Standard deviation formula confusion I'm having trouble understanding the formula for standard deviation .
I know how to calculate standard deviation,  but I cant understand some parts of the formula . I'll give you an example 
$$\sigma=\sqrt{\frac 1N \sum_{i=1}^N(x_i-\mu)^2}$$
Say we have a bunch of numbers like $9, 2, 5, 4, 15$
This part of the formula says subtract the Mean and square the result
$(x_i-\mu)^2$
The mean is 7  and when I subtract and square I get
4,  25,  4, 9, 64
This is where I get stuck - I know I have to add up all the values then divide by how many
$\displaystyle \frac 1N \sum_{i=1}^N$
but how does this part of the formula say that? 
I know the sigma means add up 
but what does the $N$ on top of sigma mean?
what does the $i=1$ at the bottom of sigma mean ? 
 A: This might be the kind of operation best explained by an example.  I'll refer you to wikipedia, but also write out an example.  The $i$ is an index variable, $i=1$ down below means you start counting with $i$ at $1$ and the $N$ on top means you stop counting when $i$ reaches $N$.  So for example,
$$\sum_{i = 1}^3 (x_i - \mu)^2 = (x_1 - \mu)^2 + (x_2 - \mu)^2 + (x_3 -\mu)^2$$
In your particular case, the $x_i$ are the sample values you have, $x_1 = 4, x_2 =25$, etc.  Does that make sense?
A: You have $9,2,5,4,15$. There are five numbers here, so $N=5$.
\begin{align}
x_1 & = 9 & & \text{In this case, $i$=1.} \\
x_2 & = 2 & & \text{In this case, $i$=2.} \\
x_3 & = 5 & & \text{In this case, $i$=3.} \\
x_4 & = 4 & & \text{In this case, $i$=4.} \\
x_5 & = 15 & &  \text{In this case, $i=5=N$.}
\end{align}
$$
\overbrace{\frac 1 N\sum_{i=1}^N (x_i-\mu)^2 = \frac 1 5 \sum_{i=1}^5(x_i-\mu)^2}^{\text{This is true because $N=5$.}} 
$$
$$
\frac 1 5 \sum_{i=1}^5(x_i-\mu)^2=\frac 1 5 \Big( (x_1-\mu)^2 + (x_2-\mu)^2 + (x_3-\mu)^2+(x_4-\mu)^2+(x_5-\mu)^2 \Big)
$$
$$
\frac 1 5 \sum_{i=1}^5(x_i-7)^2=\frac 1 5 \Big( (x_1-7)^2 + (x_2-7)^2 + (x_3-7)^2+(x_4-7)^2+(x_5-7)^2 \Big)
$$
$$
\frac 1 5 \sum_{i=1}^5(x_i-7)^2=\frac 1 5 \Big( (9-7)^2 + (2-7)^2 + (5-7)^2+(4-7)^2+(15-7)^2 \Big)
$$
A: The i at the bottom tells you where to start adding up. If the bottom of the sigma has $i=1$, then start adding the stuff to the right of the sigma, but substitute any i's in the equation for 1. After you do that i becomes 2. Add everything to the right, but substitute any i's in the equation with 2. Keep doing that until you reach a new i equal to the top part of the sigma plus one. The top part tells you when to stop adding. In this case its N. So you add until you reach the number of samples you have to add. One more thing. $x_i$ refers to the ith variable that you are considering. If you have a list like $2,3,7$, $x_3$ refers to 7. Keep that in mind when using the sigma.
A: I am not sure if I completely understand your questions but, i = 1 on the bottom and the N on the top of the sum are simply the number of terms that are in the sequence for example: $$\sum_{i = 0}^{3} i = 0 + 1 + 2 + 3 = 6$$ for your equation your given a set of numbers 9, 2, 5, 4, 15 the $x_i$ are these numbers so $x_1 = 9$, $x_2 = 2$, etc. I hope that sort of helps
