What's the difference between the mean and expected value of a normal distribution? My question might be a bit dumb but I'm confused so I'd like it if someone could clear this up for me. I've always thought that the mean of the normal distribution is equal to the expected value of the distribution but I came across an excercice in which the standarlized normal distrubtion of the sum follows chi squared of $n$ parameters ( we standarlized here by writing $\frac{X-E(X)}{\sigma}$ ).However if we standarlize with the mean $\nu=\frac{1}{n}\Sigma_{i=1}^nX_i$ we get chi squared of $n-1$ parameters ( because of the dependency ). This example made me notice that I have a misunderstanding of the concepts, so my question is what exactly is the difference between the expected value of a normal distrubtion and the mean ??
 A: The expected value of a statistic is defined as:
$$E[X] = \mu = \int_{\mathbb R}xf(x)dx$$
Where $f$ is the probability density function of $x$.
So you can see that it's simply a weighted mean.
But it's also calculated over all possible values, when you have the pdf.
Usually, this quantity is unknown, but we can estimate it.
The sample mean is an unbiased estimator of the expected value.
$$\bar x = \hat \mu = \sum_i^n x_i/n$$
Both $\mu$ and $\hat \mu$ can be described as expected values and also means, so, to answer your question, there is no difference in terminology: an expected value is a mean.
The reason you lost one degree of freedom in that Chi-squared calculation is because you used $\hat \mu$, not $\mu$. $\mu$ is an underlying, fixed, quantity, it does not depend on the sample at all. $\hat \mu$, on the other hand, is explicitly a function of the sample $x_i$s. If you use $\hat \mu$ to describe the sample in any way, then only $n-1$ degrees of freedom remain, because, if you know the values of the other $n-1$ $x_i$s, then you can calculate exactly the remaining one:
$$x_j = \bar x - \sum_i^n (1-\delta_{ij})x_i/n$$
