The results of the 1st session of JEE Main 2019 are already out. Most of the students are happy with their performance in the exam. Still, there are students who are apprehensive for the allotment of their ranks which are based on their exam scores. This confusion has probably cropped up due to the introduction of normalization by NTA. Students are still unclear about the whole concept of normalization in JEE Main 2019, as their earlier counterparts were with the earlier examinations.
Earlier, the normalization procedure included the class 12th marks and the exam scores. However, later on, CBSE scrapped off the board exam marks from the process as did not fulfill the primary goal. Then for JEE Main 2019, NTA again introduced the normalization process, but for fulfilling a different motive, i.e., to bring the entire exam at same difficulty level as the exams are conducted in different session and days.
Normalization primarily refers to equating the exam scores based on the difficulty level of the exam conducted in multiple sessions. This ensures that the students who get a difficult paper to attempt don’t lag. This also ensures that the NTA allots the ranks by following a fair method.
Why has NTA provisioned for the procedure of Normalization in JEE Main 2019?
The NTA held the JEE Main 2019 exam on different days and sessions. Hence, there is a probability of candidates getting an exam of different difficulty levels. Some students can get a simpler paper whereas others might get a tougher one. Therefore, it is obvious that the student getting an easier paper will score more. You can understand it better as compared to the student getting a tough paper.
In order to find a solution to this problem, the authorities have introduced the normalization procedure. The procedure is based on the percentile system. This process ensures that the difficulty level of the exam will not create any advantage or disadvantage to any student. That is, neither any student will get the benefit, nor they will get any disadvantage.
What is JEE Main Normalization?
JEE Main Normalization is a process in which the NTA compares a candidate’s score, in multiple sessions. The process is to ensure that the level of the exam is at the same benchmark. The exam conducting body NTA will use percentile equivalence for JEE Main 2019.
The NTA uses the normalization process in those examinations where it holds them in different and multiple sessions. This time, NTA won’t use actual marks for ranking purpose. It will use the Normalization process employing a percentile score to release the final merit list of the candidates.
How does NTA follow the JEE Main 2019 Normalization procedure?
To do the normalization process a simple predefined formula is used. By using that formula, NTA prepares the merit list of JEE Main. In preparing the merit list, it uses few terminologies. These are like percentile score, raw score, subject percentile score, merit list, and overall percentile score. On the basis of marking scheme of the examination, NTA will calculate a raw score for JEE Main 2019 exam. As per the marking scheme, the authorities provide 4 marks for a correct answer to the candidate. Similarly, for each incorrect answer, they deduct 1 mark.
The authorities employ a method to calculate the overall percentile score of JEE Main exam. They compare the raw score of a candidate against that of all the candidates. These are the candidates who have appeared in that respective session. NTA will calculate the percentile score up to 7 decimal places. Also, NTA will calculate the percentile score for each subject i.e. for Physics, Chemistry, and Mathematics. It needs the score of each subject to break the tie in the exam. The NTA calculates these scores on the basis of a predefined formula.
Finally, NTA considers the overall percentile score for preparing the merit list of JEE Main 2019 examination. It grants the AIR 1 (All India Rank 1) rank to the candidate securing maximum percentile score. On the basis of this rank, NTA counsels the candidates further, for providing admission in different institutes.
Highest Raw Score and Percentile Score
Raw score calculation is based on sessions. Hence the NTA would give the normalized percentile score of 100 to the candidate securing the highest raw score. The NTA assigns this score to the candidates for their exam held in their respective session.
What Would Happen in Case of a Tie?
In case two or more candidates secure the same percentile score then tie-breaking would follow the following order:
- The percentile score in Mathematics
- The percentile score in Physics
- The score of percentile in Chemistry
- According to Date of Birth – the older candidate will be given higher rank
Still, the tie may remain the same. Hence, the agency will give the same rank to both the candidates.
JEE Main 2019 Rank List
There may be candidates who have appeared in both the exams. Hence, the rank list is prepared on the basis of both the exams held in January and April.
The normalization procedure will impact the JEE Main rank of the candidates as the procedure is meant to bring the entire exam held in different sessions and days. Thus, no candidate will be either benefitted by the easy question paper or will be at a loss by getting a tough paper.
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