Orton's Arm Posted December 15, 2006 Author Posted December 15, 2006 Start using proper terminology, and I'll discuss it with you. The "average person" is, by definition, AT THE POPULATION MEAN, YOU MORON. THAT'S WHAT "AVERAGE" MEANS. It's impossible to discuss when you abuse the vocabulary. 867747[/snapback] This from someone who until recently thought that rolling a pair of dice would produce a binomial distribution. In case you're wondering, yes you'll have to hear about that every time you question my understanding of terminology or vocabulary. In my earlier post, the word "average" referred to the average member of the thin slice, and not the average member of the population. I think you understood that when you posted, but couldn't resist the temptation to turn this away from a statistics debate, and into a "you're an idiot because you don't understand term X" debate.
Chilly Posted December 15, 2006 Posted December 15, 2006 This from someone who until recently thought that rolling a pair of dice would produce a binomial distribution. In case you're wondering, yes you'll have to hear about that every time you question my understanding of terminology or vocabulary. 867770[/snapback] http://hyperphysics.phy-astr.gsu.edu/hbase/math/dice.html#c2
Bungee Jumper Posted December 15, 2006 Posted December 15, 2006 This from someone who until recently thought that rolling a pair of dice would produce a binomial distribution. In case you're wondering, yes you'll have to hear about that every time you question my understanding of terminology or vocabulary. In my earlier post, the word "average" referred to the average member of the thin slice, and not the average member of the population. I think you understood that when you posted, but couldn't resist the temptation to turn this away from a statistics debate, and into a "you're an idiot because you don't understand term X" debate. 867770[/snapback] No, I couldn't resist it because you consistently misuse and misunderstand words. I honestly read that and said "What the !@#$ is he saying now? An 'average' person doesn't regress toward the mean, they are the !@#$ing mean." This is entirely due to the fact that you are imprecise and completely confused - when not simply ignorant - in thought and word.
Ramius Posted December 15, 2006 Posted December 15, 2006 This from someone who until recently thought that rolling a pair of dice would produce a binomial distribution. 867770[/snapback] Sorry, rolling a pair of dice is the perfect example of a binomial distribution, as noted above by blue fire. We'll just add this to the growing epic that is "things holcombs arm doesnt know jack sh-- about" I believe we are up to roughly 3,462, but that could simply be measurement error.
Orton's Arm Posted December 15, 2006 Author Posted December 15, 2006 http://hyperphysics.phy-astr.gsu.edu/hbase/math/dice.html#c2 867792[/snapback] If you click on the "binomial distribution" link in that website, you'll see the following text: The conditions for validity of the binomial distribution are * each trial can result in one of two possible outcomes, which could be characterized as "success" or "failure". * the probability of "success", p, is constant from trial to trial. * the trials are independent. Two possible outcomes for each trial--think flipping a coin. Not rolling a pair of dice.
Chilly Posted December 15, 2006 Posted December 15, 2006 If you click on the "binomial distribution" link in that website, you'll see the following text:Two possible outcomes for each trial--think flipping a coin. Not rolling a pair of dice. 867841[/snapback] I guess their page which is entitled "Of Dice and the Binomial Distribution", and their example of a binomial distribution, http://hyperphysics.phy-astr.gsu.edu/hbase/math/binex.html must be completely wrong then!
Bungee Jumper Posted December 15, 2006 Posted December 15, 2006 If you click on the "binomial distribution" link in that website, you'll see the following text:Two possible outcomes for each trial--think flipping a coin. Not rolling a pair of dice. 867841[/snapback] Have you ever heard of a binomial expansion? Know what a binomial coefficient is?
Orton's Arm Posted December 15, 2006 Author Posted December 15, 2006 No, I couldn't resist it because you consistently misuse and misunderstand words. I honestly read that and said "What the !@#$ is he saying now? An 'average' person doesn't regress toward the mean, they are the !@#$ing mean." This is entirely due to the fact that you are imprecise and completely confused - when not simply ignorant - in thought and word. 867802[/snapback] Why I continue to debate someone who enjoys throwing around childish insults far more than he enjoys talking about stats is far beyond me. I'll put this to you again. You take a thin slice of the population, based on people's test scores the first time they took the imperfect test. Assuming the test scores you're looking for are above the population mean, the thin slice should contain a greater number of people who got lucky than people who got unlucky. Given that fact, the average I.Q. for the people in the thin slice is somewhat lower than the average test score for the people in the slice. Do you agree or disagree?
