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The process of normalization defines the process in which the distribution and calculation of marks is carried out based on the percentile count of the candidate’s score in both the State board he/she participated and the score obtained in the JEE Main 2017. This process ensures that students are neither advantaged nor disadvantaged by the difficulty of exams that they do for the Boards. This process basically works on the criteria of percentile which must not be linked to the percentage criteria as both of them are entirely different entities.

The percentile score of any candidate in **JEE Main** and Board is defined by the number of candidates who have scored aggregate marks below the score of candidate in the respective Board or **JEE Main 2017**. The basic quantities to be known for the calculation of the percentile are the number of candidates scoring below the given candidate’s score and the total number of candidates appeared for the examination. Thus we derive a formula for the calculation of the percentile score which is defined as the product of number of candidates below the candidate’s score with **100 **divided by the total number of candidates participating in the examination.

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**Percentile Score** = (100x number of candidates with aggregate marks below the candidate’s score) / Total number of candidates appearing for the examination

**Example**, A candidate has scored 60% marks in the given Board examination and there are **13170 candidates** participating in the examination. The number of candidates below him/her is **6865**, then the percentile score of the Candidate can be calculated as follows

**Percentile score for the 60% Board marks** = (100x6865) / 13170= 52.12

Thus, Percentile = 52.12

The candidate must consider the following abbreviations which shall be used further to comprehend the calculations and methodology of the normalization process

A_{0 }= Aggregate marks of candidate in JEE Main

B_{0 }= Aggregate marks of candidate in Board examination

P_{i} = Percentile marks of candidate as per the Board

B_{1} = Aggregate marks corresponding to percentile at All India level

B_{2} = Aggregate marks of scores of candidates of same board in **JEE Main**

B_{final} = normalized Board score

The calculation of percentile based on the marks obtained by the candidate appearing for **JEE Main 2017** in their specific board examination. The State Board Normalisation is the process which considers the number of candidates scoring marks below the marks of the candidate and the total number of candidates appearing for the same exam.

- The normalized Qualifying examination marks will be based on the percentile of the candidate (i.e. his/her position in the Board). The percentile score of the candidate adds up majorly to the normalisation of marks of the candidate.
- The candidates who aspiring to appear in the improvement examination to improve the qualifying examination marks are required to appear in all the five subjects for improvement.
- In the percentile system, the total marks i.e. marks obtained by the candidates in all the five subjects in a particular as mentioned in the mark sheet required for both JEE Main and JEE Advanced is considered. The marks from two different years (i.e. marks for 3 subjects from year 2015 and marks for other 2 subjects from 2016) mark-sheets are considered invalid for carrying out the percentile calculation.
- The weightage of normalized qualifying examination marks is only for declaration of the
**JEE Main Rank**which will be used for admission to all**Centrally Funded Technical Institutions (CFTIs)/ NITs/IIITs excluding IITs.** - The candidates who are appearing in the improvement examination to improve their qualifying examination marks shall receive just one chance to confirm the JEE (Main) Unit about the year of which the qualifying examination marks should be considered for the purpose of declaration of final merit.
- The candidates must follow the
**JEE Main official website**and newspapers regularly for the notification regarding the same. - The five subjects that will be taken into consideration 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) are:**

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

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

**Example**, If the candidate has scored 60% in the Board examination (B_{0}) and the number of candidates who scored below the candidate in the same board are 6770 and the total number of candidates applying for the same Board are 13370 then the percentile score can be calculated as

Percentile Score for 60% marks = (100x6770) / 13370

Thus Percentile Score (P_{i}) = 50.63

The method for calculation of the percentile requires few quantities in general to be known by the candidate.

The JEE Main Examination is conducted for Engineering and Architecture/Planning. The quantities required for calculation of Percentile are number of candidates below the candidate’s score in both JEE Main and the Qualifying Examination and the total number of candidates appearing for the same.

