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JEE Main 2024 Normalization: How scores are calculated?
Edited By Team Careers360 | Updated on Sep 20, 2023 12:59 PM IST | #JEE Main
JEE Main 2024 Normalization: How scores are calculated? - With JEE Main being a computer-based test held in multiple sessions, candidates are often confused about the JEE Main 2024 normalization process and the process through which scores are calculated to arrive at the ranks. The National Testing Agency (NTA), under whose aegis the exam is being conducted announces the relevant information regarding normalization of JEE Main and how the scores and ranks will be calculated. Other important factors like preparation of the rank list, inter-se-merit guidelines have also been announced. Given below is the JEE Main 2024 Normalization process; how scores are calculated to allot the ranks to candidates of various shifts.
Since JEE Main April 2024 will be conducted on different dates and sessions, the question of maintaining equivalence among the different sets of question papers comes. It is likely that some candidates got a tough paper while some may find theirs easy. To ensure that no candidate is at a loss, or even benefits from this, normalisation in JEE Mains will be done for the purpose of ranking. NTA will rank students on the basis of their percentile scores which will be calculated according to a pre-determined formula.
What is JEE Main Percentile Score?
NTA JEE Main 2024 Percentile scores are based on the relative performance of all those candidates who appeared for the examination. It is obtained after transforming the scores into a scale ranging from 100 to 0 for each session of examinees. Percentile score shows the percentage of candidates who scored equal to or below a particular percentile in an exam. It is the normalized score and not the raw score. After normalization, each topper of a session will get the same percentile of 100 which is the desirable one. Also, the marks between the lowest and highest JEE Main scores are also converted to the respective percentiles.
How will the JEE Main 2024 scores for Paper 1 be calculated?
The formula to be used to be used for JEE Main normalisation procedure is:
Formula - (100 x number of candidates appeared in the session with raw score equal to or less than the candidate) / total number of candidates appeared in that session
Note: To avoid bunching effect and reduce tie breaking, the JEE Main percentile scores will be calculated up to 7 decimal places.
What is normalisation in JEE Mains 2024?
Normalisation in JEE Mains 2024 is a procedure implemented to ensure fairness and equity in evaluating the performance of students across different shifts or sessions of the exam. Since JEE Main is conducted over multiple shifts to accommodate a large number of candidates, the difficulty level of the question papers may vary slightly between different shifts. JEE Main normalisation procedure helps account for these variations and ensures that no candidate is unfairly advantaged or disadvantaged due to the difficulty level of their particular shift.
The JEE Main normalisation procedure takes into consideration the principle that the overall ability of a candidate remains constant, regardless of the shift's difficulty level. It aims to create a level playing field for all candidates, regardless of the specific set of questions they face.
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What will be the Highest Raw Score and Percentile Score in JEE Main 2024?
Since the calculation is session wise, all the highest raw scores will have a JEE Main normalisation procedure based on percentile score of 100 for their respective sessions.
How to calculate JEE Main Percentile Score?
Let's take a look at how the JEE Main percentile is calculated. The JEE Main percentile score will be calculated using the following percentile formula:
Percentile Score of a Candidate = 100 x (Number of candidates who secured a raw score (or actual score) equal or less than the candidate)/(Total number of candidates who appeared in that session)
Preparation of JEE Main 2024 Result for each session
Step 1: Raw Scores Preparation
The marks obtained by each candidate in JEE Main April session 2024 for all correct answers in each subject (Maths, Physics and Chemistry) shall be summed up to arrive at the raw scores.
Step 2: Preparation of JEE Main Percentile Scores for each of the subject and total percentile score
Total Percentile (T1P): | (100 x No. of candidates from the session with raw score equal to or less than T1 score) / Total No. of candidates appeared in the session | |
Mathematics Percentile (M1P) | (100 x No. of candidates appeared from the session with raw score equal to or less than than M1 score in Mathematics) / Total No. of candidates appeared in the session | |
Physics Percentile (P1P) | (100 x No. of candidates appeared from the session with raw score equal to or less than P1 score in Physics) / Total No. of candidates appeared in the session | |
Chemistry Percentile (C1P): | (100 x No. of candidates appeared from the session with raw score equal to or less than C1 score in Chemistry) / Total No. of candidates appeared in the session | |
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Inter-se-merit to resolve tie:
When two or more candidates obtain equal percentile, then ranking will be as follows
Candidates who obtain Percentile Score in Mathematics will be ranked higher
Candidates who obtain Percentile Score in Physics will be ranked higher
Candidates who obtain Percentile Score in Chemistry will be ranked higher
Date of Birth – the older candidate will be ranked higher
- In case the tie is still unresolved, the same rank will be given to the candidates
JEE Main 2024 Rank list
The authorities shall release the final JEE Main rank list 2024 after conclusion of April session exam. This JEE Main normalisation procedure is considered to create the final merit list. This shall be used for admissions into JEE Main participating institutes.
First attempt compilation of NTA Scores:
After the conclusion of JEE Main January session on February 1, JEE Mains results will be displayed with the four NTA scores – for each of the three subjects (Mathematics, Physics, and Chemistry) and the total NTA score for that attempt. JEE Main NTA scores will be the percentile scores calculated as mentioned above.
Second attempt compilation of NTA Scores:
Similar to the January attempt, the April session will be conducted from April 1 to 15 and the result will be declared after that. The JEE Main 2024 scores will be calculated for the total and that of the three subjects individually as mentioned for the first attempt.
