Formula for standard deviation So I am studying for a final exam and I know, or thought I knew, the formula for standard deviation. I quadruple checked in my notes and the slides for the lecture and see that the formula for the standard deviation when $\sigma$ is unknown is:
$\sqrt((\Sigma d^2 - nd^-) /  n - 1)$ 
But for some reason on the final review solution sheet our professor somehow and some reason replaced $n$ in the numerator with $1/n$ and I have no clue why. So the formula is now:
$\sqrt((\Sigma d^2 - (1/n)d^-) / n - 1)$
My question is, how come we are able to use $1/n$ instead of $n$ and on the test, how can I tell to use $1/n$ as compared to $n$?
Thanks
 A: Yes your professor has a typo.  The standard deviation for paired samples is usually written as follows:
$$\sigma = \sqrt{\frac{\sum_{i=1}^n(d_i-\bar d)^2}{n-1}}$$
What you professor is doing is using the trick (or arithmetic if you want to think of it that way) that we can rewrite the numerator as
$$\sum_{i=1}^n(d_i-\bar d)^2 = \sum_{i=1}^nd_i^2-n\bar d^2$$
And thus, the formula for the standard deviation should be 
$$\sigma = \sqrt{\frac{\sum_{i=1}^n(d_i-\bar d)^2}{n-1}}= \sqrt{\frac{\sum_{i=1}^n(d_i)^2-n\bar d^2}{n-1}}$$
And finally, it should be $n$ and not $\frac{1}{n}$.
A: Sample variance is defined as
\begin{align}
&\sqrt{\frac{1}{n-1}\sum_{i=1}^n\left(d_i-\overline{d}\right)^2}\\
&=\sqrt{\frac{1}{n-1}\sum_{i=1}^n\left(d_i^2+\overline{d}^2-2d_i\overline{d}\right)} \\
&=\sqrt{\frac{1}{n-1}\left[\sum_{i=1}^nd_i^2+\sum_{i=1}^n\overline{d}^2-\sum_{i=1}^n2d_i\overline{d}\right]}\\
&=\sqrt{\frac{1}{n-1}\left[\sum_{i=1}^nd_i^2+n\overline{d}^2-2\overline{d}\sum_{i=1}^nd_i\right]}\\
&\overset{(a)}{=}\sqrt{\frac{1}{n-1}\left[\sum_{i=1}^nd_i^2+n\overline{d}^2-2n\overline{d}^2\right]}\\
&=\sqrt{\frac{1}{n-1}\left[\sum_{i=1}^nd_i^2-n\overline{d}^2\right]}\\
\end{align}
$(a)$ follows since sample mean $\overline{d}=\frac{1}{n}\sum_{i=1}^nd_i$.
Hence it is $n\overline{d}$ not $(1/n) \overline{d}$
