CBSE XII RESULTS: Top Twenty Percentile scores

CBSE declared results of Senior School Certificate Examination 2013(Class XII) of
the Board on 27th May. For all of JEE Advanced eligible candidates the important issue is whether you land in top 20 percentile or not. As per CBSE the top 20 percentile scores are

General-391 i.e. 78.20%

Obc-389 i.e. 77.80%

Sc-350 i.e.75%

St -338 i.e. 73.60%

 

Normalization: A Scary experiment

Well to all those who still don't have any idea about what this normalization is I advice just to read the comments on this post of Prof. Sanghi.
I am not a statistics expert but I feel the statisticians are not just playing with stats they are playing with careers of students who never ever in distant dream would have thought of this butcher like normalization process to suddenly appear as a gigantic monster standing between them and a premier engineering seat of NITs/IIITs. I am still not sure what exact effects it will have on final merit list of JEE MAIN but what I have found yet is very scary. A slight less though still good say 85% in CBSE Board is even threatening someone 's rightfully seat say the person scores 250 in JEE MAIN.
85% in cbse will be roghly 90-92 percentile which when mapped to JEE MAIN distribution curve would be say 130 marks . So effective score would be like 0.6*250+0.4*0.5(130+140)=204 marks.
I am not sure now what would be his ranking which like in2012 would have been around 5-6k.
Will right more as I get myself equipped with true stats.

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.

NEET ANSWER KEY

The answer key is now available at www.triumphacademy.com

JEE ADVANCED ONLINE REGISTRATIONS TO BEGIN ON 8 MAY

JEE (Advanced) - 2013 Online Registration

This year onwards, admissions to undergraduate programmes at all IITs and ISM, Dhanbad, will be through the JEE (Advanced) examination. This year, it will be held on Sunday, 2 June, 2013.

Eligibility to appear in JEE(Advanced) examination will be decided by the candidates’ score in the JEE(Main) examination, which will be known by 7th May, 2013.  Only the top 1,50,000  candidates (including all categories)  who qualify in Paper -  I   of JEE (Main) - 2013  will be eligible to appear for JEE (Advanced) - 2013.

To write JEE (Advanced) - 2013 all eligible candidates should register online at [http://jeeadvonline.iitd.ac.in/] OR [http://jeeadv.iitd.ac.in/] during 8 May to 13 May, 2013.

After online registration, candidates must pay the registration fee as applicable through challan (generated during registration process) of all SBI branches having core banking solution (CBS) on or before 14 May, 2013. The candidate will be able to pay the fees at the SBI branch one day after the online registration. The fee for GE and OBC(NCL) male candidates is Rs.1800/-. The fee for SC, ST, PwD male candidates is Rs. 900/-. There is no fee for female candidates of all categories. Registration is complete only after the payment.

After successful registration and payment of prescribed fee through bank, the candidates can download their admit card for JEE (Advanced) - 2013 from 15 May to 31 May, 2013. Only those candidates, who carry valid downloaded Admit Cards to the examination hall, will be allowed to write the examination.

At the examination hall, the downloaded admit card should be presented to the invigilators for verification.  The original admit card for the JEE (Advanced) - 2013 will be issued to the candidates in the respective examination centre on the day of examination after verifying their candidature. The original Admit Card should be carefully preserved till the admission process through JEE (Advanced) - 2013 is completed.