# Difference between Z notation and sigma

1. May 30, 2015

### Ry122

1. The problem statement, all variables and given/known data
What is the difference between this statement being expressed in z-notation and the notation shown? What are the benefits of the two and how do you do the conversion?

if X ~N(mu,sigma) show that P(|X-mu|<= 0.675*sigma) = 0.5

2. Relevant equations

3. The attempt at a solution

2. May 30, 2015

### LCKurtz

The statement that$$P(|X-\mu|\le .675\sigma)=.5$$is completely equivalent to$$P\left(\frac{|X-\mu|}{\sigma}\le .675\right)=.5$$which is equivalent to$$P\left(-.675\le \frac{X-\mu}{\sigma}\le .675\right)=.5$$Is that what you are asking?

3. May 30, 2015

### Ry122

is that due to symmetry?

4. May 30, 2015

### LCKurtz

No, it is just algebra. To get the second statement you just divide both sides of the inequality by the positive $\sigma$ and the last form just uses properties of absolute values, specifically that $|x|\le a$ is the same thing as $-a\le x \le a$.

5. May 31, 2015

### pasmith

The only normal distribution whose cdf and inverse cdf you will find tabulated is N(0,1). If $X \sim N(\mu, \sigma^2)$ then to use tables you have instead to work with $Z = (X - \mu)/\sigma \sim N(0,1)$.

6. Jun 1, 2015

### Ry122

what is function N()?

7. Jun 1, 2015

### Ray Vickson

It is exactly the same N( ) that you yourself wrote in post #1. It is actually not a function, but, rather, a name (short for "Normal" or "Normal distribution").

8. Jun 1, 2015

### WWGD

True, but doesnt the symmetry in the algebra follow from the symmetry of the normal distribution?

Last edited: Jun 1, 2015
9. Jun 1, 2015

### Ray Vickson

No:. For a general density we have
$$P(|X-\mu| \leq a) = P(-a \leq X- \mu \leq a) = \int_{\mu-a}^{\mu+a} f(x) \, dx,$$
whether or not $f$ possesses any symmetry properties.

10. Jun 1, 2015

### vela

Staff Emeritus
I don't think so. There's absolutely no information about the normal distribution that goes into the algebraic manipulations LCKurtz showed above. The same algebra would have been used if X obeyed a completely different distribution.

11. Jun 1, 2015

### WWGD

What I meant is that the original formula, in the OP is satisfied by the normal, i.e. $, P((|X- \mu|)/\sigma <0.675)$ reflects the symmetry of the normal. Obviously, $|x|< a \rightarrow -a<x<a$ follows straightforward from definition of absolute value. But I guess the OP was referring to the derivation $P(|x- \mu|/\sigma< 0.675 \rightarrow -0.675 \sigma < P(|x-\mu| )< 0.675 \sigma$, so my bad, I jumped in without reading carefully.

12. Jun 1, 2015

### HallsofIvy

It follows from the "symmetry" in the absolute value.

13. Jun 1, 2015

### WWGD

Please check my last post, since I corrected myself. I was referring to something different.

14. Jun 3, 2015

### micromass

Why would that hold? What does $\mathbb{P}(|X-\mu)|)$ even mean???

15. Jun 3, 2015

### WWGD

$\mathb {P}(|x- \mu|)$ doesn't mean anything, what I wrote was $P(|x-\mu|<\sigma (.675))$ , which means $P( -0.675 \sigma < x- \mu < 0.675\sigma )$. Anyway, I jumped in without reading carefully and I wrote something that is pointless here.

All I was trying to say (replying to no one but my own head) was that , by symmetry of the normal, $P( x-\mu< \sigma) =P(x-\mu > -\sigma )$ , which is not true for all distributions.

Last edited: Jun 3, 2015