Calculating normalized score for JEE Main 2017 requires the knowledge of percentile obtained in board, JEE Main score, corresponding JEE Main marks against the percentile for the board and all India. Marks obtained by giving 60% weightage to JEE Main score and 40% to the normalized JEE Board score is used to determine merit of candidate. JEE Main 2017 Result uses the normalization process adopted by the Central Board of Secondary Education. Marks of 5 subjects in the board exam are used in this process. Read this article to know all about normalization, calculation of percentile, and other details associated with it.

## JEE Main 2017 Normalization

The formula used for drawing the merit list of JEE Main 2017, for admission to NITs, IIITs, CFTIs, is

**C = 0.6 * A _{0} + 0.4 * B_{final}**

Here,

- A
_{0}is the aggregate marks obtained by the student in JEE Main, and - B
_{final}is the average of two components – (1) JEE Main aggregate marks corresponding to percentile at the All- India level referred to as as B_{1}, and (2) JEE Main aggregate marks corresponding to percentile among the set of aggregate scores obtained in the JEE Main by the students of that board, referred to as B_{2}.

A_{0} is known when CBSE declares JEE Main 2017 scores. Calculation of B_{1}, B_{2}, B_{final} are explained here.

*The following data is an example set only and does not reflect actual score versus percentile distribution.*

Percentile score of a Candidate in a Board or JEE (Main) will reflect what Percentage of Candidates have scored below that candidate in that Board or JEE (Main) Examination. A Percentile score is the value of below which a certain Percent of observations fall.

**Step 1 – Knowing percentile and normalization**

Student can know percentage obtained in board exams (qualifying class 12) by checking respective board exam result.

*Example* : Student (S) is studying on board ABC in which 13711 candidates have appeared. He has scored 60% marks and the number of students who have scored less than him is 6865. The number of JEE main candidates all India in this example is 11.5 lakh.

Table 1: Percentage and corresponding Percentile score of 13711 Candidates of Board ABC

Percentile (**Pi**) for student is = 100 X (6865/13711) = 50.07.

The board arranges in ascending order, marks of each candidate and calculates percentile. As an example, a table is shown above.

* Normalisation of Board Marks for JEE (Main)* – To find out normalised marks of a candidate from the Board ABC who has scored, say, 60% in the Board, first thing to do is to calculate Percentile (Pi) of the candidate with reference to the group of candidates of Board ABC who also satisfy the eligibility criteria of JEE (Main).

From the Table 2, this is 50.07

Table 2: Showing the percentage score in a particular Board, say, ABC of 13711 candidates who satisfy the eligibility criteria of JEE (Main) and their corresponding Percentile (P)

**Step 2 – Arrange JEE Main score of all candidates in ascending order and find out Percentile for each score.**

Say 11.5 lakh students appeared for JEE Main exam and the marks are declared. Having arranged the the JEE (Main) score of 11.5 lakhs candidates, their corresponding Percentile is calculated as well. The table below shows a sample data.

Table 3: JEE Main Scores and Percentile for All India (sample data)

**Step 3 – Arrange JEE Main score of all candidates of Board ABC in ascending order and find out Percentile for each score.**

Take the JEE Main scores of only those students from the same board as candidate, and arrange them in ascending order. Also calculate corresponding percentile as shown below in the sample data.

Table 4: JEE Main Scores and Percentile for board ABC (sample data)

**Step 4 – Calculate B _{1}**

To find B_{1}, find out JEE (Main) score, corresponding to the matched Percentile. This is B_{1}.

Here, B_{1} = 48.

**Step 5 – Calculate B _{2}**

To find B_{2}, find out JEE (Main) score, corresponding to the matched Percentile above. This is B_{2}.

Here B_{2} = 80.

**Step 6 – Calculate B _{final}**

B_{final} = 0.5 * (B_{1} + B_{2})

–> B_{final} = 0.5 * (48+80)

–> B_{final} = 0.5 * (128)

–> B_{final} = 64

## JEE Main 2017 Percentile and Normalization of Class 12 Marks / Qualifying Examination

The percentage of marks and percentile are totally different entities hence the candidates should not confuse with both the above terms.

Percentage is a number out of 100.

Percentile Score of a candidate in a Board or JEE (Main) will reflect how many Candidates have scored below that candidate in his/her Board or JEE (Main) Examination.

A Percentile score is the value below which a certain percent of observations fall. For example, the 40th Percentile is the value or score below which 40 Percent of the observations may be found.

