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Hard to fit the whole question in the title. I'm struggling to intuitively grasp probability density functions, but I hope I'm on to something here.

Is it correct of me to say that, when dealing with normally distributed data, the (y) value resulting from the PDF is equal to the probability of the random variable being within 0.5 standard deviations from the x value?

miroli
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  • The probability density function is not itself a probability. It doesn't have to be $≤1$ for example. If, say, you take the normal with mean $0$ and $\sigma=.1$, then the p.d.f. at $x=0$ has value $3.99$. – lulu Apr 26 '17 at 17:03
  • If it's not a probability, what is it? – miroli Apr 27 '17 at 08:13
  • It's a density. Think about uniform distributions. If you have a uniform distribution on $[0,1]$ then the density is the constant $1$. If you have a uniform distribution on $[0,\frac 12]$ the density is the constant $2$. The connection between the p.d.f. and probability is the same as the connection between ordinary density and mass...you have to integrate over a volume. – lulu Apr 27 '17 at 10:24
  • @trevorDashDash you ask "what is it?" In layman's terms it's a function whose height at any point is proportional to the probability, but every point, being infinitely narrow, has a zero probability. TFor exampl the probabiltiy the temperature is exactly 10 degrees, is zero. But hte probabiltiy it is within half a degree of ten degrees, is non-zero. So to measure a probability we integrate over a range to find the probability that our variable falls within that range. The function represents the height of the range at every point. – it's a hire car baby Jun 26 '17 at 15:00

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Not quite right.

The "$y$ value (or more conventionally, the value of $f(x)$ which in the case of a normal distribution is $ \frac1{\sqrt{2\pi} \sigma} e^{-(x-\mu)^2/2\sigma}) $ is the limit as $h$ approaches zero of $$ \frac{P(x-h < X < x+h)}{2h} $$ where $X$, of course, is a random variate selected with that normal distribution.

Your conjecture has the germ of the right idea, and would be correct if this ratio were the same (as $h\to 0$) as $$ \frac{P(x-\frac\sigma{2} < X < x+\frac\sigma{2} )}{\sigma} $$ but it is not the same.

Mark Fischler
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