Wondering why when I use a normal distribution and try and calculate the probability of $\mathbb P(Z>-2)$ I do this: $\mathbb P(Z>-2) = 1- \mathbb P(Z<-2) = 1- (1-\mathbb P(Z<2))$. Is that right?
2026-04-08 16:00:12.1775664012
Given a $Z$-score of $-2$, is $\mathbb P (Z>-2) = \mathbb P(Z<2)$? How is it equal?
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Yes, as $Z$ is a symmetric distribution around $0$, hence, $$ \mathbb{P}(Z > -2) = \mathbb{P}( Z < 2). $$