Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer
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
Ok, now I see that u did it on-the-fly. Thank you. – mickula Jan 20 '13 at 16:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.