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Scores on IIM-CAT
IIM - CAT 2022 shall involve administering a different form of the Test across each of the 4 or more different Test Sessions. In order to ensure fairness and equity in comparison of performances of the candidates across different test sessions, the scores of the candidates shall be subjected to a process of Normalization. The Normalization process to be implemented shall adjust for location and scale differences of score distributions across different forms and the scaled scores obtained by this process shall be converted into percentiles for purposes of shortlisting. For purposes of reporting, Scaled Scores for each Section (Section I: Quantitative Ability and Data Interpretation, Section II: Verbal Ability and Logical Reasoning) and Total along with the Percentile shall be published. The process of Normalization is an established practice for comparing candidate scores across multiple Forms and is similar to those being adopted in other large educational selection tests conducted in India such as Graduate Aptitude Test in Engineering (GATE).
CAT Normalization process is based on the GATE evaluation process in with multiple examinations for different Engineering specializations are conducted. Thus they apply suitable normalization process taking into account all the variations in terms of difficulty level.
Based the GATE Normalization formula, CAT 2014 Normalization formula can be derived as below:
Let Us Suppose the Normalized marks of the mth candidate in the nth slot is (Snm), then
(Snm)= (Sij – Sik)(Sjl – Slk)/(Snm – Slk) + Sik
Normalized marks (Mij) of jth candidate in ith slot, is given by
Here,
Snm = Actual Marks scored by nth candidate in the mth spot
Sij = Average Marks scored by top 0.1% from all the slots
Sik= Sum of mean and SD (Standard Deviation) score of all candidates in all the slots
Sjl = Average score of top 0.1% candidates in nth slot
Slk = Sum of mean and SD of score of all candidates in the nth slot.
Let take an example to understand the formula. Let us suppose the total number of the candidate who took CAT 2014 exam were 2 lakhs
Thus, Total Candidate = 2,00,000
No. of Slot = 4
Candidate in Each Slot = 50,000
0.1% of Candidate in Each Slot = 50
Let us Suppose, Snm = 200 out of 300
Sij = (200 + 198 + 205 + 209)/4 = 203 (Over here we are guessing 200, 198, 205, 209 are the average score of top 0.1% candidate of all 4 slot)
Sik = 125
Sjl = 210
Slk = 128
Thus, according to the formula the Normalized score would be:
(Snm) = (203-125)(202-128)/(210-128)+125 = 195.4