Calculating the variance of a population 
The following sample shows 10 different values for the voltage of a
  cable:
57, 64,  68,  66,  54,  60,  48,  46,  75,  62
Find the mean and the variance of the population.

My professor did this


Why didn't he use this formula for the variance?

I suppose that the fact that it asks for the variance of the population and not of the sample probably has something to do with it but I don't understand.
 A: First of all we have 
$$s^2=\frac{\sum\limits_{i=1}^n (x_i-\overline x)^2}{n} =\frac{\sum\limits_{i=1}^n x_i^2-n\cdot \overline x^2}{n} $$
This is the variance of a sample and a $\textrm{biased}$ estimator for the variance of the population. Now we want an $\textrm{unbiased}$ estimator $s_u^2$ for the variance of the population. This is 
$$s_u^2=\frac{\sum\limits_{i=1}^n (x_i-\overline x)^2}{n-1} =\frac{\sum\limits_{i=1}^n x_i^2-n\cdot \overline x^2}{n-1}$$
$s_u^2$ is unbiased since $\mathbb E\left(s_u^2 \right)=\mu$
A: That's because you don't have the population, just a sample with size n = 10. Now, this is important, 'n' is the size of the sample, and 'N' is the size of the population which you don't have. So, since you don't have the population itself you can't use the $\sigma^2$ formula because it requires some informations of the population as 'N' to work. Most of the time we don't have the information of the population itself, so we need to work with a sample to estimate parameters of that population, and the best estimators for the mean of the population is $\bar{X}$ and for the variance is $S^2$, because they attend some criteria that good estimators have, some examples of those criterias are unbiasnedness, consistency, efficience, minimium variance.
