$$
A: $$
\frac{\bar X - \mu}{\sigma/\sqrt{n}} \sim \mathcal{N}(0,1)
$$
$$
\frac{\bar X - \mu}{S/\sqrt{n}} \sim T_{n-1}
$$
where $T_k$ is Student's t-distribution with $k$ degrees of freedom.
So
$$
\Pr\left(\left|\frac{\bar X - \mu}{S}\right| > 1\right) = \Pr\left(\left|\frac{\bar X - \mu}{S/\sqrt{n}}\right| > \sqrt{n}\right) = \Pr(|T|>\sqrt{n})
$$
where the distribution of $T$ is Student's t-distribution with $n-1$ degrees of freedom.
I don't know any neat expression for this.  For any particular value of $n$, you can get a number from standard on-the-shelf software.
