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This is likely a duplicate, but can't find it on MSE.

Let's say I have a normally distributed population with $\mu=2.75$ and $\sigma=0.25$. If $x$ is a value in the population of interest, using the normal probability distribution function I find that

$$P(x=3)\approx 0.96788,$$

and using the normal cumulative probability distribution I find that

$$P(x\le3)\approx 0.84134.$$

At first glance, it seems counterintuitive that a value being equal to 3 is more likely to be chosen from the population than a value being less than or equal to 3.

I do not have a strong statistics background, but I understand that the normal probability density curve is continuous (while the population must be finite and hence discrete), and I suspect this may be the issue with this seeming paradox.

Now my question is:

Can someone give an elementary explanation for why this occurs using simple (practical) language?

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  • $\begingroup$ The first value is not a probability, it is a density. No single value has a positive probability of occurring (for this distribution anyway). The probability of $x$ being in some region is the integral of the density function over that region. $\endgroup$ – lulu Nov 13 '15 at 13:03
  • $\begingroup$ ...And actually, $P(x=3)=0$, which is less than $P(x\le3)$ since $P(x\le3)>0$. $\endgroup$ – Did Nov 13 '15 at 13:06
  • $\begingroup$ @lulu, then what should I take the first value to mean (practically speaking in terms of the population)? According to Wikipedia, the first value "describes the relative likelihood for this random variable to take on a given value." $\endgroup$ – John Molokach Nov 13 '15 at 13:06
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    $\begingroup$ It's not wrong so much as informal. "take on" ought to read "approach" or something like that. Here's something that seems to give a good qualitative discussion: mathinsight.org/probability_density_function_idea (Caveat: I just now found it by an online search. Looks ok on a quick read but I didn't study it carefully). $\endgroup$ – lulu Nov 13 '15 at 13:21
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    $\begingroup$ @JohnMolokach Sorry, I didn't notice you were the author of that answer. With regard to that, I've always stayed away from trying to use the density other than by integrating it. As regards your answer I do think that your reasoning is sound. $\endgroup$ – Mitchell Kaplan Nov 13 '15 at 15:53
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I will assume upvotes validate the accuracy of the following answer:

The first value is not a probability, but a density value, $D_p$ given by

$$D_p=\frac1{\sigma\sqrt{2\pi}}e^\frac{-(x-\mu)^2}{2\sigma^2}.$$

The density for a value $x=a$ should be used to compare whether a value $x_i$ is more likely to be chosen from the population if it is near (within a specified $\varepsilon$) of $a$ as opposed to some other value $b\ne a$.

In this sense, the probability is defined as an accumulation of these densities. Since the densities are distributed on a continuous curve, the accumualtion is done by an integral so that $$P(x_i\le a)=\int_{-\infty}^aD_p\,dx.$$

So for this question, a value near $x=3$ is more likely to be chosen from the population than say $x=2$, and values near the mean value of $x=2.75$ are most likely to be chosen (and this due to symmetry).

It should also be noted, and related to this question, that in the event the normal distribution is used to approximate a discrete distribution, then as the sample size $n\to\infty$, a value $x_i$ near the mean remains most likely to be chosen at random from the population, while $P(x_i)\to 0$ for all such $x_i$ in the population.

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  • $\begingroup$ "I will assume upvotes validate the accuracy of the following answer" Upvotes on MSE? The mathematical accuracy? Certainly not... :-) $\endgroup$ – Did Nov 13 '15 at 17:27
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It is unlikely but could it be the probability of being within 3 standard deviations of the mean ? If you computed statistics you generated using few samples, outliers may have more weight than they should ie 1-96.8% instead of 1-99.7%.

How did you find 96.788% ?

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  • $\begingroup$ normalpdf(3,2.75,0.25) on the TI-83, or equivalently, $$\frac4{\sqrt{2\pi e}}.$$ $\endgroup$ – John Molokach Nov 13 '15 at 13:48

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