Mean and variance after subtracting mean and dividing by standard deviation 
That is mean, variance and standard deviation.
My question is how to get from point 4 to 5 and also with the variance from point 6 to 7. 
 A: These equations represent a particularly obscure way to make some important points that everybody ought to understand.  I will therefore provide an indirect answer by highlighting the fundamentals (1-4 below), demonstrating them, and then applying them in what amounts to an equivalent proof.


*

*When you add a constant $a$ to all data $x_i$, the mean of the new values is $a$ plus the mean of the old values.  This should be obvious, because adding $a$ to each of $n$ values adds $na$ to the sum.  When the sum is divided by $n$ to get the mean, $na$ is divided by $n$ to show $na/n=a$ is added to the sum.

*When you multiply each $x_i$ by a constant $b$, the mean of the new values is $b$ times the original mean.  This truly is obvious (it's a direct application of distributive and commutative laws of arithmetic).

*When you add a constant $a$ to all data, the variance is unchanged.  This is because the variance is the average of the squared residuals, $(x_i-\bar x )^2$.  By (1), $\bar x$ increases by $a$ and that exactly cancels the addition of $a$ to each $x_i$, whence the residuals are unchanged.  Consequently the variance is unchanged.

*When you multiply all data by a constant $b$, the variance is multiplied by $b^2$.  Since (3) tells us each $x_i$ as well as their mean $\bar x$ are multiplied by $b$, the residuals $x_i - \bar x$ are also multiplied by $b$.  Consequently the squared residuals are multiplied by $b^2$ and so (exactly as in (2)) the mean squared residual is multiplied by $b^2$.
The equations in the question attempt to demonstrate that the mean and variance of $z_i$ are zero and one, respectively, when the $z_i$ are formed by standardizing the data: that is, $-\bar x$ is first added to the data (giving the residuals) and those results are divided by the square root of the variance. Call the square root $s$, so the variance is $s^2$.
Here, then, is an alternative to the equations in the question: 
By (1), the mean after the first step is $\bar x - \bar x = 0$. 
By (2), the mean remains zero upon division by the square root of the variance.  (This should remind you of step "5" in the question.)
By (3), the variance is unchanged after the first step.  
By (4), the variance $s^2$ is divided by the square of $s$ in the second step: but that just divides the variance by itself (step "7" in the question), giving $s^2/s^2=1$, QED.
A: I'm not sure if this is what you wanted, but I'll give it a shot.
First, we declare these known relationships: 
$\bar{x}=\dfrac{\sum_{i=1}^n x_i}{n}$
$\sum_{i=1}^n \bar{x}=n\bar{x}$
$s_x^2=\dfrac{\sum_{i=1}^{n} (x_1-\bar{x})^2}{n}$
So, starting with (4):
$\dfrac{\sum_{i=1}^n x_i}{ns_x} - \dfrac{\sum_{i=1}^n \bar{x}}{ns_x} = \left(\dfrac{\sum_{i=1}^n x_i}{n} \cdot \dfrac{1}{s_x} \right) - \dfrac{n\bar{x}}{ns_x} = \left( \bar{x} \cdot \dfrac{1}{s_x} \right) - \dfrac{n\bar{x}}{ns_x} = \dfrac{\bar{x}}{s_x} - \dfrac{n\bar{x}}{ns_x}$. 
We get (5). QED.
Starting with (6):
$\dfrac{\sum_{i=1}^{n} (x_1-\bar{x})^2}{ns_x^2} = \left( \dfrac{\sum_{i=1}^{n} (x_1-\bar{x})^2}{n} \cdot \dfrac{1}{s_x^2} \right) = \left( s_x^2 \cdot \dfrac{1}{s_x^2} \right) = \dfrac{s_x^2}{s_x^2}$. 
We get (7). QED.
