We are given $X \sim \mathcal{N}(\mu, \sigma^2)$ , and $\mu, \sigma$ are unknown.
The exercise is to find the probability that $\mu \in I_n$, where $I_n = \left[ \bar{X}_n - \frac{1.96}{\sqrt{n}}, \bar{X}_n + \frac{1.96}{\sqrt{n}} \right]$ is a confidence interval, and $\bar{X}_n = \frac{1}{n}\sum_{i = 1}^nX_i$ is the sample mean.
In the solution it says:
$P_{\mu,\sigma}(\mu \in I_n) = P( |\bar{X}_n - \mu| \leq \frac{1.96}{\sqrt{n}}) = 2P( \bar{X}_n - \mu \leq \frac{1.96}{\sqrt{n}}) - 1$
I don't understand the last step. Where did the $-1$ come from?
A simpler way to write it, with $\mu = 0$, $\frac{1.96}{\sqrt{n}} = c_n$ is:
$P( |\bar{X}_n| \leq c_n) = 2P( \bar{X}_n \leq c_n ) - 1$
I don't see why this is true..
$$P( | X - \mu | < a ) = P( X - \mu < a) - P( X - \mu < - a)$$
Then we use symmetricity of normal distribution: $$P(X - \mu < - a) = P(X - \mu > a) = 1 - P(X - \mu < a )$$
So finally we arrive at: $$P( | X - \mu | < a ) = P( X - \mu < a) - (1 - P(X - \mu < a)) = 2 P(X - \mu < a) - 1$$
which is exactly what you asked for.