P(x=mean) of normal PDF with low sigma - not allowed?

[nomadreid]

P(x=mean) of normal PDF with low sigma -- not allowed?

About the normal probability distribution: with formula
P(X=x) = (1/(σ*sqrt(2π))*exp(-(x-μ)²/2σ²), what happens when you look at P(X=μ) if σ<(1/sqrt(2π))? You get P(X=μ)>1, an absurdity. What is going on?

Second question is one about intuition: suppose μ=0, then why would a scale change of the horizontal axis (say by changing units from meters to kilometers) , which would also change σ, affect the probability of the mean, which it would by the formula?

 

[jk22]

the probability is 0 since what you looked at is a probability density p(X in x,x+dx) equals f(x)dx, where f is your gaussian function

 

[HallsofIvy]

For any continuous PDF, the probability that x is equal to any specific value, rather than in a given interval or set, is 0.

 

[nomadreid]

Thanks, jk22 and HallsofIvy. My reaction upon reading your replies and thinking for a couple of seconds was, "Of course. Stupid of me." So thanks for that destruction of the mental block.

 

[CompuChip]

What you can calculate is
P(X \le \mu) = \int_{-\infty}^{\mu} p(x) \, dx
where p(x) the PDF. Of course, this will give you 1/2 independent of the mean or standard deviation.