To answer this type of question we need to know the width of the airway and then apply the 1 in 60 rule. The JAR/EASA version of this questions specifies an airway width of 10 nm.
With 10 nm airway width when the aircraft is on the boundary of the airway it is 5 nm off the centerline.
The 1 in 60 rule states that for each degree of error the cross track error will be 1/60 of the range.
This can be stated as:
Cross Track Error = 1/60 x Error angle x Range
Rearranging this gives
Error angle = Cross Track Error x 60 / Range
Inserting the data provided in the question gives the following:
Error angle = 5 nm x 60 / 100 nm = 3 degrees
Using 2 degrees per dot 3 degree error produces a 1.5 dot indication.
Using a different airway width will of course give a different answer.
Last edited by keith williams; 15th Jul 2012 at 22:58.