How to calculate z- value of absolute value or modulus for normal distribution. i.e: $P(|X-a|>b)$

Question

Lets say $X\sim N(3,9)$ where variance= 9. What is $ P(|X-3|>6)$?

Answer 1

You have to solve $|X-3|>6$.

i) For $X-3>0$ we have

$X-3>6\Rightarrow X>9$

ii) For $X-3<0$ we have

$-X+3>6\Rightarrow X<-3$

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Thus $P(|X-3|>6)=1-P(-3<X<9)=1-\left(\Phi\left(\frac{9-3}{3} \right)-\Phi\left(\frac{-3-3}{3}\right)\right)$

$=1-\left(\Phi(2)-\Phi(-2)\right)=1-\left(2\Phi(2)-1 \right)$

$=2-2\Phi(2)$