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September 16, 201139 Insightful Publications

**JEE Main Normalization 2022:** JEE Main exam is conducted in two phases this year: June and July sessions. The session 1 exam of JEE Main was conducted from June 23 to 29, 2022, whereas the July Session was conducted from July 25, 2022, onwards. Since the dates are being revised multiple times, there can be variations in all the scoring evaluation procedures. It can lead to further discrepancies within the final merit list of JEE Main 2022.

In order to solve this problem, the National Testing Agency (NTA) will normalize the raw or actual score secured by the candidate to the NTA score (normalized score), which takes care of the variation in the difficulty level and other factors. The JEE Main Normalization process 2022 will be based on percentile score (Percentile Equivalence Method). This article will provide candidates with all the details about JEE Main 2022 Normalization procedure and how the marks will be calculated.

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Let us first understand the JEE Main 2022 Normalization process to understand what percentile score means.

The percentile score reflects the relative performance of the candidates. It shows how one candidate has performed with respect to the other candidates cumulatively. One should note that percentile and percentage are two different terms and have different meanings.

A percentile score is the percentage of the total marks scored by one particular candidate compared to the percentage of all other candidates who scored ‘Equal to or Less Than the score of that candidate.

This means that the cumulative scores of each session are converted into percentile scores (0 to 100), also known as normalization. Thus, the topper of each session secures 100 percentile as per this method.

Let us now see how JEE Main percentile is calculated. The percentile score for JEE Main 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 TO OR LESS than the candidate)/(Total number of candidates who appeared in that session)**

Let us understand the formula with the help of an example.

Consider the following data for a particular session of the JEE Main exam:

Number of candidates who appeared for the exam in that session | 41,326 |

Highest marks secured in that session | 275 (out of 300) |

Number of candidates who secured the highest marks (let us say Candidate A) | 1 |

Number of candidates who secured equal to or less than 275 | 41326 |

Marks secured by Candidate B in that session | 121 |

Number of candidates who secured less than or equal to 121 | 37244 |

Now, using the above formula,

Percentile score secured by the topper, Candidate A = (100 x 41,326)/41,326 = 100

This means 100 per cent of candidates secured less than or equal to 275, and no candidate secured more than 275.

Percentile score secured by Candidates B = (100 x 37,244)/41,326 = 90.124411

This means out of the 41326 candidates who appeared for this session, 90.124411 per cent of candidates secured less than or equal to Candidate B. It also means all other candidates have scored more than Candidate B.

As mentioned in the introduction, the JEE Main normalization 2022 is based on the candidates’ percentile score (or NTA score). Both subject-wise and total NTA scores are calculated for each candidate.

This reduces the chances of discrepancy in results due to variation in the difficulty level of the papers across slots. How? We will elaborate in the next section.

Let us look at how JEE Main normalization eliminates discrepancies caused by variation in difficulty level. Let us start with the following assumptions:

- Candidate A secured the highest marks among all the students who appeared in that Session 1 – scoring 280 out of 300.
- A total of 45,632 students appeared for the exam in Session 1.

- Candidate B secured the highest score among all the students who appeared in that Session 2. She scored 265 out of 300.
- A total of 45,067 students appeared for the exam in Session 2.

Let us also assume that Session 1 was easier than Session 2.

If we consider the raw scores secured by the two candidates, Candidate A would be the overall topper. But that would be unfair because Session 2 was more difficult than Session 1.

Now, let us see what happens if we normalize the scores secured by Candidate A and Candidate B using the above formula and express them in percentile.

Number of candidates who secured a raw score ‘Equal to or less’ than Candidate A **= 45,632 **

Total number of candidates who appeared in Session 1: **45,632**

Using the above formula,

*Percentile Score Obtained by Candidate A = 100 x (45632/45632) = 100*

Number of candidates who secured a raw score ‘Equal to or less than Candidate B **= 45,067 **

Total number of candidates who appeared in Session 1: **45,067 **

Using the above formula,

*Percentile Score Obtained by Candidate A = 100 x (45,067 /45,067) = 100*

So, if expressed in percentile, both Candidate A and Candidate B have secured the same scores.

