This seems just like sloppy choice of words. The sentence
Whenever we use a number other than the actual mean to calculate the variance, we will end up with a larger variance
seems to mean the term 'actual mean' to refer to the sample mean. I think this is where your confusion comes from. This is evidenced further from the sentence
the variance and standard deviation that we calculate using the sample mean will probably be smaller than it would have been had we used the population mean.
He apparently uses this intuitive motivation to show us why the sample variance is biased down, when he says
Therefore, when we use the sample mean to generate an estimate of the population variance or standard deviation, we will actually underestimate the size of the true variance
To see why this is true, define a function
$$ f(c) = \frac{1}{n} \sum_{i=1}^{n} (x_{i} - c)^2 $$
Using basic calculus, the minimizer of $f$ satisfies $f'(c) = 0$, which is equivalent to
$$ \sum_{i=1}^{n} (x_i - c) = 0 $$
therefore the minimizer is
$$ c^{\star} = \frac{1}{n} \sum_{i=1}^{n} x_{i}, $$
the arithmetic mean of the $x_{i}$'s. I'll leave it to you to check that this is a minimum and not a maximum.
In the case of data, $X_{1}, ..., X_{n}$ with sample mean $\overline{X}$, $f(\overline{X})$ is exactly the sample variance. Given what we've said above
$$ f(\overline{X}) \leq f(z) $$
for any other $z$, which includes the case where $z = \mu$, the population mean. This is why the sample variance is always less than the mean squared difference from the population mean (except in the case where $\overline{X} = \mu$, which occurs with probability zero when the $X_i$'s come from a continuous distribution).
In that situation the sampled variance should be:
$$\begin{align}S^{2}&=\frac{\sum f\left(X-\overline{X}\right)^{2}}{\left(\sum{f}\right)-1}=\frac{\sum f\left(X-\overline{X}\right)^{2}}{n-1}\\\overline{X}&=\frac{\sum{f\dot{}X}}{\sum{f}}=\frac{\sum{f\dot{}X}}{n}\end{align}$$
And there's no way that could be negative.
So, you have to:
Determine $n = \sum{f}$
Add a column $f\dot{}X$ to your table
Use $\frac{\sum{f\dot{}X}}{n}$ to compute $\overline{X}$
Add a column $\left(X-\overline{X}\right)$
Square it in another column $\left(X-\overline{X}\right)^{2}$
Multiply it by the frequency $f\dot{}\left(X-\overline{X}\right)^{2}$ in another column
Then sum the values of this last column and divide to calculate $S^{2}=\frac{\sum{f\dot{}\left(X-\overline{X}\right)^{2}}}{n-1}$
Best Answer
You are asking about the root mean squared error of the prediction. There is a Wikipedia entry for Mean squared prediction error, which is the analog for the variance. However, RMSEP will be more interpretable (just like the SD is). The calculation is just as you suspect, replacing the estimated sample mean with the true value. Note that RMSEP will include both the SD and the bias (the degree to which the sample mean deviates from the true value). Depending on what you want, it may be useful to keep those two numbers distinct.