This post says

A PDF is used to specify the probability of the random variable falling within a particular range of values, as opposed to taking on any one value.

Is it true?

this is the PDF of the standard normal distribution.

$$varphi(x) = frac{1}{sqrt{2pi}} e^{-x^2/2}$$

plug in x=0 into the formula above, I can get the probability of taking on one value.

Does that post mean the PDF could be used both for point and interval?

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#### Best Answer

The citation is true. When you plug $x=0$ to the PDF function, you do NOT get the probability of taking this particular value. The resulting number is *probability density* which is not a probability. The probability of taking exactly $x=0$ is zero (consider the infinite number of similarly-likely values in the tiny interval $xin[0,10^{-100}]$).

To further convince yourself that this $varphi(x)$ cannot be a probability, consider decreasing the standard deviation of your normal distribution from $sigma = 1$ to $sigma = frac{1}{100}$. Now, $varphi(0)=frac{100}{sqrt{2pi}}$ – much more than one. Not a probability.

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