What percentage of ppl at T-14 at/above median?? Forum

[mec30]

And it's relevent to your question because if you changed the data set to {3.8, 3.6, 3.3, 3.3, 3.3, 3.2, 3.1, 2.9} using the way he calulated the last problem you would end up with a median at 3.31x instead of 3.3.

median is wholly dependent on data set. So what I did and what you've just done are entirely different things. My calculation was not relevant to yours whatsoever.

There are only two ways I know to calculate a median - by taking average of the standard deviations of each number in the set, or by picking the middle number of the set, and when no such number exists, averaging the two middle numbers (which is to say the set has no median).

[CG614]

I think the "I.e. 3.41" was just referring to a random datapoint between 3.4 and 3.5, not actual median calculation. Regardless, the example doesnt really relate to my question. My question specifically dealt with situations where the calculated median would not be a unique value(i.e. the calculated median is 3.4 and a lot of people have 3.4 gpas.).

Not going to happen. GPAs can go out many decimal places, the likelihood of that happening is worse than the topic of this thread. And for HYS, who cares?

[CG614]

I think the "I.e. 3.41" was just referring to a random datapoint between 3.4 and 3.5, not actual median calculation. Regardless, the example doesnt really relate to my question. My question specifically dealt with situations where the calculated median would not be a unique value(i.e. the calculated median is 3.4 and a lot of people have 3.4 gpas.).

Even though this scenario is completely ridiculous, here is your answer. Median is actually defined as a value where no more than 50% is above or below said value. In your unlikely scenario, all the 3.4s would be said to be at the median and the rest would be above or below.
/thread

[zanda]

I think the "I.e. 3.41" was just referring to a random datapoint between 3.4 and 3.5, not actual median calculation. Regardless, the example doesnt really relate to my question. My question specifically dealt with situations where the calculated median would not be a unique value(i.e. the calculated median is 3.4 and a lot of people have 3.4 gpas.).

Even though this scenario is completely ridiculous, here is your answer. Median is actually defined as a value where no more than 50% is above or below said value. In your unlikely scenario, all the 3.4s would be said to be at the median and the rest would be above or below.
/thread

He was aware of this. That was his question. See thread's title.

Was it a silly question? Maybe... It wasn't as dumb as some are pretending it was.

[CG614]

I think the "I.e. 3.41" was just referring to a random datapoint between 3.4 and 3.5, not actual median calculation. Regardless, the example doesnt really relate to my question. My question specifically dealt with situations where the calculated median would not be a unique value(i.e. the calculated median is 3.4 and a lot of people have 3.4 gpas.).

Even though this scenario is completely ridiculous, here is your answer. Median is actually defined as a value where no more than 50% is above or below said value. In your unlikely scenario, all the 3.4s would be said to be at the median and the rest would be above or below.
/thread

He was aware of this. That was his question. See thread's title.

Was it a silly question? Maybe... It wasn't as dumb as some are pretending it was.

It is a dumb question in that it wouldn't happen using continuous values like GPA.

[zanda]

I think the "I.e. 3.41" was just referring to a random datapoint between 3.4 and 3.5, not actual median calculation. Regardless, the example doesnt really relate to my question. My question specifically dealt with situations where the calculated median would not be a unique value(i.e. the calculated median is 3.4 and a lot of people have 3.4 gpas.).

Even though this scenario is completely ridiculous, here is your answer. Median is actually defined as a value where no more than 50% is above or below said value. In your unlikely scenario, all the 3.4s would be said to be at the median and the rest would be above or below.
/thread

He was aware of this. That was his question. See thread's title.

Was it a silly question? Maybe... It wasn't as dumb as some are pretending it was.

His question implies that there could be more that 50% above median.

His question implies that there could be more than 50% at or above median.

[CG614]

His question implies that there could be more than 50% at or above median.

I realized that and edited my post.

[zanda]

His question implies that there could be more than 50% at or above median.

I realized that and edited my post.

Word. I do think that at a school like NYU where very few people get anything below a B or above an A- there are probably a decent number of people at the median. Is it so large that it's worth thinking about? Probably not.

[CG614]

His question implies that there could be more than 50% at or above median.

I realized that and edited my post.

Word. I do think that at a school like NYU where very few people get anything below a B or above an A- there are probably a decent number of people at the median. Is it so large that it's worth thinking about? Probably not.

a+ 2%
a 5%
a- 30%
b+ 30%
b 30%
b- 3%

Grades
a-
b+
b+
a-
a
b
b
0.00365%

Given a distribution where 90% of the grades are a- through b, the odds of getting one particular distribution is really low. Even at NYU, it is probably very unlikely.

[zanda]

sorry... not clear on where you got .00365.

you are aware that the distributions towards the middle are more likely than the outside right?

getting all A- = .3 to the 6th (we have 6 grades 1L year), while if we consider getting a B+ average...
1) getting 2 A-, 2 B +, and 2 B is more likely AND not only that but
2) there are multiple ways of getting the same average (6 B+, 1-4-1, 2-2-2, 3-0-3)

In other words there are 27 possible distributions if we simply further and say that there are only A-, B+, B. Four of these 27 distributions give a B+ average. Among these distributions are 2-2-2, which is the most common, and 1-4-1 and 3-0-3 which are somewhat common, and 6 B+ which is .3 to the 6th.

I admit that my combinatorics are rusty and that I can't think of an easy way to figure out the probability. It's possible your number is correct... I'm just not sure how you got it.

