Is a Bell Curve Discrete or Continuous
Here is a distinction between discrete and continuous random variables that I think you may find useful.
Discrete random variable: Suppose a random variable can take integer values $0, 1, 2, \dots, 10.$ A random variable $X \sim \mathsf{Binom}(10, p)$ is an example. Then we specify the distribution of $X$ by somehow giving the probability that $X$ takes each of these possible values. This can be done by making a table, making a bar chart, or giving an equation. If $p = 1/2,$ then the equation is $$P(X = k) = {10 \choose k}(1/2)^n,$$ for $k = 0, 1, \dots, 10.$
In R statistical software, I can make a table of probabilities correct to ten decimal places as follows (you can ignore the row numbers in brackets [ ].)
k = 0:10; PDF = dbinom(k, 10, .5) cbind(k, PDF) k PDF [1,] 0 0.0009765625 [2,] 1 0.0097656250 [3,] 2 0.0439453125 [4,] 3 0.1171875000 [5,] 4 0.2050781250 [6,] 5 0.2460937500 [7,] 6 0.2050781250 [8,] 7 0.1171875000 [9,] 8 0.0439453125 [10,] 9 0.0097656250 [11,] 10 0.0009765625 And here is a bar chart:
Whether we give the probabilities by formula, table, or barchart, the idea is the same: In order to specify the distribution of a discrete random variable, we need to give the probability of each possible point value.
Continuous random variable. A beta random variable can take any value in the interval $(0, 1),$ which is called the support $S_X$ of the random variable. We use a density function to specify the distribution of a continuous random variable. A density function $f(x)$ has three rules:
(a) $f(x) \ge 0,$ for all $x \in S_X.$
(b) The area under the the density function over $S_X$ must be $1.$ Using an integral, this can be stated as $\int_{S_X} f(x)\, dx = 1.$ For a beta distribution in particular it's $\int_0^1 f(x)\, dx = 1.$
(c) The probability $P(a < X \le b)$ for $a < b; a,b \in S_X$ is the area under the density function over the interval $(a,b].$ In terms of an integral, this is can be written as $\int_a^b f(x)\, dx.$
This means that we define probabilities only for intervals and the probability $P(X = a) = 0$ for any one point $a.$ Hence, intervals $(a,b), [a,b), (a,b],$ and $[a,b]$ (where $a < b)$ all have the same probability.
For example, the density function of the distribution $\mathsf{Beta}(2, 1)$ is $f(x) = 2x,$ for $0 < x < 1.$ Clearly, this definition satisfies condition (a), and a simple integration (or some elementary school geometry) shows that $\int_0^1 f(x)\,dx = \int_0^1 2x\,dx = 1,$ so that (b) is satisfied.
We can use condition (c) to find $P(0.1 < X < 0.6) = \int_{.1}^{.6} 2x \, dx = 0.35$.
Here is a graph of the density function. The area between the vertical dotted lines is $P(0.1 < X < 0.6).$
Note: For some important continuous distributions, the integration in (c) has to be done by numerical methods instead of using calculus. The normal distribution is one of those distributions, which is why we use printed normal tables, statistical calculators or software for finding areas under a normal curve.
Distinctions. Colloquially speaking, discrete distributions 'live on' a finite or countable number of discrete points. For $X \sim \mathsf{Binom}(10, .5),$ if you want $P(3 \le X \le 5) = 0.5683594.$ then you have to get the probability of this 'interval' by adding the probabilities of the three relevant points (3, 4, and 5).
By contrast, continuous distributions 'live on' intervals. If $Y \sim \mathsf{Norm}(70, 7)$ and you ask for $P(Y = 77),$ then the technical answer is $0.$ If this distribution is intended to describe the masses (in kg) of collegiate make swimmers, then maybe by $P(Y = 77)$ you really mean $P(76.5 < Y \le 77.5).$ Now, that is an interval and you can use normal tables or software to find $P(76.5 < Y \le 77.5) = 0.0346.$
diff(pnorm(c(76.5, 77.5), 70, 7)) [1] 0.03456723 harrisonwharyince.blogspot.com
Source: https://math.stackexchange.com/questions/3174171/why-do-we-associate-discrete-values-to-data-from-a-continuous-distribution
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