Half you jokers are below average

[Beano's Fourth Chin]

But, I bet if asked, the vast majority would consider themselves above average.

http://www.damninteresting.com/?p=406

[Howzit]

Competence is overrated.

[jk the sooner fan]

and mediocrity is highly underrated....

[BoogercountySooner]

My name is Abi Normal!!:D

[85Sooner]

My name is A.B. Normal!!:D


fixed

[Howzit]

and mediocrity is highly underrated....

Hell, I aspire to mediocrity.

[GDC]

Well, it's kind of cold here today.

[Vaevictis]

Ironically, the author is obviously below average:


... by definition, descriptive statistics says that it is impossible for a majority of people to be above average.

Actually, it doesn't. Take, for example, the following data set:

{1, 10, 10, 10, 10, 10, 10}

The average is 8.7ish. Six values are above average.

It's entirely possible for a majority to be above average if a there is a group of sufficiently small outliers at the bottom end.

[OKC Sooner]

http://www.subgenius.com/Graffix/dobbs.jpg

[OKC Sooner]

It's entirely possible for a majority to be above average if a there is a group of sufficiently small outliers at the bottom end.

So what you're saying is, taking whorns into account, sooner fans are freakin' geniuses.

I can live with that. :texan:

[Vaevictis]

So what you're saying is, taking whorns into account, sooner fans are freakin' geniuses.

C'mon. You don't need stats to tell that. You can tell just by looking at them.

[OKC Sooner]

C'mon. You don't need stats to tell that. You can tell just by looking at Hornfans.com.

fixed :D

[Vaevictis]

I don't think that my statement needed to be fixed at all. That should not, however, be interpreted to mean that I disagree with your statement ;)

[OKC Sooner]

You are correct, sir!!! :)

[jk the sooner fan]

Hell, I aspire to mediocrity.


our team motto could be "we're number 3! we dont try at all"

[Howzit]

Ironically, the author is obviously below average:



Actually, it doesn't. Take, for example, the following data set:

{1, 10, 10, 10, 10, 10, 10}

The average is 8.7ish. Six values are above average.

It's entirely possible for a majority to be above average if a there is a group of sufficiently small outliers at the bottom end.

I would assume that the author is using something akin to a Normal Distribution for the population. The Normal Distribution is also called The Law of Really Really Big Numbers (or something like that ;) ) because it assumes a very large data size, so the author is probably safe with the statement about average definitions.

[OKC Sooner]

2+2=4, except for very large values of 2

[Vaevictis]

I would assume that the author is using something akin to a Normal Distribution for the population. The Normal Distribution is also called The Law of Really Really Big Numbers (or something like that ;) ) because it assumes a very large data size, so the author is probably safe with the statement about average definitions.

I'm thinking you're meaning the "Law of Large Numbers", which only states that as the size of a sample increases, the mean of the sample will approach the mean of the population. Statistically speaking, if you increase the sample size, all you're doing is reducing the range of uncertainty (margin of error) with respect to how well your sample represents the population.

Really, he's not safe at all. Just take my set and add an infinite number of 10's to it. The mean will asymptotically approach ten; the majority of the values will *still* be higher than than the mean. Increasing the sample size does you no good if the population really is skewed that way.

[Howzit]

I'm thinking you're meaning the "Law of Large Numbers", which only states that as the size of a sample increases, the mean of the sample will approach the mean of the population. Statistically speaking, if you increase the sample size, all you're doing is reducing the range of uncertainty (margin of error) with respect to how well your sample represents the population.

Really, he's not safe at all. Just take my set and add an infinite number of 10's to it. The mean will asymptotically approach ten; the majority of the values will *still* be higher than than the mean. Increasing the sample size does you no good if the population really is skewed that way.

Ah, the old 'infinite number of 10's' trick. You must be an academic. ;)

[Vaevictis]

Ah, the old 'infinite number of 10's' trick. You must be an academic. ;)

Nah, but having to do 10 hours a week of probability-related homework has the tendency to make me sound like one ;)

[olevetonahill]

Im way above nurmal ;)

[Beano's Fourth Chin]

Ironically, the author is obviously below average:



Actually, it doesn't. Take, for example, the following data set:

{1, 10, 10, 10, 10, 10, 10}

The average is 8.7ish. Six values are above average.

It's entirely possible for a majority to be above average if a there is a group of sufficiently small outliers at the bottom end.

There's more than one way to measure central tendency. In this particular case, you have to assume that he's talking about the median. That poor schlup in the middle. Half will be above average, half will be below.

In reading comprehension/intrepretation, then, you would be considered below average. Your attempt to prove your above averageness simply proves the author's point.

[proud gonzo]

heh :D

[Howzit]

There's more than one way to measure central tendency. In this particular case, you have to assume that he's talking about the median. That poor schlup in the middle. Half will be above average, half will be below.

In reading comprehension/intrepretation, then, you would be considered below average. Your attempt to prove your above averageness simply proves the author's point.

Isn't the name for the theorem dealing with the Normal Distribution function the Theorem of Central Limits? And doesn't it say, in a nutshell, that if the sample is big enough everything approximates a normal distribution?

[OKC Sooner]

"There are three types of lies: lies, damn lies, and statistics." ~ Benjamin Disraeli

[Beano's Fourth Chin]

Isn't the name for the theorem dealing with the Normal Distribution function the Theorem of Central Limits? And doesn't it say, in a nutshell, that if the sample is big enough everything approximates a normal distribution?

No clue. My 30 seconds of research on wikipedia simply didn't cover this.

[BajaOklahoma]

Does all of this mean that if I think I am below average then I am actually above average?
I hated stats.
TIA.

[Al Gore]

Let's bring out the math nerds......

[Howzit]

No clue. My 30 seconds of research on wikipedia simply didn't cover this.

