Why CUET-UG ‘normalises’ marks, and how it differs from other scoring systems

More than a week after the announcement of the CUET-UG (Common University Entrance Test-Under Graduate) results, students continue to have doubts over the assessment and scoring system followed by the National Testing Agency (NTA) in preparing “normalised” marks that will be considered by the universities in carrying out admissions.

The NTA and the University Grants Commission (UGC) maintain that the normalisation of marks is the best method available to offer a level playing field to all candidates who took the entrance exam, saying it was drawn up by a committee of experts from the Indian Statistical Institute, IIT Delhi, and Delhi University.

What is ‘normalisation’?

The CUET-UG was held on multiple dates and multiple sessions between July 15 and August 30. As a result, there were different sets of question papers for every subject. Naturally, there were variations in the level of difficulty of question papers on the same subject. Moreover, in the case of CUET-UG, there were apprehensions that those taking the test in the initial set of dates may not be able to meet the performance level of candidates who appeared in dates fixed in August. The normalisation formula is supposed to address these concerns by ensuring that candidates are neither benefitted nor disadvantaged due to any such factor.

Is this method applied in the case of all entrances?

Yes, normalisation is used to compare the performances of students on a common scale even in the case of entrances such as JEE Main (engineering). However, there is a difference. In the case of JEE Main, the percentile scores are normalised. Before going further, we need to know what a percentile is. Simply put, percentile denotes the relative performance of a candidate compared to those who took the test in the same session. A 100 percentile indicates that 100% candidates of a particular session have scored equal to or less than the topper. A student with a percentile score of 70 would indicate that she has scored more than 70% of the candidates of her session.

How is CUET-UG different?

In the case of JEE Main, the raw marks obtained by the students are first transformed into a scale ranging from 0 to 100 for each session of examinees. As a result, the toppers of two different sessions may have different raw scores, but both their percentile scores will be 100. The same applies to each category of percentile. But in the case of CUET-UG, raw scores are converted into percentile scores and then into normalised scores. But universities will consider only the normalised raw score while preparing merit lists. This method has been shown to be accurate for estimating normalised marks of candidates when the tests are held in multiple sessions with varying difficulty levels in a given subject, according to the NTA.

What is the mathematical formula used for a normalised percentile?

To calculate the normalised percentile score of a candidate, three variables are required: the raw score of the candidate (A), the total number of candidates with a score equal to or less than her raw score (B), and the total number of candidates in that particular session (C). The mathematical formula applied is B divided by C multiplied by 100. For example, take the case of a student who appears in a session with 41,326 other candidates. Say, she obtains 121 marks, scoring more than 37,244 other candidates of her session. Her normalised percentile score will be 37,244/41,326X100, which comes to 90.1224411. Percentile scores will be calculated up to 7 decimal places.

And what is the formula to calculate a normalised raw score?

In tabulating the marks, the NTA first sorted the candidates in a descending order based on their percentile scores first. Then the raw marks of the students were listed against their percentiles. A third metric called ‘interpolated mark’ was calculated for each candidate using a method called linear interpolation. To understand its meaning, consider a person covering a certain distance at a uniform pace. Now, through linear interpolation one can find out the possible time that she would take to cover one third of the distance (or two third or any other length of the total distance). This mathematical concept was applied to find the interpolated marks for every student, assuming that every student appeared in the same paper twice. Their actual raw marks were then divided by the interpolated marks, and the normalised marks arrived at.


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