CBSE Finally announces the Board Marks Normalization Details

Central Board of Secondary Education (CBSE) announced the normalization procedure for class 12 board exam marks in preparing all India JEE Main 2013 merit list. In view of large scale variations in the examination system in country, the JEE Interface Group (JIG) decided that for this year (2013), 50% of Boards marks be normalized by equating percentile amongst different boards/examining bodies and anchoring them to All India JEE-Main percentiles, and 50% be normalized by equating each Board's/examining Body's percentile with JEE(Main) percentile marks of respective Boards /examining bodies. The detailed procedure for normalization of Board marks will be as follows:

1. Note down the aggregate marks (A0) obtained by each student in JEE- Main.

2. Compute the percentile (P) of each student on the basis of aggregate marks in his/her own board (B0) computed from the list of five subjects specified (each marked out of 100). The percentile is to be computed among all students of the board whose subject combinations meet the eligibility criteria of JEE-Main. 

3. Determine the JEE- Main aggregate marks corresponding to percentile (P) at the All- India level. Regard this as B1.

4. Also, determine the JEE- Main aggregate marks corresponding to percentile (P) among the set of aggregate scores obtained in the JEE- Main by the students of that board. Regard this as B2.

 The normalized board score of the candidate will be computed as:

 B (final) = 0.5 * (B1 + B2)

 For the purpose of admission to CFTIs where it has been decided to use the JEE Mains performance and the Normalized Board performance in the 60:40 ratio, the composite score for drawing the merit list will be computed as: 

C = 0.6 * AO + 0.4 * B (final),

 Five subjects to be used for normalization: 

 Physics 

 Mathematics 

Any one of the subjects Chemistry, Biology, Biotechnology and Computer Science One language Any subject other than the above four subjects.

 In respect of 3, 4 and 5, the best mark in a given category will be chosen. 

 For any query please write to jeemainnormalisation2013@gmail.com .

Example to understand. Student A of CBSE has secured 100 marks in JEE (Main) but has got 99.9 percentile in board. His final likely score will be 0.6*100+0.4*.5(345+345)=60 +138 = 198. Another student B of CBSE has has got 250 marks in JEE(Main) and 97 percentile in CBSE.His final likely score will be 0.6*250+0.4*.5(145+155)=150 +60 = 210.