How to find value of $\vert$$z$$\vert$ in normal distribution Given that for a standard normal variable $Z$,$p(0<z<0.8) =0.2881$
The value of  $p($$\vert$$z$$\vert$ $\geq$$0.8)=?$

I already know how to find $p(z$$\geq$$0.8)$ which is equal to $0.21186$.
 
But I dont know how to find that of the above question.
 A: Normal is symmetric about $z=0$, so $p(z \geq 0.8) = p(z \leq -0.8)$.  Also, because its symmetric, $p(|z| \geq 0.8) = p(z \geq 0.8) + p(z \leq -0.8)$.  
A: I would start by thinking about the fact that the sum of probability over the whole domain is 1. Given the way you have written the problem, I write the sum of probability as:
$1 = p(z<-0.8) + p(-0.8<z<0)+p(0>z>0.8)+p(z>0.8)$.
The quantity of interest is 
$p(\left | z \right |>0.8)$
which I will rewrite as 
$p(\left | z \right |>0.8) = p(z<-0.8) + p(z>0.8)$.
So going back to the sum of probability, I can make the following substitution
$1 = p(-0.8<z<0)+p(0>z>0.8)+p(\left | z \right |>0.8)$.
Because the distribution is defined to be a standard normal distribution and thus symmetric about zero, the following is true,
$p(-0.8<z<0)=p(0>z>0.8)$.
Using this information, I again rewrite the sum of probabilities,
$1 =  2\,p(0>z>0.8) + p(\left | z \right |>0.8)$
and solve for the quantity of interest
$p(\left | z \right |>0.8) = 1 - 2\,p(0>z>0.8)$.
Hopefully that makes sense, I tried to do this in terms of the quantities on hand. However, integrating over the probability density function (pdf) is the simplest path to the answer:
$p(\left | z \right |> a) = 1 - \int_{-a}^a$pdf$(x)\,dx$
In this case, the pdf is a a standard normal distribution, and $a = 0.8$.
