# Thread: Abs Value & Normal Distribution

1. ## Abs Value & Normal Distribution

I'm pretty sure that this is correct?

$\displaystyle P(|z|=1) = P(z>1)+P(z<-1) = 2\times P(z>1)$

2. Originally Posted by pickslides
I'm pretty sure that this is correct?

$\displaystyle P(|z|=1) = P(z>1)+P(z<-1) = 2\times P(z>1)$

The probability of what you've posted is equal to zero:

$\displaystyle P(|z|=1) = P(z=1)+P(z=-1) = 0 + 0 = 0$

since you're dealing with a continuous variable.

What is correct is that

$\displaystyle P(|z| {\color{red}>} 1) = P(z>1)+P(z<-1) = 2\times P(z>1)$

3. And we're assuming this is a st normal?
(sorry didn't notice the title)
You only need this to be a distribution that is symmetric about 0.

4. Originally Posted by mr fantastic

What is correct is that

$\displaystyle P(|z| {\color{red}>} 1) = P(z>1)+P(z<-1) = 2\times P(z>1)$
$\displaystyle \Rightarrow P(|z|< 1) = P(z<1)-P(z<-1) = 2\times P(0<z<1)$ ?

and $\displaystyle P(|z|< 1) + P(|z|> 1) = 1$ or am I just spiralling out of control now?

5. Originally Posted by matheagle
And we're assuming this is a st normal?
(sorry didn't notice the title)
You only need this to be a distribution that is symmetric about 0.
Yep, going with all these assumptions, just never bumped into a problem with the abs wraped around the Z before.

6. Originally Posted by pickslides
$\displaystyle \Rightarrow P(|z|< 1) = P(z<1)-P(z<-1) = 2\times P(0<z<1)$ ?

and $\displaystyle P(|z|< 1) + P(|z|> 1) = 1$ or am I just spiralling out of control now?
No, these are false...

What you did in a first time was right, except for the red part corrected by Mr F :

$\displaystyle \mathbb{P}(|Z|{\color{red}>}1)=\mathbb{P}(Z<-1)+\mathbb{P}(Z>1)$

Why so ? Because |Z|>1 means "Z<-1 OR Z>1"
Since the event "Z<-1" and the event "Z>1" cannot happen at the same time, $\displaystyle \mathbb{P}(|Z|>1)=\mathbb{P}(\{Z<-1\} \cup \{Z>1\})=\mathbb{P}(Z<-1)+\mathbb{P}(Z>1)$

(this is the formula of P(A u B)=P(A)+P(B) when A and B are disjoint)

7. Originally Posted by pickslides
$\displaystyle \Rightarrow P(|z|< 1) = P(z<1)-P(z<-1) = 2\times P(0<z<1)$ ?

and $\displaystyle P(|z|< 1) + P(|z|> 1) = 1$ or am I just spiralling out of control now?
Originally Posted by Moo
No, these are false...

[snip]
No moo, they're true

8. Thanks Mr F, I think you have helped to consolidate my understanding.