How 'rare' is no-slipping? I have come across a lot of questions that say something like:

A ball rolls down a hill without slipping...

But I have done the maths and found that a ball would only 'not slip' if the friction is precisely:
$$mgI\sin\theta \over mR^2 +I$$
Where $I$ is the moment of inertia, and $\theta$ is the angle made to the horizontal by the plane. A friction of this value would never appear exactly which makes me thing that no-slipping is an idealist case. So here is my question:
Am I right in saying that no-slipping (i.e. when $\dot x =R \dot \theta$) is very uncommon and to have no-slipping on an incline plane you must have friction of exactly the value I have given? 
 A: I think you have misunderstood how friction works here. The friction you have written down is (typically) the minimum friction needed for the 'non slip' to occur. 
Imagine a very faint slope. You will only need a small amount of friction to avoid slipping. As you make the slope steeper more friction is needed for the 'non slip'. Eventually the friction will not be enough and the ball will begin to slip (as well as roll).
So, no, non-slipping is not as idealized as you suggest.
A: When a tire rolls without slipping it does not leave skid marks.
A ball sitting on a level surface has a force mg downward. It does not move because the surface exerts an upward reaction force just strong enough to hold it still. 
Friction is like this. When a ball rolls down a hill, the sum of the gravity and reaction forces is parallel to the surface in a downhill direction. The ball accelerates in that direction. Friction acts on the contact point on the bottom. Friction exerts a backward force just strong enough to keep the contant point still.
