Normalization puts premium on Board results, makes JEE (Main) redundant Engineering Entrance
The normalization process worked out for admission to National Institutes of Technology (NITs) and other central government technical educational institutions is turning out to be more complicated and cumbersome than what the CBSE had claimed.
The way normalization is being worked out it could result in a student with high score in JEE (Main) and not so low percentage in Class XII (say 90%) could have a rank much lower than a friend who scored less in JEE(Main) but had a higher percentage in the Board results.
For admission, it was decidedto use the JEE (Main) performance and the normalized Board performance in the 60:40 ratio. But after the JEE (Main)results were out, the JEE Interface Group has come up with the novel normalization formula, which is as follows: C= 0.6 X Ao + 0.4 X Bfinal, where (Ao) component is aggregate marks obtained byeach student in JEE (Main) and Bfinal component corresponds to the Board percentile. The final rank of the student in the JEE (Main) will be decided by C.
The normalization process decided by the Joint EntranceExamination (JEE) Interface Group for the Class XII markscomponent is as follows: 50% of Board marks be normalized by equating percentile among different Boards and anchoring them to All India JEE (Main) percentiles, and 50% be normalized by equating each Board's percentile with JEE (Main) percentile marks of respective Boards.
Many students and parents are pointing out that the normalization system is not working out. The moot point is if a student scores 95%, just 2% more than his friend who gets 93% in the Board it would mean he has scored 10extra marks (CBSE: 5 subjects each for maximum marks 100). But this two per cent marks will be resulting into a big difference in JEE marks mapping, which won't be uniformly distributed.
For instance, a student Tom gets first rank in JEE (Main). He scored 345/360. Tom alsoscored 90% in CBSE class XII examination. Let's make a fairassumption that these marks will probably correspond to about 93 percentile in the boards. Now, for the calculation of 40% equivalent from Boards: If there are 12 lakh people giving JEE (Main),this person will be allotted marks equivalent to 93 percentile of the JEE (Main) ranks. The official cut-off declared for JEE (Advanced) is 113 with a rank of 75,000 for general category. Probably this year 93 percentile in JEE (Main) wouldcorrespond to about 84,000 rank (7% of 12 lakh applicants) and corresponding marks would be about 113 only.
So Tom will get B final component (B1). JEE (Main) aggregate marks corresponding to percentile at the All India level) + B2: (JEE-Main aggregate marks corresponding to percentile among the set of aggregate scores obtained in the JEE (Main) by the students of thatboard) in the formula approximately as 0.4X113=45.2 marks. (This is on the assumption that B1 and B2 components remain almost same in the case of CBSE). So total marks of Tom will be C =0.6x345+0.4x11 0=207+45.2=262. 2
Now if a student Peter gets 200 marks in JEE (Main) and 97.5% in boards then probably he is at 99.98 percentile and correspondingto this he will get 330 marks in JEE (Main). (If we map 99.98percentile to JEE main percentile then marks for B-final component would be 330. His B final component willbe 0.4x330=132. So total marks of Peter will be C=0.6x200+0.4x33 0=120+132=252.
The error is self-evident. Tom with high JEE (Main) score but slightly less percentage in CBSE gets only45.2 marks in the Board component whereas Peter with 200 marks in JEE (Main) but 97.5% in CBSE Board gets132 in the Board component, a massive difference of 86.8 marks. Someone gets 90% in CBSE Board and ranks first inJEE (Main) might not even figure in top final ranking of NITs and other central government educational institutions. "This is just stupid," a parent says.
The way normalization is being worked out it could result in a student with high score in JEE (Main) and not so low percentage in Class XII (say 90%) could have a rank much lower than a friend who scored less in JEE(Main) but had a higher percentage in the Board results.
For admission, it was decidedto use the JEE (Main) performance and the normalized Board performance in the 60:40 ratio. But after the JEE (Main)results were out, the JEE Interface Group has come up with the novel normalization formula, which is as follows: C= 0.6 X Ao + 0.4 X Bfinal, where (Ao) component is aggregate marks obtained byeach student in JEE (Main) and Bfinal component corresponds to the Board percentile. The final rank of the student in the JEE (Main) will be decided by C.
The normalization process decided by the Joint EntranceExamination (JEE) Interface Group for the Class XII markscomponent is as follows: 50% of Board marks be normalized by equating percentile among different Boards and anchoring them to All India JEE (Main) percentiles, and 50% be normalized by equating each Board's percentile with JEE (Main) percentile marks of respective Boards.
Many students and parents are pointing out that the normalization system is not working out. The moot point is if a student scores 95%, just 2% more than his friend who gets 93% in the Board it would mean he has scored 10extra marks (CBSE: 5 subjects each for maximum marks 100). But this two per cent marks will be resulting into a big difference in JEE marks mapping, which won't be uniformly distributed.
For instance, a student Tom gets first rank in JEE (Main). He scored 345/360. Tom alsoscored 90% in CBSE class XII examination. Let's make a fairassumption that these marks will probably correspond to about 93 percentile in the boards. Now, for the calculation of 40% equivalent from Boards: If there are 12 lakh people giving JEE (Main),this person will be allotted marks equivalent to 93 percentile of the JEE (Main) ranks. The official cut-off declared for JEE (Advanced) is 113 with a rank of 75,000 for general category. Probably this year 93 percentile in JEE (Main) wouldcorrespond to about 84,000 rank (7% of 12 lakh applicants) and corresponding marks would be about 113 only.
So Tom will get B final component (B1). JEE (Main) aggregate marks corresponding to percentile at the All India level) + B2: (JEE-Main aggregate marks corresponding to percentile among the set of aggregate scores obtained in the JEE (Main) by the students of thatboard) in the formula approximately as 0.4X113=45.2 marks. (This is on the assumption that B1 and B2 components remain almost same in the case of CBSE). So total marks of Tom will be C =0.6x345+0.4x11 0=207+45.2=262. 2
Now if a student Peter gets 200 marks in JEE (Main) and 97.5% in boards then probably he is at 99.98 percentile and correspondingto this he will get 330 marks in JEE (Main). (If we map 99.98percentile to JEE main percentile then marks for B-final component would be 330. His B final component willbe 0.4x330=132. So total marks of Peter will be C=0.6x200+0.4x33 0=120+132=252.
The error is self-evident. Tom with high JEE (Main) score but slightly less percentage in CBSE gets only45.2 marks in the Board component whereas Peter with 200 marks in JEE (Main) but 97.5% in CBSE Board gets132 in the Board component, a massive difference of 86.8 marks. Someone gets 90% in CBSE Board and ranks first inJEE (Main) might not even figure in top final ranking of NITs and other central government educational institutions. "This is just stupid," a parent says.
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