Originally posted by pcx_demolition017
last [year] an ext 1 student had an internal assessment of around 30% and yet gained an aligned mark of 90 (or 45). the difference between him and the top guy was apparently around 50-60%. can this actually happen, coz it just doesn't seem likely.
You people can really make your figures confusing sometimes. I'll just convert everything to percentages and work from that, okay?
At first glance, the scenario seems possible. Let's assume that it is, and for the sake of simplicity, that there were only three students in the class - the third student was ranked last (with an unknown mark).
So:
[1] Joe Bloggs was ranked 2/3 with an internal mark of 30%.
[2] The top guy was ranked 1/3 with an internal mark of 80%.
[3] Joe ended up with a moderated mark of 90%.
We can deduce that:
[4] The difference between the internal marks of Joe and the top guy was 50% (from [1] and [2]).
[5] The bottom internal mark was less than or equal to 30% (from [1]).
[6] The top exam mark was above 90% (from [3]) - and probably far above, so let's say 100% (for simplicity).
We can also deduce from those deductions that:
[7] The difference between the moderated marks of Joe and the top guy was equal to 10% (from [3] and [6]).
Finally, we define some variables:
B<sub>i</sub> = the bottom internal mark
B<sub>e</sub> = the bottom exam mark
J<sub>e</sub> = Joe's exam mark
D<sub>i</sub> = the difference between Joe's internal mark and B<sub>i</sub>
D<sub>i</sub> = 30% - B<sub>i</sub>
D<sub>e</sub> = the difference between Joe's exam mark and B<sub>e</sub>
D<sub>e</sub> = J<sub>e</sub> - B<sub>e</sub>
Now, the moderating procedure should generally preserve the relative differences between students. Therefore in order for the 50% difference between Joe and the top guy to be compressed to a difference of 10%, D<sub>i</sub> must have been compressed by the same factor to become D<sub>e</sub>. Other factors would fail to preserve the relative differences between students.
i.e. -
50/10 = D<sub>i</sub> / D<sub>e</sub>
5 = D<sub>i</sub> / D<sub>e</sub>
D<sub>e</sub> = D<sub>i</sub> / 5
Substituting for D<sub>e</sub>:
J<sub>e</sub> - B<sub>e</sub> = D<sub>i</sub> / 5
J<sub>e</sub> = (D<sub>i</sub> / 5) + B<sub>e</sub>
Substituting for D<sub>i</sub>:
J<sub>e</sub> = [(30% - B<sub>i</sub>) / 5] + B<sub>e</sub>
Now, we want information regarding the situation where J<sub>e</sub> = 90%.
Substituting, simplifying and rearranging the above:
B<sub>e</sub> = (420 + B<sub>i</sub>) / 5
B<sub>e</sub> = 78 + 0.2 * B<sub>i</sub>
We impose the following constraint (from [5]):
0 ≤ B<sub>i</sub> ≤ 30%
We can use this to define a valid range for B<sub>e</sub>:
78% ≤ B<sub>e</sub> ≤ 84%
So, given that the bottom internal mark falls within the range (0%, 30%), Joe will receive a moderated mark of 90 where the bottom exam mark falls within the range (78%, 84%), and both marks satisfy the equation B<sub>e</sub> = 78 + 0.2 * B<sub>i</sub>.
I'm sorry for including all of that maths. I got a bit carried away. I (effectively) finished my uni exams today, and had far too much spare time. I think I'll go out.
In short... yes.
The bottom exam mark just needed to be 78 more than one fifth of the bottom internal mark. The class of three students is really a special case (and could have been determined much more simply, but derivations are more enlightening), but it's possible. A number of other factors can also come into play to limit those ranges further (e.g. enforcing the requirement that the conversion curve is monotonic over the range of the internal marks - if a stationary point occurs in that range, student's ranks would be changed).
You could have also tested it yourself (
www.boredofstudies.org/moderate.php).