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?