Difference between Z notation and sigma

• Ry122
In summary: 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.
Ry122

Homework Statement

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

Ry122 said:

Homework Statement

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

The Attempt at a Solution

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?

is that due to symmetry?

Ry122 said:
is that due to symmetry?

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##.

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 $Z = (X - \mu)/\sigma \sim N(0,1)$.

what is function N()?

Ry122 said:
what is function N()?

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").

LCKurtz said:
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##.

True, but doesn't the symmetry in the algebra follow from the symmetry of the normal distribution?

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WWGD said:
True, but doesn't the symmetry in the algebra follows from the symmetry of the normal distribution?

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.

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.

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.

WWGD said:
True, but doesn't the symmetry in the algebra follow from the symmetry of the normal distribution?
It follows from the "symmetry" in the absolute value.

HallsofIvy said:
It follows from the "symmetry" in the absolute value.

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

WWGD said:
##P(|x- \mu|/\sigma< 0.675 \rightarrow -0.675 \sigma < P(|x-\mu| )< 0.675 \sigma ##

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

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

##\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:

What is Z notation and sigma?

Z notation and sigma are two mathematical notations used for specifying and representing formal specifications of software systems. They are used to describe the behavior and properties of a system and are commonly used in software engineering and computer science.

What are the main differences between Z notation and sigma?

The main difference between Z notation and sigma is their underlying mathematical foundations. Z notation is based on set theory and first-order logic, while sigma is based on predicate calculus. Z notation is also more expressive and allows for more precise specifications compared to sigma.

Which notation is more commonly used in software engineering?

Z notation is more commonly used in software engineering due to its ability to provide more detailed and precise specifications. It also has a larger community and more tools and resources available for its use.

Can Z notation and sigma be used together?

Yes, Z notation and sigma can be used together. In fact, many software engineers use both notations in combination to provide comprehensive specifications for their systems.

Which notation should I use for my software specifications?

The choice between Z notation and sigma depends on the level of detail and precision required for your specifications. If you need to provide a more detailed and formal specification, Z notation would be a better choice. However, if your specifications are less complex, sigma may be sufficient. It is also important to consider the resources and tools available for each notation before making a decision.

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