Compilation of NTA Score for JEE Main Result and Merit List/ Ranking for Paper 1
The ranking or merit list for JEE Main 2024 will be compiled as follows
In case of candidates who will appear for both the attempts the best of the two NTA JEE Main scores will be used. Here the total NTA score will be used for ranking purpose and not those of the subjects.
In case of candidates who will appear for only one exam, the four JEE Main 2024 NTA scores of that attempt will be used for the preparation of the merit list
The final NTA scores will be used to rank the candidates and for the declaration of the JEE Main 2024 result. In case of any tie, the inter-se-merit guidelines mentioned above in this article will be used to resolve it.
JEE Main 2024 Rank Declaration:
The ranks derived through the method explained above will be used for the purpose of admissions to NITs, IIITs and GFTIs.
The JEE Main 2024 All India rank and All India Category Rank will be declared once the NTA scores are compiled by the authorities for April session.
How will the JEE Main 2024 scores for Paper 2 be calculated?
Raw marks for the first attempt will be announced in February 2024 for students who appeared for the exam held in January.
The actual marks for the April exam will be announced soon after the conclusion of exam.
The JEE Main Paper 2 marks will not be normalized as the exam was held in a single shift so there will be no question of equivalence of the question papers and variation in difficulty levels
Rank List preparation for JEE Main 2024 Paper II
The raw marks for both attempts will be considered for candidates who will give both the attempts and the better of the marks considered.
For candidates who appeared for only one attempt, the raw marks obtained for the same will be used for preparing JEE Main rank list.
The NTA JEE Main 2024 Rank lists will be prepared as per the marks obtained as mentioned above
Inter-se-Merit will be resolved as follows:
- Candidates who scores more marks in Aptitude Test will be ranked higher
- In case the tie is unresolved, Candidates who scores more marks in Drawing Test will be given the higher rank
- Lastly the age of the candidates i.e. the older candidate will be awarded the higher rank
- If the tie is still unresolved, the candidates will be given the same rank
Frequently Asked Question (FAQs)
1. How does normalization work in JEE Main?
Normalization involves a formula that takes into consideration the raw scores of candidates and the statistical characteristics of the entire exam. It accounts for the differences in difficulty levels and ensures that candidates are evaluated fairly across all sessions.
2. Why is normalization necessary?
JEE Main is conducted in multiple shifts to accommodate a large number of candidates. Since different shifts may have slightly varying difficulty levels, normalization ensures that the final scores accurately reflect a candidate's performance rather than the difficulty of their particular session.
3. How is the normalized score calculated?
The normalized score is calculated using a formula that considers the mean, standard deviation, and raw scores of candidates in each session. The details of the formula are usually provided by the exam conducting authority.
4. Is normalization applicable to both Paper 1 and Paper 2 of JEE Main?
Yes, normalization is applicable to both Paper 1 (for B.E./B.Tech) and Paper 2 (for B.Arch/B.Planning) of JEE Main.
5. How does normalization affect the ranking?
Normalization ensures that candidates' ranks are determined based on their relative performance across all sessions. It prevents an advantage or disadvantage based on the difficulty level of their specific session.
6. Is normalization beneficial for candidates?
Normalization aims to provide a fair evaluation of candidates' abilities, regardless of the session they appeared in. It eliminates any potential advantage or disadvantage that may arise due to the varying difficulty levels of different sessions.
7. Is normalization used for all engineering entrance exams?
Normalization is a specific method used by some entrance exams, including JEE Main, to ensure fairness in evaluating candidates across different sessions. Not all entrance exams employ normalization.
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Have a question related to JEE Main ?
Hello aspirant,
Based on past year's trends, I think you need to be at least in the 93 percentile to get into ECE, ME, or EE, or maybe Mining or Metallurgical, to get admission in BIT Sindri. You roughly need ranking under 80,000 in JEE Mains for admission in this college.
Thank you
Hope this information helps you.
Sajal Trivedi 04 May,2024
You need pay again and raise dispute with bank in case od error
nathgabhilash 05 May,2024
Jee mains sc category 83.7 percentile rank 13161 hai konsi nit milega
chandansarkar781 04 May,2024
You can do two things either wait for few hours or till tomorrow or you can do payment again because in that case previous money will be refunded shortly
shikhartiwari1206 05 May,2024
I am also a parent looking for admission process...the information I got regarding caste certificate is that we have get a fresh certificate and central caste certificate is different from the state one
atmajasaraikar 04 May,2024
Popular JEE Main Questions
5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.820C. If Na2SO4 is 81.5% ionised, the value of x (K for water=1.860C kg mol−1) is approximately : (molar mass of S=32 g mol−1 and that of Na=23 g mol−1)
Option: 1 15 g
Option: 2 25 g
Option: 3 45 g
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What is A?
Option: 1
Option: 2
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A capacitor is made of two square plates each of side 'a' making a very small angle between them, as shown in the figure. The capacitance will be close to :
Option: 1
Option: 2
Option: 3
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Among the following compounds, the increasing order of their basic strength is 
Option: 1 (I) < (II) < (IV) < (III)
Option: 2 (I) < (II) < (III) < (IV)
Option: 3 (II) < (I) < (IV) < (III)
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As shown in the figure, a battery of emf is connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = tc (tc is the time constant of the circuit ) is :
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As shown in the figure, a particle of mass 10 kg is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed of the particle at B is x m/s. (Take g = 10 m/s2 ) The value of 'x' to the nearest is ___________.
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