Is the Standard Deviation Symmetrically Centered?

[jk22]

Suppose i have a random variable X and its standard deviation dx. We could write the average with error like $$<X>\pm dx/2$$.

But how do we know it is centered or not ? It could be +1/4 -3/4 for example.

 

[mfb]

If you just know the standard deviation, you cannot know if the uncertainty is symmetric. If it is (significantly), which is an unlikely case, more information should be given, at least +x-y separately, but ideally the full likelihood distribution.

 

[jk22]

Could we compute $$\int^{<X>}x^2P (X=x)dx$$ for the minus sign and above the average for the plus sign ?

 

[mfb]

That would be odd. Better use confidence intervals.

 

[Dale]

Could we compute $$\int^{<X>}x^2P (X=x)dx$$ for the minus sign and above the average for the plus sign ?

The thing you should compute is called skewness:

https://en.m.wikipedia.org/wiki/Skewness#Definition

 

[Hornbein]

Suppose i have a random variable X and its standard deviation dx. We could write the average with error like $$<X>\pm dx/2$$.

But how do we know it is centered or not ? It could be +1/4 -3/4 for example.

You could look at the data and see whether it looks centered. Or you could know something about the situation that tells you whether to expect it to be centered or not.

Confidence intervals can be any shape you desire. You just have to have a good reason to choose such a shape.

Many people just assume everything is normally distributed. Sometimes it is, sometimes it isn't.