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Normalized mark of $j^{th}$ candidate in $i^{th}$ session, $\widehat{M}_{ij}$ is given by

$\begin{equation*} \widehat{M}_{ij} = \dfrac{\bar{M}^g_t - M^g_q}{\bar{M}_{ti} - M_{iq}} \left( M_{ij} - M_{iq} \right) + M^g_q \end{equation*}$

where

${M}_{ij}$ is the actual mark scored by $j^{th}$ candidate in $i^{th}$ session,

$\bar{M}^g_t$ is the average mark of the top 0.1% students considering all sessions,

$M^g_q$, is the sum of the mean and standard deviation marks considering all sessions, 

$\bar{M}_{ti}$, is the average mark of the top 0.1% students in $i^{th}$ session,

$M_{iq}$, is the sum of the mean and standard deviation marks in $i^{th}$ session.

I hope most people won't be getting anything from this which is given in GATE website. So, let me explain you, if $x$ is the actual mark obtained by you in GATE, it'll be normalized to 

$$\alpha x + \beta$$,

where $\alpha$ and $\beta$ are two constants for the session you appeared in. (From 2014 on wards GATE CS&IT has 3 sessions).

Now, $\alpha$ and $\beta$ can be derived from the previous formula given. For better understanding I'll explain with the end result of normalization in GATE 2015. 

Session 1: Normalized mark = 1.0638 * AM + 1.81

Session 2: Normalized mark = 1.066 * AM - 5.172

Session 3: Normalized mark = 0.941 * AM + 2.270

Just looking at this shows that Session 2 people are doomed. Or we can say session 2 was the easiest paper and people did well in it. But based on feedback, session 2 paper was indeed the toughest, though no one can prove this. The statistics collected are as shown:

https://docs.google.com/spreadsheets/d/1Ql6D0uD-ljPxielCzpxyXrhGvt3o4T4csYMzalPvTIk/pubhtml

Just a bit of solving shows the following

  • M2 = M1 + 6.58, Mi is the sum of mean and SD of session i.
  • T2 = T1 + 6.58, Ti is the mark of top 0.1% in session i.

This do justify the +7 mark difference session 1 people got compared to session 2. But the problem also starts here. 

If you see the spreadsheet, the average mark for top 0.1% is around 65 marks. 

115k students appeared in GATE CSE and it can be assumed they are divided equally to 3 sessions- 38k each. Now, 0.1% would mean 38 students from each session. 

Marks of the first 38 students of session 2 is more than that of session 1 students by 6.5 marks on average. 

Does this mean session 2 was easier scoring?

No. because the top mark was more than 15 marks higher than the average of top 100, and a few toppers can really swing the average their way. 

  • Avg. mark of top 38 in session 2 = 66
  • Avg. mark of top 38 in session 1 = 59.5
  • I got 60 marks in set 1, I get normalized to 58.8
  • I got 60 marks in set 2, I get normalized to 65.6 (6.8 marks difference)

Now, lets take 5 students who got around 75 marks in session 2 (there were as seen in the spreadsheet) and swap them with 5 students who got 55 in session 1 (assuming they got allotted to another session). 

  • Avg. mark of top 38 in session 2 = 64.8
  • Avg. mark of top 38 in session 1 = 62.13
  • I got 60 marks in set 1, I get normalized to 59.75
  • I got 60 marks in set 2, I get normalized to 62.9 (just 3.15 difference)

So, depending on how a few select people get allotted to different sessions my normalized mark is greatly affected. 6 marks is all it takes for me to get to the best IIT and to not getting any IIT. And this much mark I can loose depending on how a select brilliant students take exam in which session. I don't think this is justifiable. So, I request to consider the following for GATE 2016

  1. Do away with normalization and do the exam in one session
  2. Make normalization formula better by using different formula for different mark range
  3. Or at least just remove normalization and take the responsibility to make all the exam papers of equal difficulty. It is way better than losing 7 marks simply like that.
by Veteran (432k points)
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0
This point is really valid. Top 0.1% is a handful of students, their average might not correlate with the difficulty of their set. I think taking top 1% is a better idea.
0
This year go rank predictor is showing normalization of about 5.5.I thought it should go around 7 seeing the difference between the two sessions.
0
@Aditya Relative difference is like 7+.
0
Sir,But my original marks in set 1 is 62 and normalized is showing 67.6.So normalized is around 5.5 right?
0
yes.. but those getting 67 in set 2 is normalized to around 64..
0
Also this increase is even more for those who scored 70+
0
Understood sir.Thanks:)
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