Process for Calculation of Dux and First Place in Years 11 and 12.
A students top five subjects will be considered to determine the highest academic performing students at each year level. For a subject to be considered it must be at the appropriate year level and also must include at least four achievement standards. Unit standards grades are not considered in the calculations to determine the highest performing academic students. Subjects that are at the appropriate level and were studied in previous years may also be included in a student’s top five subjects.
The following process is used to calculate a numerical value to rank student academic performance:
NOTE – all achievement standards must be considered in the process:
Where
5 = Excellent
3 = Merit
2 = Achieved
0 = Not Achieved
E.g. Calculus has one internal worth four credits and through practice externals an estimated grade has been derived for three externals that are worth a total of 17 credits.
If a student achieved excellence in all four standards, then total number of credits (21) is multiplied by the excellence gdn. (21*5=105).
E.g. One student gets E on Internal (AS 90637) two Merits (AS 90635 worth 6 credits) and (AS 90636 worth 6 credits) and 1 Achieved (AS 90638 worth 5 credits) on practice externals.
E = 4 * 5
M = 6 * 3
M = 6 * 3
A = 5 * 2
Total = 20 +18 +18+ 10 = 66
This is out of a possible 105 (see above).
So student has got 66/105 = 0.6286
I need to see the following information from all teachers of seniors and for all senior students they teach.
If some students’ grades look like they are missing, Inquiries will be made with Teachers about clarification of student’s grades.
SLT and Senior Academic Dean will determine 1st Place in Year and DUX with this information. Discretion of Principal might need to be used. In a situation where students are equal, consideration will take place where credits were gained first or as a resubmission / resit.
Name of AS 90637 AS 90636 AS 90637 AS 90638 Students Total Subject
Student (I) 4 credits (E) 6 credits (E) 6 credits (E) 5 credits Total Available Calculated Value
JOE E=20 M= 18 M= 18 A=10 66 105 .6286
ERANA E=20 E= 30 M=18 E =25 93 105 .8857
NEVILLE E=20 M =18 E=30 M = 15 83 105 .7905
SOPHIE N=0 E=30 E=30 E=25 85 105 .8095
POSSIBLE SUBJECT TOTAL CALCULATION
90637 = 4 * 5 = 20
90636 = 6 * 5 = 30
90637 = 6 * 5 = 30
90638 = 5 * 5 = 25
Total possible = 105
Sophie was a late enrolment. In this case we would have to organise for her to do internal assessment.
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