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I can't wrap my head around the idea they are both equal. I mean shouldn't we have $P(-Z > 1.5)$ which is not equal to $P(Z < 1.5)$?

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  • $\begingroup$ 16 minutes. $ $ $\endgroup$
    – Did
    Commented Aug 4, 2013 at 23:10
  • $\begingroup$ @Did: Um, what? $\endgroup$ Commented Aug 5, 2013 at 1:38

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The normal distribution is symmetric about zero, that is why $P(Z < -1.5) = P(Z > 1.5)$ (the two regions of interest are the tails of the distribution). As the normal distribution is a probability distribution, $P(Z \leq 1.5) + P(Z > 1.5) = 1$ (the two regions cover all of the possibile values for $Z$), so $P(Z > 1.5) = 1 - P(Z \leq 1.5)$. As there is zero probability that $Z$ will be precisely $1.5$, $P(Z \leq 1.5) = P(Z < 1.5)$ so $P(Z > 1.5) = 1 - P(Z < 1.5)$.

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  • $\begingroup$ wow, ok makes sense now $\endgroup$ Commented Aug 4, 2013 at 23:07
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    $\begingroup$ In fact, the results are correct for any continuous random variable whose density function is symmetric about $0$, and the assumption of normality is not required in this instance. $\endgroup$ Commented Aug 5, 2013 at 3:12
  • $\begingroup$ Of course. I didn't use the normality assumption anywhere. $\endgroup$ Commented Aug 5, 2013 at 3:39

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