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The probability density of the normal distribution is : $$ f(x) = \frac{1}{\sqrt{2\pi}\sigma} \cdot e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}$$

And for a random variable $X$ such that : $X \sim \mathcal N (0, 1)$ then we have : $$f(x) = \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{1}{2}x^2}$$

Yet what is the intuition behind this formula ? Why is there the number $\pi$ and $e$ in the formula of the probability density of the normal distribution ?

Moreover, in a lot of my exercices on the normal distribution it's always saying at the beginning of the exercice :

Let $X$ be a random variable that follow a normal distribution.

Yet, how can we know that a random variable follows a normal distribution ? For example let's say we are studying some process, and more spefically the behaviour of a random variable. Then how can we know, and from which properties of this random variable we can say that it follows a normal distribution ?

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  • $\begingroup$ The number $\pi$ is there just to make sure that you have a valid probability function. The number $e$ can be viewed as the result of taking limit of some probability function, for example if the sample size in binominal distribution goes to infinity, its limiting probability is a normal pdf. Central Limit Theorem is probably the most important reason why normal distribution is so ubiquitous. It also has a lot of nice properties and is convenient for the computational (thus pedagogical) purpose. $\endgroup$
    – SiXUlm
    May 19, 2017 at 21:15

2 Answers 2

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One reason that random variables are often (approximately) normal is because of the Central Limit Theorem. Essentially, a random quantity is likely to have approximately a normal distribution if its deviations from the average are relatively small and are the cumulative result of many independent influences, each of which is very small.

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Robert's answer is good! And I'll provide an intuition about how we arrive at this distribution (your first question):

We could view normal distribution is the limit case of a binomial distribution as sample size $N$ goes to $\infty$

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