Linear Regression Point Estimates Suppose we construct the linear relation (using least squares)
$$\text{Weight} = \text{Height}\cdot b + c$$
As I recall from school 30 years ago, Weight is normally distributed with the mean of E(Weight) and Var(Weight).  The model gives us our point estimate of Weight, but what is the estimate of Var(Weight)?
Also, how might you go about extending this to a model with multiple independent variables?
 A: Note that the regression model is a model for the conditional distribution of $y$. So if you're modelling Weight, it doesn't tell you about $E(\text{Weight})$ or $\text{Var}(\text{Weight})$, but the conditional expectation and conditional variance (i.e. $E(\text{Weight}|\text{Height})$ and $\text{Var}(\text{Weight}|\text{Height})$).
If you construct a linear model you estimate variance about the model from the model fit. 
Specifically, the usual estimate of $\sigma^2 = \text{Var}(Y|X=x)$ is 
$$\hat{\sigma}^2=s^2=\frac{1}{n-p}\sum_i(y_i-\hat{y}_i)^2\,.$$ 
In the above equation, $p$ is the number of predictors including the constant, so for simple linear regression, $p=2$. This equation applies just as well to cases where $X$ contains several variables.
The square root of that variance estimate, the residual standard deviation (sometimes called 'residual standard error'), $s$, is available in the output of any decent regression package.
Here's some output from R, for example (I cut out a few extraneous lines):
>  summary(lm(dist~speed,cars))
[...]
Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) -17.5791     6.7584  -2.601   0.0123   
speed         3.9324     0.4155   9.464 1.49e-12 
---    
Residual standard error: 15.38 on 48 degrees of freedom   
Multiple R-squared:  0.6511,    Adjusted R-squared:  0.6438 
F-statistic: 89.57 on 1 and 48 DF,  p-value: 1.49e-12

As we see on the third-last line, $s=15.38$. (The data set cars comes with R.) 
Or we can compute it by hand:
> r <- residuals(lm(dist~speed,cars))
> sqrt(sum(r^2)/48)   # n=50, p=2, so we divide by 48
[1] 15.37959

You can do similar calculations in your favourite stats program, or even something like Excel.
