# Normal probability distribution with absolute value of X

Random variable X has a normal distribution N(30,5)

find $P(|X| > 25)$

Having this I started to solve it normal way: $$P(|X| > 25) = 1 - P(|X| \le 25)$$

Now, normalize: $$1-P(|X| \le 25) = 1 - P (\frac{|X|-30}{5} \le \frac{25-30}{5})$$

$$1-(P(x<-25)+P(X > 25)) = 1 -(P( \frac{x-30}{5} < \frac{25-30}{5})+1-P( \frac{x-30}{5} \le \frac{25-30}{5}))$$

but solving it I obtained probability of... 1.8. So something is definitely wrong. Could you point me my mistake, please?

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how did you get $1.8$? – Ilya Jan 20 '13 at 12:06
I believe I wrongly interpeted absolute value in just added part of my solution. – mickula Jan 20 '13 at 12:12

You shall not normalize as you are dealing with the module, so a better way is $$P(|X|>25) = 1-P(|X|\leq25) = 1-P(-25\leq X\leq25)=$$ here we put $\xi$ to be a standard normal r.v. - i.e. we do a normalization $$= 1-P\left(-25\leq5\xi+30\leq 25\right) = 1-P(\xi\in[-11,-1])$$ and the latter probability you can find easily. Note that here I assumed (as you also did) that $5$ is the deviation of $X$ and not the variance.
just simply got one more question: when normalizing why didn't you touch -25 and 25? Shouldn't it be $1-P( \frac{-25-30}{5} ... )$ and same for 25? – mickula Jan 20 '13 at 15:17