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I'm in high school statistics and I'm unsure about distributions. Is the probability that I select an exact value in a distribution equal to zero? For example, with a normal distribution what is this equal to:

$$P(z = 0.6) \quad\quad Z \sim \mathcal N(0,1)$$

If we're looking at an integration definition of the normal density curve, we would have:

$$\int_{0.6}^{0.6} f(x) \mathrm dx = 0$$

Where $f(x)$ is the density curve. But is the probability really $0$ that you select a datum with exactly that specific value from the population? Is it because the normal distribution is smooth and continuous and not of discrete samples?

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  • $\begingroup$ That's exactly right. It should be noted that it's unclear as to whether any real life examples of these models actually exist, or if they are just approximations. But yes, you are technically correct. $\endgroup$ – Don Thousand Sep 13 '18 at 17:22
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Yes, it's zero! A single point has "measure zero" with respect to a continuous real measure space, like the unit interval or the real line.

This concept extends to more dimensions, too. So if you had a target (unit disk) and you wanted to hit an exact bullseye, the probability would be zero. Also, the probability of hitting exactly on a ring is zero, because the circle is thin in the real plane, so it's "measure zero".

The exact mathematical definition uses limits and open sets for the underlying topology of the measure space. Put simply, if the set of points $S$ is the subset of a set of measure $\epsilon$ for arbitrary $\epsilon > 0$, then $S$ is measure zero.

If you want to learn more, this gets into measure theory. Probability spaces are a type of measure space (with the whole space being a finite measure of $1$).

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For the theoretical mathematical object $\mathcal N(0,1)$, your answer is yes, as you showed.

It is a completely different question whether there is anything in the real world corresponding to $\mathcal N(0,1)$.

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