[CG614]

sorry... not clear on where you got .00365.

you are aware that the distributions towards the middle are more likely than the outside right?

getting all A- = .3 to the 6th (we have 6 grades 1L year), while if we consider getting a B+ average...
1) getting 2 A-, 2 B +, and 2 B is more likely AND not only that but
2) there are multiple ways of getting the same average (6 B+, 1-4-1, 2-2-2, 3-0-3)

In other words there are 27 possible distributions if we simply further and say that there are only A-, B+, B. Four of these 27 distributions give a B+ average. Among these distributions are 2-2-2, which is the most common, and 1-4-1 and 3-0-3 which are somewhat common, and 6 B+ which is .3 to the 6th.

I admit that my combinatorics are rusty and that I can't think of an easy way to figure out the probability. It's possible your number is correct... I'm just not sure how you got it.

Yeah, it was just the probability of that one distribution, but you are right. Distributions towards the mean are more likely, and hence toward the median. Been a while since my stats class.. ha

[cahesu]

Hey sorry but I'm really rusty too...

can you tell me how you got 27 distributions if we assume 6 courses and 3 possible grades?

[zanda]

Hey sorry but I'm really rusty too...

can you tell me how you got 27 distributions if we assume 6 courses and 3 possible grades?

I have to admit I wrote them out.

There is 1 possibility with 6 A-s. 2 possibilities with 5 A-s (1 B+ or 1 B). 3 Possibilities for 4 A-s (2 B+, 2 B, or 1 of each). The pattern goes like that (4 possibilities for 3A-s, 5 possibilities for 2, 6 for 1, 7 for 0.

By distribution I meant total number of each category, such that BBBBBB+ is the same distribution as BBBB+BB.

Correction- that makes 28, not 27.

[09042014]

http://ocw.mit.edu/courses/mathematics/ ... pring-2005

For the love of science please do this.

[270910]

http://ocw.mit.edu/courses/mathematics/ ... pring-2005

For the love of science please do this.

That's a nice picture showing the union of events you have there DF.

[let/them/eat/cake]

I think this is my favorite thread in a long, long time. the initial bandwagon clusterfuck of responses was GD priceless.

[zanda]

I think in our simplied example it would be approximately 15% of the class with exactly a B+ average... I got 729 possibilities total (3 to the 6th), and 114 possibilities with a B+ average (1 with all B+, 17 3-0-3, 30 1-4-1, and 66 2-2-2)... my counting could be off, but I think that's close-ish. keep in mind this is an oversimplified curve where 1/3 get A-, 1/3 get B+, and a 1/3 get B. Another problem with it is that the grades aren't truly independent... someone with 5Bs has less than a 1/3 chance of getting an A- in the 6th class... although near the middle they're probably fairly close to independent since we're only allowing for 3 different grades in our example anyway.

[chicagolaw2013]

Jesus holy fucking Christ my ass.

So epic I almost just died.

[arstech]

This entire thread can be summed up in three bullet points:

- In a data set such as {4.0, 3.3, 3.3, 3.3, 2.6}, there is no way of choosing a median that divides the class equally.

- OP was inquiring about the extent to which this occurs.

- betasteve needs to retake Junior High arithmetic (to learn math) as well as kindergarten (to learn how to play well with others)

[CG614]

This entire thread can be summed up in three bullet points:

- In a data set such as {4.0, 3.3, 3.3, 3.3, 2.6}, there is no way of choosing a median that divides the class equally.

- It can be divided evenly. 1 above median, 1 below, and 3 at median.

- OP was inquiring about the extent to which this occurs.

- OPs question was initially aimed at HYS, once OP realized HYS doesn't rank, they tried to make it about GPA. This doesn't make sense because there will never be a large enough clump at median to ever care about. Statistically it is VERY unlikely, given the nature of the data.

- betasteve needs to retake Junior High arithmetic (to learn math) as well as kindergarten (to learn how to play well with others)

-betasteve was correct in his math.

*How about those three bullet points?

[PKSebben]

.

[arstech]

It can be divided evenly. 1 above median, 1 below, and 3 at median.

Obviously I meant divided in two, per his original statement:

Just no. Look, the class is either going to have an odd number of students or an even number of students. If there are an even number of students, then the school will pick the lowest student in the top half, and that will be above median. The rest will be below median, and the division will be equal. If, on the other hand, there are an odd number of students, then it is true that the division will not be exactly 50/50. But, it will only be off by 1 student. That's just the way a median works.

[ResolutePear]

This is exactly why stats should be a required course for law school.

Just for fuck's sake; you learn about median(the basics) in 4th grade and some more in basic Algebra.


Oh, and TROLOLOLOLOL.

[mallard]

This entire thread can be summed up in three bullet points:

- In a data set such as {4.0, 3.3, 3.3, 3.3, 2.6}, there is no way of choosing a median that divides the class equally.

- OP was inquiring about the extent to which this occurs.

- betasteve needs to retake Junior High arithmetic (to learn math) as well as kindergarten (to learn how to play well with others)

No, that's not right. The OP was wondering about a situation like {4.0, 3.8, 3.6, 3.5, 3.4, 3.4, 3.4, 3.4, 3.3}.

[mec30]

OPs question was initially aimed at HYS, once OP realized HYS doesn't rank, they tried to make it about GPA. This doesn't make sense because there will never be a large enough clump at median to ever care about. Statistically it is VERY unlikely, given the nature of the data.

That might be true in UG where many people are taking different class loads with a different number of units per class (and consequently a different GPA weighting), but it would be much more likely after 1L where everyone has the same schedule and the curve is tightly controlled.

And the OP specifically referenced grades in his first post in addition to what would be considered median (in any sense of the word).