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?
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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