**JEE Main Normalization 2020:** The National Testing Agency (NTA) is going to conduct the JEE Main September Session (postponed from April) from 1st to 6th September 2020. Earlier in January, NTA conducted the JEE Main January Session on January 7th, 8th and 9th 2020. As you can see, the exam is conducted on multiple days with two shifts every day. As a result, there is variation in the difficulty level of the exam across various dates and slots. Students attempting a tough paper are likely to score less than those who got an easy paper. This can result in discrepancies in the JEE Main result and Merit List.

In order to overcome this problem, NTA will normalize the raw or actual score secured by the candidate into the NTA Score (normalized score) which takes care of the variation in the difficulty level and other factors. The JEE Main Normalization procedure will be **based on Percentile Score (Percentile Equivalence Method)**. In this article, we will provide you with all the details about JEE Main 2020 Normalization Procedure and how the marks will be calculated. Read on to find out.

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## JEE Main Normalization 2020: Percentile Score Calculation

Before getting into the details of JEE Main Normalization 2020, let us first see what Percentile Score means.

### JEE Main Normalization: What Is Percentile Score

Percentile scores, basically, tells you how you have performed relative to all other candidates who appeared for the exam. It is not the same as your percentage score.

While the percentage score tells you the percentage of the total marks or maximum marks for the exam that you have secured, the percentile score indicates what percentage of the total number of candidates appearing for the exam have scored EQUAL TO OR LESS THAN the score you secured.

This means, in each session, the marks secured by the candidate (say, out of 300 in the case of JEE Main Paper 1) is converted into a scale ranging from 100 to 0. The topper for each session will, thus, secure a percentile score of 100.

### JEE Main Normalization: How Is Percentile Score Calculated

The Percentile Score for JEE Main will be calculated using the following 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 elaborate on it with an example.

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

Number of candidates who appeared for the exam in that session | 41326 |

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

Number of candidates who secured the highest marks (let’s 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 41326)/41326 = 100

This means 100 percent of candidates secured less than or equal to 335 and no candidate secured more than 335.

Percentile Score secured by Candidates B = (100 x 37244)/41326 = 90.124411

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

### JEE Main Normalization Based On Percentile Score

As mentioned in the introduction, the JEE Main Normalization is based on the percentile score (or NTA score) secured by the candidates. 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.

### How Does JEE Main Normalization Eliminate Discrepancies Caused By Variation Of Difficulty Level?

So, let us elaborate it with an example.

Let’s start with the following assumptions:

**Session 1:**

- a. Candidate A secured the highest marks among all the students who appeared in that Session 1. He scored 280 out of 300.
- b. A total of 45632 students appeared for the exam in Session 1.

**Session 2:**

- a. Candidate B secured the highest score among all the students who appeared in that Session 2. She scored 265 out of 300.
- b. A total of 45067 students appeared for the exam in Session 2.

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

Now, if we consider the raw scores secured by the two candidates, of course, Candidate A would be the overall topper.

But that would be unfair because Session 2 was more difficult than Session 1.

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

**In Session 1:**

No. of candidates who secured a raw score EQUAL TO OR LESS than Candidate A **= 45632 **

Total number of candidates who appeared in Session 1: **45632**

Using the above formula,

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

**In Session 2:**

No. of candidates who secured a raw score EQUAL TO OR LESS than Candidate B **= 45067 **

Total number of candidates who appeared in Session 1: **45067 **

Using the above formula,

**Percentile Score Obtained by Candidate A = 100 x (45067 /45067) = 100**

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

Let us assume the following scenarios as well:

**In Session 1:**

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

**In Session 2:**

- a. 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, 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’s not happening because the variation in difficulty level was not considered.

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

- a. Percentile Score of Candidate C = 100 (
) = 99.9978086*45631/45632* - b. Percentile Score of Candidate D = 100 (
) = 99.9978086*45631/45632* - c. Percentile Score of Candidate E = 100
= 99.9934257**(45629/45632)** - d. Percentile Score of Candidate F = 100
*(*= 99.9977811**45066/45067**)

Now, the merit list considering the Percentile Scores is as under:

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 arised if we had considered the raw scores of the candidates who appeared for the exam in different sessions having different difficulty level, have been eliminated.

The NTA follows the same procedure to calculate the JEE Main percentile scores (both total and subject-wise) and determine the ranks.

### How Is Merit Determined In JEE Main?

The NTA released the JEE Main January Result in this manner. For the September session too, the JEE Main Result will comprise the subject-wise and overall NTA Score (Percentile Score).

For candidates who appeared in both January and September sessions, the best of the two 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:

a. NTA score in Mathematics,

b. NTA score in Physics,

c. NTA score in Chemistry,

d. Candidates with less proportion of negative responses,

e. 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 Here**

### Jee Main Normalization (Percentile): How To Calculate Score PDF?

Check out the Normalization procedure PDF:

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

Now you have all the details on the JEE Main Normalization procedure and you also know how the NTA Score is calculated. With the release of the JEE Main Result 2020 you will get the NTA Normalized Score and your qualifying status for the JEE Advanced 2020.

Only the top 2,50,000 candidates will be eligible for JEE Advanced and candidates will have to qualify this eligibility criterion. If you are qualified for JEE Advanced then, start your JEE Advanced preparation right away.

Take **JEE Advanced Mock Tests** and **JEE Advanced Practice Test** to make sure that you are well prepared for the upcoming JEE Advanced 2020.

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*We hope this detailed article on JEE Main Normalization helps you. If you have any queries or confusions, feel free to ping us. You can also drop a comment below. We will get back to you.*

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