**Percentile Score** =(100x number of candidates with aggregate marks below the candidate’s score) / Total number of candidates appearing for the examination

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The final percentile (B_{final}) is calculated by calculating the average of the aggregate marks corresponding to percentile at All India level (B_{1}) and the aggregate marks of scores of candidates of same board in JEE Main (B_{2}). The formula created for the same can be written as:

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

The Composite Score is the criteria used for admissions in the CFTI’s which includes JEE Main performance of the candidate and the normalized Board Performance of the Candidate in 60:40 ratio. The formula used for the calculation is as follows:

C= 0.6 x A_{0 }+ 0.4 x B_{final}

The performance of the candidate in the **JEE Main** examination is determined by the calculation of the normalized score (B_{final}) in the **JEE Main examination**. This entity is required for the further processes in the Admission of the Candidate in various CFTI’s (Centrally Funded Technical Institutes), NIT’s (National Institute of Technology) and IIIT’s (Indian Institute of Information Technology) excluding the IIT’s. The JEE Main score obtained after the calculations aids the candidate to qualify for the **JEE Advanced 2017**.

**Basic entities to be known during the calculation process of the normalized score for JEE Main 2017 are as follows:**

A_{0 }= Aggregate marks of candidate in **JEE Main**

B_{0 }= Aggregate marks of candidate in Board examination

P_{i} = Percentile marks of candidate as per the Board

B_{1} = Aggregate marks corresponding to percentile at All India level

B_{2} = Aggregate marks of scores of candidates of same board in **JEE Main**

B_{final}= normalized Board score

**Thus the final formula to be used for the process is**

B_{final}= 0.5 x (B1+B2)

The steps involved calculation of the **JEE Mains** require the above entities. The candidate must be aware of the mentioned marks which are the basics of the method involved in bringing out the score. The steps that follow for the derivation of the final **JEE Main** score required are listed below:

- The initial step that commences the calculation based process is the calculation of Percentile score of the candidate based on the Aggregate marks of the candidate in his/her respective Board examination. The percentile score obtained at this step is abbreviated as P
_{i} - Post calculation of the initial percentile, you need to calculate the aggregate percentile marks corresponding to the same percentile in the JEE Mains 2017. The formula used shall be the same. This helps you obtain the entity B
_{1} - Now the step to obtain B
_{2 }follows after calculating B_{1}. The aggregate percentile marks corresponding to the candidate’s percentile score calculated in his/her respective board examination is calculated. - The received entities at the end helps calculate the final JEE Main score of the candidate.

The process can be easily illustrated using an example which shall help you interpret the methodology. Consider the case that follows:

**Example**, Consider if the candidate has scored 60% aggregate in his/her respective Board examination with 6865 candidates scoring below the candidate out of 13711 appearing for the exam. (The information given shall be continued later)

**Step 1 Calculation of Percentile Score considering the given quantities**

Percentile Score for 60% marks = (100x6865) / 13711

Percentile Score (P_{i}) = 50.07

Now consider if the number of candidates participating in the JEE Main 2017 are 1153667 and the candidates scoring below the candidate calculating the score be 577633.

**Step 2 Calculation of Score applicable for the similar percentile as obtained in step 1 for the JEE Main Rank**

P_{i} = 50.07

Let us consider the Score of the candidate in JEE Main to be 48

Now calculating the Percentile score of the candidate corresponding 48 marks with candidates below the existing candidate be 577633 and total number of candidates 1153667

Percentile corresponding 48 marks = (100 x 577633) / 1153667

P= 50.07

Now consider if the number of candidates of similar board exam participating in the **JEE Main 2017** are 13711 and the candidates below in scores than the existing candidate are 6865.

**Step 3 Calculation of Score applicable for the similar percentile comparing the candidates appearing for JEE Main belonging to the Same Board**

P_{i} = 50.07

Let us consider the score of the candidate in JEE Main among the Candidates of same Board be 80

Now calculating the Percentile score of candidates corresponding 80 marks with the candidates below the existing candidate be 6865 and the total number of candidates who appeared for the same Board are 13711.

Percentile corresponding 80 marks = (100 x 6865) / 13711

P = 50.07

Hence as concluded from the above steps, B_{1} = 48 and B_{2 }= 80

**Step 4 Calculation of Final Normalised Score **

Now we derive,

B_{1} = 48 (from step 2)

B_{2} = 80 (from step 3)

Now calculating, the Final JEE Main Score (B_{final})

B_{final} = 0.5 x (48+80)

= 0.5 x 128

B_{final }= 64

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