Given i.i.d random variable with mean $\mu$ and variance 1, $X_n = \frac{1}{n}(X_1+\cdots+X_n)$, approximate the probability by using CLT 
Given i.i.d random variable with mean $\mu$ and variance 1, $\bar{X}_n =
 \frac{1}{n}(X_1+\cdots+X_n)$,
  use the CLT to approximate the following probability:

$$P(|\bar{X}_n - \mu| \ge \frac{2}{\sqrt {n}})$$
My attempt:
$$P(|\bar{X}_n - \mu| \ge \frac{2}{\sqrt {n}}$$
$$=P(\sqrt n|\bar{X}_n - \mu| \ge 2)  \;\;\;\text{by multiplying both sides by $\sqrt n$}$$
$$=P(|\sqrt n \bar{X}_n - \sqrt n\mu| \ge 2)$$
Let $Z = \sqrt n \bar{X}_n - \sqrt n \mu$, by CLT, we know that $\sqrt n \bar{X}_n - \sqrt n \mu \rightarrow N(,)$
This is the part where I'm not sure how to figure out what  $\sqrt n \bar{X}_n - \sqrt n \mu \rightarrow$ to
Is it $N(\mu, 1)$ since we are given i.i.d random variable with mean $\mu$ and variance 1 or $N(0, 1)$ by the definition of CLT?
If anyone could help me in determining how to find what  $\sqrt n \bar{X}_n - \sqrt n \mu \rightarrow$ to or in general how to determine by CLT the approximation normal distribution for questions similar to this that would be really appreciated.
 A: Suppose $X_1,\ldots,X_n$ are uncorrelated (a weaker assumption than independence) and all have the same expected value $\mu$ and the same variance $\sigma^2<\infty$ (a weaker assumption that identical distribution).  Then
$$
\operatorname{E}\left( \frac{X_1+\cdots+X_n} {\sqrt n} - \mu \right) = 0 \text{ and } \operatorname{var}\left( \frac{X_1+\cdots+X_n} {\sqrt n} - \mu \right) = 1.
$$
You don't need to know anything about the normal distribution, let alone the central limit theorem, in order to show that much.
If in addition $X_1,\ldots,X_n$ are normally distributed and independent, then so is $\dfrac{X_1+\cdots+X_n} {\sqrt n} - \mu$.  And if they are not normally distributed but are independent, then the distribution of $\dfrac{X_1+\cdots+X_n} {\sqrt n} - \mu$ approaches a normal distribution as $n\to\infty$, and that normal distribution has the same expected value and the same variance. 
A: EDIT:
Since last I read this, there has been a change in the value of $\mu$.
So, I will let $\bar X = \bar X_n$. I think you can determine that $\mu = E[\bar X] $, and $\sigma^2 = \text{Var}[\bar X] = \frac{1}{n}$. Then
\begin{align*}
P\left[|\bar X - \mu| > \frac{2}{\sqrt n}\right]
&= P\left[\bar X-\mu > \frac{2}{\sqrt n} \cup\bar X-\mu <\frac{-2}{\sqrt n}\right]\\
&=P\left[\frac{\bar X - \mu}{\sqrt{1/n}}>\frac{(2+\mu)-\mu}{\sqrt{1/n}\sqrt{n}}\right]+P\left[\frac{\bar X - \mu}{\sqrt{1/n}}<\frac{(-2+\mu)-\mu}{\sqrt{1/n}\sqrt{n}}\right]\tag{1}\\
&=P\left[\frac{\bar X - \mu}{\sigma}>2\right]+P\left[\frac{\bar X - \mu}{\sigma}<-2\right]\\
&=P(Z>2)+P(Z<-2)\\
&=0.04550026
\end{align*}
where $(1)$ is true since the events are disjoint, and $Z$ is a standard normal distribution. The CLT comes into play since for large $n$
$$\frac{\bar X -\mu}{\sigma}\to Z.$$
