Published On: Wed, Oct 3rd, 2018

JEE Main 2019: A New Normalisation Procedure Introduced by NTA

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National Testing Agency, or NTA, is all set to conduct JEE Main 2019 from 6th January, 2019 to 20th January 2019 for the very first time. With the incoming of NTA, many significant changes have been introduced in the exam. The most major one being that the exam will be conducted in a significantly higher number of shifts, and in online mode only. Further, two attempts per year are also going to be a big relief for students. But a big issue with multiple shifts is that there is always a chance of one set of students getting benefit over the others due to easier paper, and vice versa. Even till last year, students have been complaining about the anomaly in difficulty levels of online and offline papers.

Well, to eliminate that, NTA is introducing a new type of normalization for JEE Main 2019. Normalization is not a new concept with JEE Main, as earlier CBSE used to normalize the Class 12 scores across boards to give all candidates an even chance. However, since that criterion is no more alive, the normalization here would be the actual JEE Main 2019 result of candidates.

With this, what NTA wants to make sure is that no candidate gets any undue gain/loss due to a comparatively easy/tough paper, as compared to other shifts. Let us understand what exactly this normalization is and how it will work. Take a look below: –

What is ‘Normalization’ under JEE Main 2019?

NTA calls this new normalization system as ‘Percentile based Normalization.’ Well, this outrightly means that candidates may no longer be able to see their ranks in the result, rather, the percentile score will prevail.

In relation to such exams, the percentile depicts the percentage of candidates (from the total set of candidates appearing for the exam) who scored less than or equal to a particular candidates. So let’s say this year again someone scores a perfect 360, and no one else is able to do that, his/her percentile will be 100.0000000, as all the candidates appearing in the exam have a score lower than this candidate.

Percentile based Normalization: How does it work?

Thankfully, NTA has released a detailed document explaining how this system of normalization will work. So from that, we will be explaining how this system will actually work.

Well, as the first step, NTA will calculate the percentile scores of all the candidates. This will be calculated for all 4 separate score, i.e. Score in Physics, Score in Chemistry, Score in Mathematics and the Overall Score.

The usual formula used to calculate percentile score for a particular raw score (i.e. original score in JEE Main 2019 exam) is: –

“100 * Number of candidates having score less than or equal to the particular score / Total number of candidate appearing in that shift.”

NTA has also clearly stated the formula it will be using, which is exactly the similar to the above one, and is mentioned below: –

After getting these percentiles for all candidates in all shifts, NTA will combine these and prepare a common merit list.

How the scores of all the shifts will be equated?

NTA has created a simple tie breaker method which will tell who will be getting a higher place if two people have the same overall percentile. In the case, the person with higher percentile in Mathematics will be given a higher percentile. If percentile in Mathematics is same, the percentile score in Physics will be considered. If this is also same, then percentile of Chemistry will be seen. If everything is the same, then the candidate with higher age will be given higher rank.

Let us take an example for this. Suppose a candidate, let’s call her P, appeared for her exam on January 6, and scored 331 marks which were highest in her shift, and thus she got a 100th percentile in her shift. Her percentile in Mathematics, Physics and Chemistry were (approximately) 100, 99.97 and 99.89. Another candidate, calling her C, scored 335 in her shift on January 15, and thus got a 100th percentile overall, with percentiles of 99.98, 100 and 99.95 in Mathematics, Physics and Chemistry. Now since Candidate P has a higher percentile in Mathematics, she will be given a percentile higher than Candidate C, despite both of them being the toppers of their shift.

What will happen in case a candidate attempts the exam twice?

Those who will attempt the exam twice, will get to see their better percentile (amongst the two attempts) in the final list. So if you scores approximately 99.36 in the January attempt and 99.54 in the second attempt, your final percentile will be considered as 99.54.

Normalization is certainly a better news for candidates as they do not need to worry about difficulty level of other shifts. It is widely used in other exams like GATE and CAT, and has proved successful. With such a clear and fair method in place, candidates just need to focus on their preparation for now.

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