Let us assume the following scenario:

- Both Candidate C and Candidate D secured 2
^{nd}position with a raw score of 272 - Candidate E secures the 3
^{rd}position with a raw score of 255

- Candidate F secures the 2
^{nd}position with a raw score of 255 (same as the raw score secured by the 3rd position holder of Session 1)

Now, if we consider the raw scores to compile the merit list/rank list, then the ranks of Candidates A, B, C, D, E, and F are as under:

Candidate A (280) | 1^{st} Position |

Candidate B (265) | 3^{rd} Position |

Candidate C (272) | 2^{nd} Position |

Candidate D (272) | 2^{nd} Position |

Candidate E (255) | 4^{th} Position |

Candidate F (255) | 4^{th} Position |

Here, both Candidate E (of Session 1) and Candidate F (of Session 2) secured the same score, and hence have been given the same rank. But Session 1 was easier than Session 2. So, Candidate F should have been the one with a better rank than Candidate E. But that is not happening because the variation in difficulty level was not considered.

So, let us look at the Percentile scores of the candidates now:

- Percentile Score of Candidate C = 100 (
) = 99.9978086*45,631/45,632* - Percentile Score of Candidate D = 100 (
) = 99.9978086*45,631/45,632* - Percentile Score of Candidate E = 100
= 99.9934257*(45,629/45,632)* - Percentile Score of Candidate F = 100
*(**45,066/45,067**)*= 99.9977811

According to the Percentile Scores, the JEE Main rank list/merit list is as follows:

Candidate A (350) | 1^{st} Position |

Candidate B (310) | 1^{st} Position |

Candidate C (334) | 2^{nd} Position |

Candidate D (334) | 2^{nd} Position |

Candidate E (305) | 4^{th} Position |

Candidate F (305) | 3^{rd} Position |

Here, Candidate F gets a better rank than Candidate E. So, the discrepancy that would have arisen if we had considered the raw scores of the candidates who appeared for the exam in different sessions having different difficulty levels has been eliminated.

In order to determine the JEE Main ranks and percentile scores (both total and subject-wise), the NTA uses the same procedure.

The NTA will release the JEE Main February Result in the manner mentioned below. For the March, April, and May sessions too, the JEE Main Result will comprise the subject-wise and overall NTA Score (Percentile Score).

For candidates who appeared in more than one session, the best overall (total) NTA Scores will be considered.

After that, NTA will compile the JEE Main Merit List and declare the All India Ranks.

In case, two or more candidates secure the same total NTA score, NTA will consider the following factors (in the same order) to determine their ranks:

– NTA score in Mathematics

– NTA score in Physics

– NTA score in Chemistry

– Candidates with less proportion of negative responses

– Candidates older in age

If the tie exists even after applying these, candidates will be given the same rank.

**Check Out JEE Main Percentile Vs Rank for Free**

Candidates can check the Official JEE Main Normalization PDF released by NTA, which has all the important information pertaining to how NTA will compile the scores for multi-session papers.

The counselling and admission procedure will be based on the final merit list of the candidates.

All the details about the JEE Mains Normalization procedure and how the NTA score is calculated have been mentioned in this article. With the release of the JEE Main result 2022, the candidates will get the NTA Normalized score and qualifying status for the JEE Advanced.

Only the top 2,50,000 candidates will be eligible for JEE Advanced. Take JEE Advanced mock tests, and JEE Advanced practice questions to ensure that you are well prepared for the upcoming JEE Advanced 2022.

Here are the most frequently asked questions about JEE Main Normalization Process 2022:

**Q1: What is the normalization process in JEE Mains?****Ans**: Normalization is introduced to compare scores secured by candidates across multi-session papers. NTA employs percentile equivalence for normalization in JEE Mains.

**Q2: How is normalization done in JEE Mains?****Ans:** We have explained the normalization procedure in the above article.

**Q3: When will the JEE Mains 2022 results be announced?****Ans: **NTA released the JEE Main June session results on July 11, 2022. The result for the July session are announced.

**Q4: When will the cutoff list for JEE Main 2022 be released?****Ans: **The cutoff list of JEE Main 2022 have been announced along with the final results.

**Q5: Who conducts JEE Mains counselling?****Ans:** The Joint Seat Allocation Authority (JoSSA) conducts counselling for JEE Mains. JEE mains 2022 counselling is expected to be held online mode. In general, JoSSA conducts JEE Mains Counseling in seven rounds.

JEE Main 2022 Percentile vs Rank | JEE Main 2022 Marks vs Rank |

JEE Mains 2022 Toppers | JEE Main Cut-off 2022 |

JEE Main 2022 Question Paper | JEE Main 2022 Answer Key |

NIT Cut-off | JEE Main 2022 Analysis |

*We hope this article on JEE Main 2022 Normalization helps you. If you have any queries regarding JEE Main 2022, do reach out to us. *

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