Intuition behind $2\Phi(x)-1$ When using the central limit theorem to calculate for $n$ I come across having to use $2\Phi(x)-1$ to find this. However, I'm unsure why I have to use this and what it means and how it's derived. It just seems to be rather common as an qpproach to follow when trying to derive $n$, what's the explanation behind $2\Phi(x)-1$?
where symbol $\Phi(x)$ denotes the cumulative distribution function of a standard normal variable.
For example:
Suppose that a random sample of size $n$ is to be taken from a distribution for which the mean is $\mu$ and the standard deviation is $3$. Use the central limit theorem to determine approximately the smallest value of $n$ for which the following relation will be satisfied:
$$Pr(|\bar{X}_n-\mu|<0.3)\ge 0.95$$
By following the $Z$ distribution: $Z = \frac{\sqrt(n)(\bar{X}_n-\mu)}{\sigma}$ I can get: $Pr\left(\frac{\sqrt(n)(\bar{X}_n-\mu)}{3}<0.3\right)=Pr(|\bar{X}_n-\mu|<0.1\sqrt{n})\ge 0.95 $
For the next step I do: $Pr(|\bar{X}_n-\mu|<0.1\sqrt{n})\approx2\Phi(0.1\sqrt{n})-1\ge 0.95$
Then re-arranging to find $n$.
So what's the intuition behind using $2\Phi(x)-1$?
 A: For a standard normal random variable $Z \sim N(0,1)$ and a positive real number $\alpha$,
\begin{align}
P(|Z| \leq \alpha) &= P(-\alpha \leq Z \leq \alpha)\\
&= \Phi(\alpha) - \Phi(-\alpha)\tag{1}\\
&= \Phi(\alpha) - [1 - \Phi(\alpha)]\tag{2}\\
&= 2\Phi(\alpha) - 1 \tag{3}
\end{align}
where about the only place where any sort of intuition might have been used is in recognizing that symmetry of the standard normal pdf about $0$ allows us to recognize (or intuit) that $\Phi(-\alpha)$, the area under the pdf to the left of the point $-\alpha$, must equal $1-\Phi(\alpha)$, the area under the pdf to the right of the point $\alpha$, and so $(2)$ follows from $(1)$.  In particular, note that tables of $\Phi(\cdot)$ show that $\Phi(1.96) = 0.9750$ and so
$$P(|Z| \leq 1.96) = 2\times 0.9750 - 1 = 0.95.$$
Since the right side of $(3)$ is an increasing function of $\alpha$, choosing $\alpha$ to be larger that $1.96$ will certainly give us a probability larger than $0.95$.
Next, given a random sample $X = (x_1, x_2, \ldots,x_n)$ from a distribution with known mean $\mu$ and known standard deviation $\sigma$, it is known that the sample mean $\bar{X}$ is a random variable with mean $\mu$ and standard deviation $\dfrac{\sigma}{\sqrt{n}}$ and so genuflections in the general direction of the CLT allows us to claim that
$$\frac{\bar{X} - \mu}{\sigma/\sqrt{n}} = \frac{\sqrt{n}(\bar{X} - \mu)} {\sigma} \sim Z \sim N(0,1)$$ or at least when $n$ is "large".
