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
- Do away with normalization and do the exam in one session
- Make normalization formula better by using different formula for different mark range
- 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.