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 ?
Best Answer
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.