JEE Main Normalization Procedure: The exam of JEE Main 2020 will be conducted by National Testing Agency, as per the announcement by MHRD Ministry. With the introduction of the newly formed exam conducting body, major changes has been done at the procedure of JEE Main 2020. Every year around 14 lakh candidates appear for the exam from various State and Central Boards across the country. The selection of the candidate is through the marks secured in JEE Main and through the counselling.

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JEE Main 2020 Normalization Procedure

The often conflicted issue is the difficulty level of the exam which varies in each session which results in some of the candidates ending up with the difficult set of paper. It is very much natural that the candidate with the more difficult set of an exam will scoreless in the exam in comparison to candidates getting an easier set of the question paper.

JEE Main 2020 exam holds in the month of January and April and the Application Form releases in September and February respectively. Candidates should make sure that as soon as the JEE Main 2020 Admit card is made available, they download the same. After the exam is over the JEE Main 2020 Result will be released for Paper I and Paper II online.

JEE Main Rank Calculator 2020

After the declaration of JEE Main 2020 January exam, candidates can check their expected rank from the formula provided below. The candidate will have to multiply their percentile to 100 which will be divided by 100. The provided score will be multiplied again with the total number of candidates who appeared for the January exam. The candidate must note that the below-provided formula will only provide the estimated rank of January session 2020. The rank might vary after the April exam. The final rank will be declared by the National Testing Agency (NTA) after JEE Main April 2020 exam.

Find the Expected Rank of JEE Main January 2020 Exam.   

Details of JEE Main Normalization Procedure 2020

The National Testing Agency (NTA) releases the JEE Main result for session 1 (January) & session 2 (April) in the form of a normalized score. This will help in resolving issues of discrepancy in the difficulty level of each session. To understand how the normalization procedure takes place, read further.

For Example, there two students, candidate A, and candidate B.

  1. Candidate A has taken JEE Main 2020 in shift 1 one and scored a raw mark of 340 out of 360. He is the highest scorer among all the shift 1 candidates where 56789 candidates appeared.
  2. Candidate B has taken JEE Main 2020 in shift 2 and scored 310 out of 260. She is the highest scorer among all the shift 2 candidates where 54689 candidates appeared.

Now, considering only the raw scores of these two candidates to prepare the merit list, The candidate A will be the topper with a higher rank. However, if shift 2 was more difficult than shift 1 then this will be unfair for candidate B who had to take a more difficult paper and scored less.

This is where JEE Main Normalization Procedure comes in. Let us see how the ranks change with the normalization procedure.

Shift 1: Total candidates who appeared = 56789

Candidates who scored equal to or less than the topper of shift 1 = 56789

Percentile Obtained = 100 (56789/56789)

Percentile Obtained = 100

Shift 2: Total candidates who appeared = 54689

Candidates who scored equal to or less than the topper of shift 2 = 54689

Percentile Obtained = 100 (54689/54689)

Percentile Obtained = 100

This shows us that although both candidates had a different score, based on their difference of difficulty level, both got the same percentile which is 100.

Now let us take 4 more candidates C, D, E, and F.

  1. Candidate C & D have a raw total score of 330 and secured the 2nd position in shift 1.
  2. Candidate E has a raw total score of 304 and secured the 3rd position in shift 1.
  3. And candidate F has a raw total score of 304 and secured the 2nd position in shift 2.

If the raw score of candidates will be considered, the merit list will be something like this:

Candidate A 3401st Position
Candidate B 3103rd Position
Candidate C 3302nd Position
Candidate D 3302nd Position
Candidate E 3044th Position
Candidate F 3044th Position

If we consider these ranks of the candidate, then it will be unfair for Candidate F to be on the 4th position as shift 2 was difficult than shift 1 and also his rank is equal to candidate E. This happens because of the difference in the difficulty level of the various shifts held.

