Bored so came up with formula to predict if you will get into MIT

Colleges & Universities A-Z Massachusetts Institute of Technology
1

<p>Yeah so this is a just for fun formula that I came up with to predict your chances at getting into MIT (or any school) using a standard normal curve and an exponential model to predict percentages from there. The first part is pretty straight forward, but the parts after are my experimental twists.</p>

<p>Take your math and CR scores and divide by 20 (if the school looks at the writing section, add in the writing section and then divide by 30). Do the same with your two highest SAT scores. Take your weighted GPA and divide by the highest weighted GPA in your class and multiply that by 80. Add those numbers together. Now use the table below for your weighted academic ranking (this table is for MIT; it will vary from school to school)</p>

<p>230-240 = 5
220-229 = 4
210-219 = 3
200-209 = 2
199 and less = 1</p>

<p>Rank your extracurriculars and awards in a similar 1-5 scale. Most should have EC/awards ranging in the 2-4 area. I am still working on how to judge this part, but 5 should be national/international, 4 state/province/big district, 3 local/school, 2 no recognition/leadership at any level but participation in local/state level, 1 negligible ECs (again, this applies to MIT; schools of less rigor would probably have more lenient scoring).</p>

<p>Essays also should go on a 1-5 scale, but again, this is very subjective. Grade essays on adherence to topic, creativity, alignment with the school's qualities.</p>

<p>Please, feel free to use decimals for greater accuracy.</p>

<p>Square the three scores and add them, then take the square root of them. Divide by square root of 75 (highest possible score) and then multiply by 5. This gives you your raw overall score.</p>

<p>The distribution is assumed to be a standard normal distribution, with 5 being 3 standard deviations away from the mean (score of 3). Calculate your z-score by taking your raw score, subtract 3, then multiply by 1.5 (standard deviation is 2/3). Now I feel that a linear model of acceptance rate would be quite unrealistic. So I used an exponential model (with e as the base). Use the following formula to determine your chance of acceptance (as a percentage):</p>

<p>e^((ln(%accepted)+(zscore)ln(100/%accepted)/3)</p>

<p>For MIT, %accepted would be around 10. Check it out and see what you think.</p>

2

<p>hahaha nice</p>

3

<p>I only have a 8% chance, going to go withdraw my apps :(</p>

4

<p>can you just make an Excel file with those formulas programmed? :)</p>

5

<p>47.8%</p>

<p>Don’t know how the heck you came up with all these, but good job; it was quite interesting =p</p>

6

<p>49.9%. Really interesting.</p>

7

<p>Hey man what about the ACT? People from the Midwest get in too, you know :P</p>

8

<p>22.3%</p>

<p>Better than expected haha</p>

9

<p>33.57% .</p>

10

<p>Hmm, I forgot about ACT. Just use a conversion chart to SAT for that. Also, this is supposed to be a raw type of chances. I still have to work out the whole URM/special cases/recruited athletes into the formula…but for the run of the mill applicant (hooked extracurricular/awards wise or normal unhooked) this seems to still work out fine. When I did mine out it was in the 40s somewhere. What I figure is you have a good shot if you are in the top %admitted (zscore of ~1.3 for MIT) and if you are in the top .3% (zscore ~2.75), then you have an extremely high chance of admission.</p>

11

<p>Although my chances are higher than what I think they are even when I lowballed my GPA and ECs, I still don’t like the idea of it. It seems a bit faithless to quantify a subjective decision.</p>

12

<p>Its not quantifying a decision. It is coming up with a rough estimate of what your real chances are using your academic stats, scaled rating of your extracurriculars and awards (with a bit in between if you fit in between a category), and a subjective rating of your essays and recommendations which are rated by a certain set of criteria. When people see 10% admitted, that doesn’t necessarily mean you personally have a 10% shot. Some people have a higher shot than others. Yes, I understand that the top 10% do not automatically get in, that’s why my formula returns a percent chance of you getting in. It is not something that decides whether you are in or not; I find it to be a more useful tool than a chance thread.</p>

13

<p>Where did you get your weightings from for the various components, by the way? Was it just from logic, or did you do research, or what?</p>

14

<p>haha, my chance seems to high to be true…70.6%??? I legitimately received a 5 for academics, gave myself a 3 for essays which should be about what they’ll end up as since they’ve been corrected two or three times…</p>

<p>I gave myself a 4 for extracurrics, which I’m not sure about; but, I’m pretty high up in ASB at my school, have put on a few strong performances at math competitions, and Coach Dill of the Men’s Volleyball team insists that his recs to the admissions office are taken seriously.</p>

<p>That being said, my z-score with 5, 4, 3 was 1.62…which puts me in the top 5-6%? I’m happy of course, but really hesitant to get my hopes up -_-</p>

15

<p>eevvaann22 I think you made a mistake with your calculation. It should be 34.773% by my formula, not 70.6%.</p>

<p>As to developing the formula…</p>

<p>I originally watched an MIT admissions video and went to a Splash class at MIT on admissions. From the MIT admissions video, I found that what they do is that they rank the students on a scale, with one axis being academics ranging from 1-5 and the other being extracurriculars/awards. I added in essays and recs. Then I figured that the distribution of students applying probably follows a normal curve, with most applicants falling in the 3 range and very few being at the extremes. Using that I went on an assumption that chances of admissions does not increase linearly, from looking at Princeton admissions statistics. It probably rises/falls the most at the extremes and is little affected near the mean. Since the normal curve is based on a natural exponential function, I also based my % chances of acceptance on a natural exponential function.</p>

16

<p>Interesting. Well, one thing I’ve learned in my classes is that if you set out with a set of data and try to come up with some sort of statistic that fits, you can slap a bunch of logical (or random) coefficients on your prediction formula and get stuff that is kind of legit for some data points and stuff that is extremely, EXTREMELY inaccurate for other data points. While your formula is cute, I’d urge you to use some real-life examples and try to alter the coefficients (or the relationships between the variables) if you’re actually interested in creating something that will predict admission or rejection. Otherwise, it is a cute project, but I’m unconvinced of its accuracy.</p>

17

<p>Oh I completely understand that. I just made this up late one night last week and then posted it the next day. I plan on modifying it, but I think that right now, for a first draft, it works out pretty well. I will have to do some more research on the topic to further develop my model.</p>

<p>EDIT: What I could do is ask people who are applying to MIT on this board to test drive the formula so that I can see if there is a correlation between higher scores and admissions. They would just have to provide the ratings on essays/recs and I would do the rest of the work for calculating their scores.</p>

18

<p>

</p>

<p>My EC actually did explain something similar to me, so yes.</p>

<p>I don’t think it’s very accurate regardless. If I’m reading this correctly three 5s would mean a chance of 100%, which is obviously unrealistic. Also, fix those parentheses. Is it</p>

<p>e^((ln(%accepted)+(zscore)ln(100/%accepted))/3)</p>

<p>or e^(ln(%accepted)+(zscore)ln(100/%accepted)/3)?</p>

19

<p>e^(ln(%accepted) + ((z-.01)(ln(100/%accepted)))/3)</p>

<p>(See I added that little -.01 to the z so you never get to 100). Dividing by three is only for the last ln part, not for the whole thing, so my original one was correct (I think). The two people who got into MIT who used it reported high 40s for chances. I do agree that I need more data to work with, which I will start looking for after I submit my own app to MIT and after I finish my linear algebra course (ends a few days after Nov 1st).</p>

20

<p>

</p>

<p>Were you in my class?! </p>

<p>LOL.</p>