normal distribution with absolute value If $N(1,4)$, find $P(X<0)$ and $P(|x|>4)$.
I already have the answers which are $0.3085$ and $0.07302$ respectively.
please note this is not a homework.
Can someone explain in detail how to reach the answer? I tried multiple times but every time I am not getting it. 
 A: In general case
$$ \Pr(|X| > x) = \Pr(X < -x \cup X > x) \\
   = \Pr(X < -x) + \Pr(X > x) $$
and since normal distribution is symmetric,
$$ \Pr(X < -x) = \Pr(X > x) $$
what leads to
$$ \Pr(X < -x \cup X > x) = 2 \times \Pr(X > x) $$
You are interested in $\Pr(|X| > x)$ if you want to learn something about tails of distribution, e.g. when you want to learn about probabilities of observing values that are anomalous, outlying, or extreme. If you want to imagine this graphically, we are looking at the two tails of normal distribution (grayed).

You can also easily compute the opposite probability
$$ \Pr(|X| < x) = \Pr(-x < X < x) \\
   = \Pr(X < x) - \Pr(X \le -x) \\
   = \Pr(X > -x) - \Pr(X \ge x) \\
   = 1 -  \Pr(X < -x \cup X > x) $$ 
that is, the "common area" (colored in orange) of the two cases.

Check also a thread about single- vs two-tailed tests that discusses practical application of such interval probabilities.
A: The 'two tailed' probabilities referred to here Explaining two-tailed tests correspond to the absolute value in your question. P(abs(x)>4) is the same as P(X < - 4) + P (X > 4). Just translate X to a standard Normal variable (subtract the mean and divide by the standard deviation) and look up the corresponding probability. 
Or, in R: 
For the absolute value one:
pnorm(-4,mean= 1,sd = 2) + (1-pnorm(4,mean = 1,sd = 2))
For the other one:
pnorm(0,mean = 1,sd = 2)
