CAT 2017 NEWS
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The CAT 2017 Exam is set to be conducted on November 26, 2017. Marks will be given according to the marking scheme of CAT 2017. CAT scores will be adjusted and normalized through “equating”. IIM Bangalore will use normalization of scores across for these two sessions of CAT 2017. Normalization of CAT 2017 scores is implemented to adjust for location and different sessions.
Before you start, please note that the exact formula that the IIMs may use for normalizing the CAT 2016 scores is not yet known, and any formula that is mentioned within this article is given solely for the purpose of illustrating the fundamental concepts of normalization and the process that is likely to be used by the IIMs.
The CAT official website says that CAT 2017 will be conducted in two sessions. In two sessions, two different test forms will be administered. Across different test sessions, the scores of the candidates will be subjected to a process of Normalization.
However, these differences in mean and standard deviation (also referred to as differences in location and scale) can be resolved by using different mathematical approaches. The final CAT Result is released after normalization of scores.
The Normalization is a type of percentile equivalence method. For reporting purposes, Scaled scores for each section (Section I: Verbal Ability and Reading Comprehension (VARC), Section II: Data Interpretation and Logical Reasoning (DILR), and Section III: Quantitative Ability (QA)) and Final Scores along with the Percentile scores are published.
A candidate's percentile denotes the percentage of candidates scoring lower than that particular candidate. It is calculated as: Percentile = ( 1 - All India rank( No. of candidates in that subject) x 100%
The CAT score normalization process formula as devised by the experts would be as follows-
Normalized mark (︿Mij) of jth candidate in i th slot, is given by
︿Mij = Mgt - MgqMti - Miq ( Mij - Miq ) + Mgq
Mij is the actual marks obtained by the jth candidate in the ith slot,
Mgt is the average marks of the top 0.1 % candidates in all slots,
Mgq is the sum of mean and standard deviation of marks of all candidates in all slots,
Mti is the average of marks of top 0.1 % candidates in the ith slot,
Miq is the sum of mean and standard deviation of marks of all candidates in the ith slot.
Total Test takers = 2,00,000
slots = 2
Test takers in each slot = 1,00,000 [ .1% of it = 100]
Mij = 170 out of 300 ( is the actual marks obtained by the jth candidate in the 2nd slot)
Mgt = [195 + 190 ]/2 = 192 ( is the average marks of the top 0.1 % candidates of both the slots)
here, 195 and 190 are the average of the top .1% aspirants of all slots
Mgq = 112 (is the sum of mean and standard deviation of marks of all candidates in both the slots)
Mti = 190(is the average of marks of top 0.1 % candidates in the 2nd slot)
Miq = 110(is the sum of mean and standard deviation of marks of all candidates in the 2nd slot.)
So, according to the given formula, normalized score = (192-112)(170-110)/(190-110) + 112 = 172