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GATE Normalization Process: It is a process used for converting raw marks into scores. After the evaluation of answers, the raw marks (actual marks obtained in GATE) will be considered for calculating GATE Score based on qualifying marks. Read the following article to know more about GATE Normalization Process, Calculation of Scores and Qualifying Marks.
As GATE being a major exam, lacs of candidates appear for it and it is not possible to conduct the exam in one go. So, GATE exam is held in multiple sessions. Therefore, it is quite natural that there will be variations in the level of difficulty of the exam. To arbitrate the difference in the level of difficulty, the normalization process is applied.
The following papers in GATE 2020 will be conducted in multiple sessions-
The basic presumption behind the concept of normalization is that "in all multi-session GATE papers, the distribution of abilities of candidates is the same across all the sessions". Now, based on the following parameters, and taking into account various normalization methods, the committee has arrived at the following formula for calculating the normalized marks for the multi-session papers.
Normalization mark of Jth candidate in the ith session M̂ij
Mij : is the actual marks secured by the jth candidate in the ith sessions
M̅gt : is the average marks of the top 0.1% of the candidates considering all sessions
Mgq : is the sum of mean and standard deviation marks of the candidates in the paper considering all sessions
M̅ti : is the average marks of the top 0.1% of the candidates in the ithsessions
Miq : is the sum of the mean marks and standard deviation of the ith session
Further, the graph can help understand the relationship between the two scores better.
Graph showing the linear relationship between "actual marks" and "normalized marks" of a candidate, in a multiple-session subject (CE, EE, ME, EE or CE) of GATE.
Once the answers are checked and assessed, the normalized marks of a candidate will be calculated corresponding to the actual marks secured by the candidate in the examination and the GATE 2020 Scorecard will be devised based on the normalized marks. The papers which are actually conducted in a single session only, the actual marks secured by the candidate will be used for calculating the GATE 2020 score.
After the normalization of marks, the final GATE 2020 score will be calculated using the following formula;
In this formulae,
M : marks secured by the candidate in the individual papers (actual marks for single session papers and normalized marks for multi-session papers)
Mq : is the qualifying marks for general category candidate in the paper
M̅t : is the mean of marks of top 0.1% or top 10 (whichever is larger) of the candidates who appeared in the paper (in case of multi-session papers including all sessions)
Sq: 350, is the score assigned to Mq
St: 350, is the score assigned to M̅t
In the GATE 2020 formula, Mq is normally 25 marks (out of 100) or +σ, whichever is larger. In this case, μis the mean and σ is the standard deviation of marks of all the candidates who appeared in the paper.
GATE 2020 scorecards can be downloaded by candidates from different categories after the declaration of results.
The candidates should have a brief idea about the qualifying marks for different streams and papers in GATE 2020 on the basis of previous year’s cut off. The cut off marks for individual papers have not varied too much for individual papers in last few years.
However, with no.of candidates varying in every session, there is no fixed pattern or margin for GATE score. The table below contains a separate cut off list for different categories.
As expected the qualifying marks are higher for the general candidates for all the subjects thus requiring them to score better marks in GATE 2020 in order to get qualified in all the individual subjects.
|Computer Science (CS)||25.00||22.50||16.60|
|Civil Engineering (CE)||26.90||24.20||17.9|
|Electrical Engineering (EE)||29.1||26.1||19.4|
|Mechanical Engineering (ME)||34.7||31.2||23.1|
|Electronics and Communication (EC)||25.0||22.5||16.6|
|Production Engineering (PI)||32.0||28.8||21.3|
|Instrumentation Engineering (IN)||37.0||33.8||24.7|
|Aerospace Engineering (AE)||25||22.5||16.6|
|Chemical Engineering (CH)||37.8||34||25.1|
|Geology and Geophysics (GG)||32.4||29.1||21.5|
|Architectural Engineering (AR)||43.9||39.5||29.2|
|Agricultural Engineering (AG)||25.0||22.5||16.6|
|Petroleum Engineering (PE)||39.8||35.8||26.5|
|GATE Paper Code||No of Candidates Appeared||GATE Paper||Qualifying Marks FOR (General)||Qualifying Marks FOR (OBC-NCL)||Qualifying Marks FOR (SC/ST/PwD)|
|EC||172714||Electronics and Communication||25||22.5||16.67|
|CS||115425||Computer Science and IT||25||22.5||16.67|
Ans: The difficulty level for all the papers is determined by analyzing density functions of that particular stream. So if we assume that the distribution of marks is denser in the region 50-60 for session 1 for a certain paper and in the same paper for session 2 it is 40-50, then, in that case, the session two will be considered easier.
However, the students should not worry as the marks get normalized through a pre-defined method to calculate individual score and rank. One can, therefore, say that the calculation of score is genuinely fair irrespective of the difficulty level of different sessions.
Ans: Yes, it is true. In case a candidate has appeared for the easier session then their normalized marks will be lesser than the actual marks they had secured.Similarly, If they have appeared for the difficult session then their normalized marks will be greater than the marks they have actually secured. Let’s take an example:
Mij : is the actual marks secured by the jth candidate in the ith session
M̅gt :is the average marks of the top 0.1% of the candidates considering all sessions
Mgq :is the sum of mean and standard deviation marks of the candidates in the paper considering all sessions
M̅ti :is the average marks of the top 0.1% of the candidates in the ith session
Miq :is the sum of the mean marks and standard deviation of the ith session
Average marks of all the candidates in all the sessions (of a particular subject, say CSE) is 30, average marks of top 0.1% is 80, assuming that the candidate has appeared in the easier session, average marks of top 0.1% in that session is 85, average marks of all candidates in your session is 35 and if candidate has secured 70 marks, then:
|GATE Expected Cutoff for NITS||GATE Round Wise CCMT Cutoff||GATE Paper Wise Cutoff|
Ans: The difficulty level does vary to a certain extent and that is the reason some of the students prefer a difficult session as there are chances of getting additional marks for an individual paper after the marks are normalized and vice-versa.
However, the selection of candidates for a particular slot is done through a computerized random process and hence beyond the control of any individual.So the aspirants should focus on preparation for individual papers rather than worrying about the difficulty level since it is completely a matter of fate.
Ans: Min limit: (0.9)*your marks; Max limit: (1.15)*your marks. However, there is no predefined limit. Therefore one cannot presume that if someone is getting an easier set, then their marks are going to be decreased by 10%. It also depends on the other factors/variables in that section. The above formulas are meant to provide a rough boundary only.
Ans: A candidate’s percentile indicates the percentage of candidates scoring lower than that individual candidate in the GATE exam. It is calculated through this formula:
AIR RANK= It is the all India rank of a particular candidate.
N= It is the total no.candidates appearing in that particular stream.