New Delhi:The normalization process worked out for admission to the National Institutes of Technology (NITs) and other central government technical educational institutions is turning out to be more complicated and cumbersome than what the HRD ministry and CBSE had thought.
Normalization could result in a student with high score in JEE (Main) and not so low percentage in Class XII (say 90%) getting a rank much lower than his friend who scored less in JEE (Main) but had a higher percentage in the Board results. Now, many more students and parents are thinking of moving court. A parent from NIT, Kurukshetra, said, “The normalization formula is skewed. For admission, it was decided to 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 normalization formula which would put many students on disadvantage.” The formula is C = 0.6 X Ao + 0.4 X B final, where (Ao) component is aggregate marks obtained by each student in JEE (Main) and B final component corresponds to the Board percentile. The final rank of the student in the JEE (Main) will be decided by C.
The normalization process for the Class XII marks component is 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 are pointing out that the normalization system is not working out. The moot point is if a student scores 95%, 2% more than his friend who gets 93% in the Board it would mean he has scored 10 additional marks. But this 2% will result in a big difference in JEE marks mapping.
For instance, Tom, gets first rank in JEE (Main). He scored 345/360. Tom scored 90% in CBSE class XII. Let’s make a fair assumption 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 taking JEE (Main), Tom 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 and corresponding marks would be about 110 only.
So Tom will get B final component which consists of B1 (JEE (Main) 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 students of that Board) in the formula approximately as 0.4X110=44 marks. (This is on the assumption that B1 and B2 components remain almost same in case of CBSE). So, total marks of Tom will be C =0.6x345+0.4x110 =207+44=251
Now, if Peter gets 200 marks in JEE (Main) and 97.5% in boards then probably he is at 99.98 percentile and corresponding to this he will get 330 marks in JEE (Main). If one maps 99.98 percentile to JEE main percentile then marks for Bfinal component would be 330. His B final component will be 0.4x330=132. So total marks of Peter will be C =0.6x200 +0.4x330=120+132=252.
The error is self-evident. Tom with high JEE (Main) score but slightly less percentage in CBSE gets only 44 marks in the Board component whereas Peter with 200 marks in JEE (Main) but 97.5% in CBSE gets 132 in the Board component, a huge difference of 86.8 marks.
Someone gets 90% in CBSE Board and ranks first in JEE (Main) he might not even figure in top final ranking.
“This is just stupid,” parent of a student says.
Normalization could result in a student with high score in JEE (Main) and not so low percentage in Class XII (say 90%) getting a rank much lower than his friend who scored less in JEE (Main) but had a higher percentage in the Board results. Now, many more students and parents are thinking of moving court. A parent from NIT, Kurukshetra, said, “The normalization formula is skewed. For admission, it was decided to 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 normalization formula which would put many students on disadvantage.” The formula is C = 0.6 X Ao + 0.4 X B final, where (Ao) component is aggregate marks obtained by each student in JEE (Main) and B final component corresponds to the Board percentile. The final rank of the student in the JEE (Main) will be decided by C.
The normalization process for the Class XII marks component is 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 are pointing out that the normalization system is not working out. The moot point is if a student scores 95%, 2% more than his friend who gets 93% in the Board it would mean he has scored 10 additional marks. But this 2% will result in a big difference in JEE marks mapping.
For instance, Tom, gets first rank in JEE (Main). He scored 345/360. Tom scored 90% in CBSE class XII. Let’s make a fair assumption 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 taking JEE (Main), Tom 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 and corresponding marks would be about 110 only.
So Tom will get B final component which consists of B1 (JEE (Main) 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 students of that Board) in the formula approximately as 0.4X110=44 marks. (This is on the assumption that B1 and B2 components remain almost same in case of CBSE). So, total marks of Tom will be C =0.6x345+0.4x110 =207+44=251
Now, if Peter gets 200 marks in JEE (Main) and 97.5% in boards then probably he is at 99.98 percentile and corresponding to this he will get 330 marks in JEE (Main). If one maps 99.98 percentile to JEE main percentile then marks for Bfinal component would be 330. His B final component will be 0.4x330=132. So total marks of Peter will be C =0.6x200 +0.4x330=120+132=252.
The error is self-evident. Tom with high JEE (Main) score but slightly less percentage in CBSE gets only 44 marks in the Board component whereas Peter with 200 marks in JEE (Main) but 97.5% in CBSE gets 132 in the Board component, a huge difference of 86.8 marks.
Someone gets 90% in CBSE Board and ranks first in JEE (Main) he might not even figure in top final ranking.
“This is just stupid,” parent of a student says.
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