The Percentile of a Candidate will be calculated as

**100 X Number of candidates in the ‘group’ with aggregate marks less than the candidate / Total number of the candidates in the ’group’**

**Example:** Suppose in a particular Board:

No of Candidates Registered =13918 and No of Candidates Appeared = 13711

a. A Candidate who has scored 50% marks in the Board and 2218 candidates have scored below him; his Percentile score will be calculated as follows

Percentile score for 50% marks in the Board = (2218 x 100)/(13711) = 16.18

b. A Candidate who has scored 60% marks in the Board and 6865 candidates have scored below him; his Percentile score will be calculated as follows

Percentile score for 60% marks in the Board = (6865 x 100)/(13711) = 50.07

c. A Candidate who has scored 90% marks in the Board and 13615 candidates have scored below him; his Percentile score will be calculated as follows

Percentile score for 90% marks in the Board = (13615 x 100)/(13711) = 99.30

With these examples, it is clear that percentage of marks obtained by a candidate (50%, 60% or 90%) is different from the percentile score (16.18, 50.07 or 99.30).

The normalized Qualifying examination marks will be based on the position of the candidate in the Board (i.e. his/her percentile score). The percentile score a candidate gets in his/her board will mainly contribute to his/her normalized marks.

The candidates, willing to appear in the improvement examination to improve the qualifying examination marks, will have to appear in all the five subjects for improvement.

In the percentile system, the total marks (of all five subjects) obtained/shown on the mark-sheet of a particular year are required in JEE (Main) and also in JEE (Advanced). The marks from two different years (i.e. marks for 3 subjects from year 2015 and marks for other 2 subjects from 2017) mark-sheets cannot be considered for percentile calculation.

**For JEE (Advanced) 2017:** Candidates can check more information about JEE (Advanced) from official website jeeadv.iitb.ac.in or Click Here for JEE (Advanced) 2017 criteria.

The weightage of normalized qualifying examination marks is only for deciding ranking of JEE (Main) which will be used for admission to all Centrally Funded Technical Institutions (CFTIs)/ NITs/IIITs but excluding IITs.

The candidates who are appearing in the improvement examination to improve their qualifying examination marks will get only one chance to inform JEE (Main) Unit as to which year qualifying examination marks to be considered for the purpose of declaration of final merit. The candidates are advised to regularly visit JEE (Main) website and newspapers for the notification regarding above.

The five subjects will be taken into account for calculation of percentile and normalization of qualifying examination marks for paper 1(B.E./B. Tech.) and paper 2 (B. Arch./B. Planning)of JEE (Main) which are:

- Language,
- Physics,
- Mathematics,
- Any one of (Chemistry, Biology, Biotechnology, Technical Vocational Subject),
- Any other subject.

**Note:** If a candidate has appeared in six subjects in the qualifying examination, the subject (fifth or sixth) with better marks will be considered.

In order to calculate normalized qualifying examination marks following data is considered.

Marks of all the students in that Board whose subject combinations meet the eligibility criteria of JEE-Main.

Marks of all the students in JEE (Main) – 2017.

## Normalization Procedure Adopted in JEE(Main)-2017 for Admission to NITs/IIITs/CFTIs on the Basis of Class 12th Qualifying examination marks

The detailed procedure for normalization of Qualifying examination marks is as follows:-

i. Note down the aggregate marks (A_{0}) obtained by each student in JEE- Main.

ii. Compute the percentile (P) of each student on the basis of aggregate marks in his/her own board (B_{0}) computed from the list of five subjects specified (each marked out of 100). The percentile is to be computed among all students of the board whose subject combinations meet the eligibility criteria of JEE-Main. The variable B0 is only a base for calculating percentile (P), which is further used to get corresponding JEE (Main) marks.

iii. Determine the JEE- Main aggregate marks corresponding to percentile (P) at the All- India level. Regard this as B_{1}.

iv. Also, determine the JEE- Main aggregate marks corresponding to percentile (P) among the set of aggregate scores obtained in the JEE-Main by the students of that board. Regard this as B_{2}. The normalized board score of the candidate was computed as:

**B _{final} = 0.5 * (B_{1} + B_{2})**

For the purpose of admission to CFTIs where it has been decided to use the JEE Mains performance and the Normalized Board performance in the 60:40 ratio, the composite score for drawing the merit list was computed as:

**C = 0.6 * A _{O} + 0.4 * B_{final}**

Five subjects to be used for normalization:-

- Physics
- Mathematics
- Any one of the subjects Chemistry, Biology, Biotechnology and Technical Vocational subject
- One language
- Any subject other than the above four subjects.

In respect of 3, 4 and 5, the best mark in a given category will be chosen.

**If you have any query regarding this topic please post your queries HERE**