JEE Main 2021 January session is expected to be scheduled in 2nd week of January 2021. NTA (National Testing Agency) used the Normalisation process or the Percentile Scoring based mechanism to calculate the scores for the candidates. This process or formula helps to calculate the percentile of the candidate rather than the raw total score. Check JEE Main 2021 Exam Dates
Quick Links:
As NTA conducts JEE Main in multiple sessions in a span of a few days, they require a process that can judge the capability of students regardless of the session they gave the exam in. This is because the papers in each session are different and the difficulty of the papers could vary even though NTA tries their best to set papers of equal difficulty. So, to be fair to all the candidates that attempt a paper harder than the other sets, this process is used to practice a fair way of calculating the eligibility of a candidate to the various colleges accepting JEE Main 2021 marks.
Normalisation process makes sure that the candidates that appeared for the harder paper are not disadvantaged by getting lower marks than all the other while the candidates with the easier paper, benefitted by their higher marks total. Being a just method, NTA uses this method to compile a percentile for all candidates.
Must Read:
The method is based on the relative performance of a candidate in the exam. Rather than taking the raw score as the final result of a candidate, each candidate gets a percentile which means the percentage of candidates that appeared for the session that have got equal to or below the marks of the particular candidate.
Example, Candidate X got a percentile of 93%, then 93% of the people that appeared for the same session as X have either got the same marks as him/her or below. The marks will be from a scale of 100 to 0 in each session and the highest scorer getting a percentile of 100 and the lowest getting 0 percentile. To avoid any confusion and overlapping difficulties, NTA calculates the percentile to 7 decimals. This will help make the merit list for the colleges.
To calculate the percentile score of a candidate, a person would require the details of the candidates which are:
The formula is :-
100 × number of candidates that appeared in the session and scored either equal to or lesser than the candidate / total number of candidates in the session
To further illustrate this formula, let us use it with a set of hypothetical numbers.
The first table has the highest scores secured in 4 sessions of the exam. The papers are out of 360 with Maths, Chemistry and Physics each having a paper of 120 marks.
Session | Highest raw score secured | Total number of candidates that gave the exam in the session | Total number of candidates who scored equal to or lesser than the Candidate | Formula | Percentile |
---|---|---|---|---|---|
Session 1 | 320 | 33457 | 33457 | (33457/33457)*100 | 100 |
Session 2 | 300 | 41289 | 41289 | (41289/41289)*100 | 100 |
Session 3 | 345 | 31562 | 31562 | (31562/31562)*100 | 100 |
Session 4 | 338 | 29196 | 29196 | (29196/29196)*100 | 100 |
As the candidates in this scenario are all the highest scorers, the total number of candidates that appeared for the session and the total number of the candidates who scored equal to or lesser than the candidate will be the same. Thus the percentile of the highest scorer will always be 100.
The next table has the lowest scores secured in the 4 sessions. In this table the lowest rankers are mentioned with the same number of candidates appeared. The formula used is the same.
Session | Lowest raw score secured | Total number of candidates that gave the exam in the session | Total number of candidates who scored equal to or lesser than the candidate | Formula | Percentile |
---|---|---|---|---|---|
Session 1 | -28 | 33457 | 1 | (1/33457)*100 | 0.002989 |
Session 2 | -30 | 41289 | 1 | (1/41289)*100 | 0.002422 |
Session 3 | -15 | 31562 | 1 | (1/31562)*100 | 0.003168 |
Session 4 | -32 | 29196 | 1 | (1/29196)*100 | 0.003425 |
The two tables mentioned above considered only the total marks, but the session result that is published will normally have the total raw score along with the raw scores of Maths, Chemistry and Physics also mentioned. This result will then also mention the percentile score for each of the above mentioned raw scores (Total, Maths, Chemistry and Physics)
The next table will consist of a hypothetical (same data of a total number of candidates mentioned in the tables above with the same highest raw score and lowest raw score) result sheet of the combined 4 sessions and the raw scores of Maths, Chemistry and Physics mentioned. The column for Date of Birth (DoB) has not been mentioned.
Candidate (from any of the 4 sessions) | Total Raw Score | Math Raw Score | Chemistry Raw Score | Physics Raw Score | Total Percentile Score | Math Percentile Score | Chemistry Percentile Score | Physics Percentile Score |
---|---|---|---|---|---|---|---|---|
A20260598 (Session 1) | 320 | 120 | 119 | 81 | 100 | 100 | 100 | 98.9945689 |
C27958347 (Session 3) | 345 | 118 | 110 | 117 | 100 | 100 | 99.79405614 | 100 |
A21548691 (Session 1) | -28 | -19 | 17 | -26 | 0.0029890 | 0.0029890 | 0.0029890 | 0.0029890 |
D26589412 (Session 4) | -32 | 15 | -30 | -17 | 0.0034250 | 0.0034250 | 0.0034750 | 0.0034450 |
C20847563 (Session 3) | 245 | 85 | 89 | 71 | 94.9496230 | 94.9544389 | 94.9553260 | 84.7538179 |
D20135694 (Session 4) | -10 | 5 | -13 | -2 | 1.92492122209 | 2.7709275 | 0.0308261 | 1.4693794 |
B20156489 (Session 2) | 299 | 110 | 78 | 111 | 99.7844462 | 99.7941340 | 86.7155901 | 99.8013999 |
D28657190 (Session 4) | 56 | 12 | 25 | 19 | 68.4511577 | 67.9031374 | 68.4648582 | 67.4407453 |
A21054834 (Session 1) | -19 | 4 | -16 | -7 | 2.4180291 | 2.4778073 | 2.1191380 | 2.3858983 |
B21078452 (Session 2) | 154 | 56 | 45 | 53 | 73.9131488 | 81.1765361 | 70.5466347 | 76.3108818 |
The final merit list considers the Total Percentile Score.
The Normalisation Process is a process that has a lesser chance for repetitions or overlapping of rank to occur in a session, but at times it is possible for a tie to occur. For those times, NTA has also come up with ways that act as tie-breakers. The order for the tie breakers to decide between two candidates is:
This is the process by which the merit list is prepared and as JEE Main 2021 will be conducted in two sessions, a candidate may take the exam both the times. NTA creates the merit list considering the candidate's best from the two.
*The article might have information for the previous academic years, please refer the official website of the exam.