Why is $P(|\bar{X}_n- \frac{1}{2}|≥0.1)=2*P(\bar{X}_n≥0.6)$? Given $X_1,...,X_n \sim Uni(0,1)$. Let $\bar{X}_n=\frac{1}{n}\sum_{i=1}^n X_i$ be the sample mean.
Also have been given: 
$E(\bar{X}_n)=\frac{1}{2}$, $Var(\bar{X}_n)=\frac{1}{12n}$
According to CLT: $\bar{X}_n \sim N(E(\bar{X}_n), Var(\bar{X}_n))$ 
Then why is $$P(|\bar{X}_n- \frac{1}{2}|≥0.1)=2*P(\bar{X}_n≥0.6)$$
I know that
$$|\bar{X}_n- \frac{1}{2}|≥0.1$$
$$= \bar{X}_n- \frac{1}{2}≥0.1 \text{ or } \frac{1}{2}-\bar{X}_n ≥ 0.1$$
$$= \bar{X}_n ≥ 0.6 \text{ or } \bar{X}_n ≤ 0.4$$
 A: This is not true in general. However, if $\bar{X}_n$ is symmetric about $\tfrac12$, so that
$$
\Pr(\bar X_n > \tfrac12 + y) = \Pr(\bar X_n < \tfrac12 - y)
$$
then it does hold:
\begin{align}
\Pr\left( \lvert \bar X_n - \tfrac12 \rvert \ge \varepsilon \right)
&= \Pr\left( \bar X_n \ge \tfrac12 + \varepsilon \right) + \Pr\left( \bar X_n \le \tfrac12 - \varepsilon \right)
\\&= 2 \Pr\left( \bar X_n \ge \tfrac12 + \varepsilon \right)
\end{align}
(where the first equality holds because the events $( \bar X_n \ge \tfrac12 + \varepsilon )$ and $( \bar X_n \le \tfrac12 - \varepsilon )$ are disjoint, and the second by the assumed symmetry property).
This is the case if $\bar X_n \sim \mathcal N(\tfrac12, \sigma^2)$. Note, though, that the CLT only says this holds asymptotically, not for any particular $n$. But in your case of $X_n = \frac1n X_i$, $X_i \stackrel{\text{iid}}{\sim} \mathrm{Unif}(0, 1)$, note that each $X_i$ has this symmetry property about $\tfrac12$, and so $\bar X_n$ does as well.