An above average individual would have invested at least a full minute.

[GottaHavePride]

This ex-math-nerd says "Hey, look! It's beer-thirty!"

[Al Gore]

I thought average was 6"?????

[slickdawg]

I thought average was 6"?????


For white guys......



;)

[MamaMia]

The art of being able to surround yourself with the best people to get the job[s] done right is what makes one above average. :D

[Al Gore]

For white guys......



;)peemed.....

[slickdawg]

peemed.....

indeed!

[Al Gore]

I'm glad you could enjoy.....

[Vaevictis]

There's more than one way to measure central tendency. In this particular case, you have to assume that he's talking about the median. That poor schlup in the middle. Half will be above average, half will be below.

With respect to my having to assume he's talking about the "median", I need do no such thing when he uses the word "average".

You know that poor schlup in the middle? He's called the "median." The "average" is the sum of values divided by the number (aka, the "mean"). They are not the same thing.

In reading comprehension/intrepretation, then, you would be considered below average. Your attempt to prove your above averageness simply proves the author's point.

[Beano's Fourth Chin]

With respect to my having to assume he's talking about the "median", I need do no such thing when he uses the word "average".

You know that poor schlup in the middle? He's called the "median." The "average" is the sum of values divided by the number (aka, the "mean"). They are not the same thing.

In reading comprehension/intrepretation, then, you would be considered below average. Your attempt to prove your above averageness simply proves the author's point.

Apparently, you don't understand. The point of this thread is that you all are all below average and don't know it. Not me. I already know it.

[Ike]

the fact that they examined the performance of people grouped according to their percentile score on various exams indicates that yes, there most certainly MUST be an average that is at the 50th percentile, regardless of what the actual scores are. the distribution of people in each percentile grouping should by definition be flat, where the distribution of people with a certain test score does not have to be.

[Vaevictis]

Isn't the name for the theorem dealing with the Normal Distribution function the Theorem of Central Limits? And doesn't it say, in a nutshell, that if the sample is big enough everything approximates a normal distribution?

The central limit theorem has nothing to do with medians, it has to do with the tendency of a large number of iid random variables to assume a normal distribution -- a distribution which, among other things, has half of the probabilities less than the mean and half greater than.

Of course, the fact that a sample tends to take on the normal distribution doesn't mean that the populations do. :)

I'd be more specific, but my normal-distribution-fu is rather weak.

[Beano's Fourth Chin]

The Internet is a veritable all-you-can-eat buffet of such misplaced confidence. Online, individuals often speak with confident authority on a subject, yet their conclusions are flawed. It is likely that such individuals are completely ignorant of their ignorance. Cough.

[Vaevictis]

Does all of this mean that if I think I am below average then I am actually above average?

According to my brief reading of the study, it suggests that the more competent you are, the less likely you are to overestimate your competence, and that the more incompetent you are, the more likely you are to overestimate it.

You sir, would be an outlier in this data set :D

[Vaevictis]

It is likely that such individuals are completely ignorant of their ignorance. Cough.

Please, enlighten me. We're talking about math here, so it's entirely possible for you to show (prove) that I'm wrong.

The person in the article claimed that it's impossible for the majority of values to take a value greater than the average (aka, the mean, defined as the sum of values divided by the number of values); I challenged his assertion with the following counter example:

{1,10,10,10,10,10,10}

Average (mean): 61/7 ~= 8.714
Number of values greater than the average: 6/7 ~= 85.7% (a majority)

So: Show me where I'm wrong. Is my definition of the "average" (in this context, statistics) wrong? Is my arithmetic wrong?

(Do note that the author of the article has since changed his article to say "absurdly improbable" as opposed to "impossible", suggesting that he did in fact intend "average" to be the mean value I'm speaking of.)

[Ike]

Please, enlighten me. We're talking about math here, so it's entirely possible for you to show (prove) that I'm wrong.

The person in the article claimed that it's impossible for the majority of values to take a value greater than the average (aka, the mean, defined as the sum of values divided by the number of values); I challenged his assertion with the following counter example:

{1,10,10,10,10,10,10}

Average (mean): 61/7 ~= 8.714
Number of values greater than the average: 6/7 ~= 85.7% (a majority)

So: Show me where I'm wrong. Is my definition of the "average" (in this context, statistics) wrong? Is my arithmetic wrong?

(Do note that the author of the article has since changed his article to say "absurdly improbable" as opposed to "impossible", suggesting that he did in fact intend "average" to be the mean value I'm speaking of.)


in this case though, the 'mean' that they are talking about is not a mean score, but a mean percentile, which can be a very different monster. now, if the scores are distributed as in the set that you give, there is one person in the zeroth percentile, and 6 people in 14th percentile, and thus they are all in fact 'below average', or below the 50th percentile.

now, they probably aren't all actually below average, so if you happen to get a distribution such as yours, then the only information you should really take away from it is that you happened to design a pretty crappy test.

[Skysooner]

Please, enlighten me. We're talking about math here, so it's entirely possible for you to show (prove) that I'm wrong.

The person in the article claimed that it's impossible for the majority of values to take a value greater than the average (aka, the mean, defined as the sum of values divided by the number of values); I challenged his assertion with the following counter example:

{1,10,10,10,10,10,10}

Average (mean): 61/7 ~= 8.714
Number of values greater than the average: 6/7 ~= 85.7% (a majority)

So: Show me where I'm wrong. Is my definition of the "average" (in this context, statistics) wrong? Is my arithmetic wrong?

(Do note that the author of the article has since changed his article to say "absurdly improbable" as opposed to "impossible", suggesting that he did in fact intend "average" to be the mean value I'm speaking of.)


The author's assertion is totally incorrect. It is called a negatively skewed distribution.