If we change these scores to percentile scores, the following will be the result:

Percentile of candidate A = 100 (56789/56789) = 100

Percentile of candidate B = 100 (54689/54689) = 100

Candidate C percentile= 100 (56788/56789) = 99.99823909

Percentile of candidate D = 100 (56788/56789) = 99.99823909

Percentile of candidate E = 100 (56786/56789) = 99.99471728

Candidate F percentile= 100 (54688/54689) = 99.99817147

Based on this percentile score, the ranks awarded will be something like this:

Candidate A 3401st Position
Candidate B 3101st Position
Candidate C 3302nd Position
Candidate D 3302nd Position
Candidate F 3043rd Position
Candidate E 3044th Position

This is how the normalization procedure will take place and candidates will be awarded a rank after their score is normalized and changed into a percentile score.

National Testing Agency is using the policy of ‘JEE Main Normalization Procedure based on percentile Score’ in order to overcome such problem of the unequal gain of marks. The Normalization Procedure was utilized earlier also for equalization of class 12th marks with the purpose of bringing all the boards on the same level. Now the normalization Procedure will be used for JEE Main score which will be based on Percentile Score in order to ensure that the level of exam is equal for every candidate appearing for JEE Main 2020. The Normalization will be done to make sure there is no uneven difficulty level in the exam.

NTA’s New Percentile Score Based On JEE Main Normalization

Percentile Score: Percentile Score is the marks obtained by the candidate and is transformed into a scale ranging from 0 to 100. The purpose is to compare the scores of the candidates in multiple sessions. Percentile Score will be the normalized score of the examination.

JEE Main Merit List as prepared by NTA according to the normalized scores

NTA Normalized Score

How will the Procedure of JEE Main 2020 Normalization work?

  • Candidates should know that the raw scores of candidate secured in JEE Main 2020 on the basis of the marking scheme will be calculated.
  • Each correct answer will add up the 4 marks and negative marking of 1 marks will be for each incorrect answer.
  • The overall score of JEE Main 2020 will be calculated by comparing the raw score of the candidate with the raw score compared by all the candidates.
  • The percentile secured by the candidate means the percentage of the candidates who have secured less than you in JEE Main exam or the percentage of candidates from whom you have secured above in JEE Main exam.

To understand this procedure go through the example explained:
In the case where  Arun’s percentile is 95 in an exam taken by 1000 students, it means that 95% of 1000 students i.e. 950 students have scored below Arun.
How will the Result of JEE Main 2020 be prepared?

  • The result of JEE Main 2020 will be prepared on the basis of raw score and the percentile score of Total and all the three subjects (Physics, Chemistry, Mathematics).

Total Percentile (T1P)

Let T1, M1, P1, C1 be the raw scores in Total, Mathematics, Physics, Chemistry of a candidate and T1P, M1P, P1P, C1 be the Percentile Scores of Total, Mathematics, Physics, Chemistry of that candidate.

Compilation and display of Result for multi-session Papers of JEE Main 2020

Since the examination of JEE Main 2020 will be held twice this year by National testing Agency, it is important for the aspirants that they are aware of the procedure and the calculation of the Result by the examination authorities. Given the difficulty level of this exam, the scores of the candidates from both the attempts will be taken for the final result and the preparation of the Merit List. Explaining this procedure further, we are providing a detailed explanation of the Result calculation and the preparation of Merit List of JEE Main 2020 in the points below.

Compilation of Result for First attempt :

Since the first attempt will be conducted in multi-sessions, NTA scores will be calculated corresponding to the raw marks obtained by a candidate in each session as per the above procedure. The calculated NTA scores for all the sessions will be merged for the declaration of result. The result shall comprise the four NTA scores for each of the three subjects (Mathematics, Physics, and Chemistry) and the total for the first attempt.

Compilation of Result for Second attempt:

Similarly, the second attempt will be conducted in multi-sessions, NTA scores will be calculated corresponding to the raw marks obtained by a candidate in each session as per the above procedure. The calculated NTA scores for all the sessions will be merged for the declaration of result. The result shall comprise the four NTA scores for each of the three subjects (Mathematics, Physics, and Chemistry) and the total for the second attempt.

Compilation of Result and Preparation of Merit List / Ranking:

The four NTA scores for each of the candidates for the first attempt as well as for the second attempt will be merged for the compilation of result and preparation of overall Merit List / Ranking. Those appeared in both the attempts; their best of the two NTA scores will be considered further for preparation of Merit List /Ranking as explained in the following example:

NTA Combined Normalized Score

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