I hope this is not a very complicated question, but I have problems finding the answer. Quite often, I need to calculate something like this: $P(-a<\bar{x}<a) = p$, where $\bar{x}$ refers to the sample mean and p is given. The standard approach is to use the CLT as follows:
$P[(-a-\mu)\sqrt{n}/\sigma) < Z < (a-\mu)\sqrt{n}/\sigma] = p$, where $n, \mu, \sigma$ are given. From here, knowing that Z is distributed $N(0,1)$, we can use "the table" and find the appropriate values for a.
Is there anyway to do it through Mathematica? In particular, I need to find a.
Probability[Abs[x] < 1, x \[Distributed] NormalDistribution[]] // N
Change the1
to whatever number you need. $\endgroup$