A discrete version of the normal distribution

• A
In summary, there is a function for the normal distribution that is given by a certain formula. The question is how the integrals of this function can be equal to their corresponding sums. It turns out that this is a well-known result, where the normal distribution is the limit of a binomial distribution as the number of trials approaches infinity. This has been discussed in a previous post on a different platform.
TL;DR Summary
How can the result of an integral of a normal distribution be the same as the result of a sum?
I have the following function for the normal distribution:
$$\displaystyle f \left(x \right) = \frac{1}{2}~\frac{\sqrt{2}~e^{-\frac{1}{2}~\frac{\left(x -\mu \right)^{2}}{\sigma ^{2}}}}{\sigma ~\sqrt{\pi }}$$
How can the following integrals be equal to their sums?
$$\displaystyle \int_{-\infty }^{\infty }\!1/2\,{\frac { \sqrt{2}}{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}\,{\rm d}x=\sum _{x=-\infty }^{\infty }1/2\,{\frac { \sqrt{2}}{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}},$$
$$\displaystyle \int_{-\infty }^{\infty }\!1/2\,{\frac {x \sqrt{2}}{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}\,{\rm d}x=\sum _{x=-\infty }^{\infty }1/2\,{\frac {x \sqrt{2}}{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}$$
and
$$\displaystyle \int_{-\infty }^{\infty }\!1/2\,{\frac { \left( x-\mu \right) ^{2} \sqrt{2} }{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}\,{\rm d}x=\sum _{x=-\infty }^{\infty }1/2\,{\frac { \left( x-\mu \right) ^{2} \sqrt{2}}{\sigma\, \sqrt{\pi } }{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}$$

Last edited:
Looks like the integral and first two moments of a normal distribution.

What is a discrete version of the normal distribution?

A discrete version of the normal distribution is a mathematical model used to describe the probability distribution of a discrete random variable. It is based on the continuous normal distribution, but instead of having an infinite number of possible outcomes, it only has a finite number of possible outcomes.

How is a discrete version of the normal distribution different from the continuous version?

The main difference between the two is that the discrete version only has a finite number of possible outcomes, while the continuous version has an infinite number of possible outcomes. Additionally, the probabilities in the discrete version are represented by a probability mass function, while in the continuous version they are represented by a probability density function.

What are the applications of a discrete version of the normal distribution?

A discrete version of the normal distribution is commonly used in various fields, such as statistics, economics, and engineering, to model the probability of discrete events. It is also used in quality control and process control to monitor and improve processes.

How is a discrete version of the normal distribution calculated?

The calculation of a discrete version of the normal distribution involves determining the mean and standard deviation of the data, and then using these values to calculate the probabilities of each possible outcome using the probability mass function formula. These probabilities can then be used to create a probability distribution table or graph.

What are the limitations of a discrete version of the normal distribution?

One limitation is that it can only be used for discrete random variables, meaning it cannot be applied to continuous data. Additionally, it assumes that the data is normally distributed, which may not always be the case in real-world scenarios. It also does not account for outliers or extreme values in the data.

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