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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asked Aug 4, 2013 at 22:50
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$\begingroup$ 16 minutes. $ $ $\endgroup$– DidAug 4, 2013 at 23:10
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$\begingroup$ @Did: Um, what? $\endgroup$– Michael AlbaneseAug 5, 2013 at 1:38
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1 Answer
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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 $P(Z > 1.5) = 1 - P(Z < 1.5)$.
answered Aug 4, 2013 at 23:00
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1$\begingroup$ 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. $\endgroup$ Aug 5, 2013 at 3:12
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$\begingroup$ Of course. I didn't use the normality assumption anywhere. $\endgroup$ Aug 5, 2013 at 3:39