Ramius Posted December 15, 2006 Posted December 15, 2006 If you click on the "binomial distribution" link in that website, you'll see the following text:Two possible outcomes for each trial--think flipping a coin. Not rolling a pair of dice. 867841[/snapback] Now you cant even understand a binomial distribution. Take a look at it moron. There are 2 possible outcomes, success or failure. In a coin flipping case, the 2 possible outcomes are NOT heads or tails. The binomial distribution says that whatever you designate, say heads, is a success, and that other outcomes (tails in the coin case) are failures. When throwing a pair of dice, designate the outcome you are looking for to determine the probability of that outcome as a success, and all other outcomes are failures. Hence, if i want the probability of throwing a 2-4, rolling a 2, or 3, or 4 is considered a success, while rolling anything else is considered a failure. See, 2 outcomes when throwing a pair of dice, success or failure. The fact that you cant comprehend what they were saying about a pair of outcomes, success and failure, and that they didnt mean an event with only 2 possible final outcomes, such as heads/tails, shows what kind of ignorant !@#$tard you are in the field of mathematics.
Bungee Jumper Posted December 15, 2006 Posted December 15, 2006 I guess their page which is entitled "Of Dice and the Binomial Distribution", and their example of a binomial distribution, http://hyperphysics.phy-astr.gsu.edu/hbase/math/binex.html must be completely wrong then! 867850[/snapback] Hyperstats is an authoritative source. Hyperphysics is not. Get it?
Chilly Posted December 15, 2006 Posted December 15, 2006 Hyperstats is an authoritative source. Hyperphysics is not. Get it? 867854[/snapback] Oh, my bad, I thought it was the word hyper which made a source credible.
Orton's Arm Posted December 15, 2006 Author Posted December 15, 2006 I guess their page which is entitled "Of Dice and the Binomial Distribution", and their example of a binomial distribution, http://hyperphysics.phy-astr.gsu.edu/hbase/math/binex.html must be completely wrong then! 867850[/snapback] The example is correct, as applied by the authors of the website. They were interested in the probability of a single event (in this case, rolling a 3). They categorized each observation into one of two categories: 3 and not-3. This is consistent with the definition of a binomial distribution. Bungee Jumper, on the other hand, was using the rolling of a pair of dice as an approximation for the normal distribution. He categorized his possible results into eleven categories; and that's not the binomial distribution he said it was.
Ramius Posted December 15, 2006 Posted December 15, 2006 The example is correct, as applied by the authors of the website. They were interested in the probability of a single event (in this case, rolling a 3). They categorized each observation into one of two categories: 3 and not-3. This is consistent with the definition of a binomial distribution. Bungee Jumper, on the other hand, was using the rolling of a pair of dice as an approximation for the normal distribution. He categorized his possible results into eleven categories; and that's not the binomial distribution he said it was. 867858[/snapback] see, you're wrong.
Bungee Jumper Posted December 15, 2006 Posted December 15, 2006 The example is correct, as applied by the authors of the website. They were interested in the probability of a single event (in this case, rolling a 3). They categorized each observation into one of two categories: 3 and not-3. This is consistent with the definition of a binomial distribution. Bungee Jumper, on the other hand, was using the rolling of a pair of dice as an approximation for the normal distribution. He categorized his possible results into eleven categories; and that's not the binomial distribution he said it was. 867858[/snapback] You're an idiot.
Orton's Arm Posted December 15, 2006 Author Posted December 15, 2006 see, you're wrong. 867859[/snapback] You're an idiot. 867860[/snapback] Too bad for you guys that saying things doesn't make them so. Like it or not, a binomial distribution has a very specific definition; which Bungee Jumper was unaware of until I provided the Hyperstats link and the textbook quote. You are required to classify your results into only two categories to use a binomial distribution; instead of using eleven categories as Bungee Jumper was doing in his so-called "binomial distribution."
Bungee Jumper Posted December 15, 2006 Posted December 15, 2006 Too bad for you guys that saying things doesn't make them so. 867864[/snapback] No. You being an idiot is what makes you an idiot. Not me saying it. Ever hear of a binomial coefficient? Probably not...
Ramius Posted December 15, 2006 Posted December 15, 2006 Too bad for you guys that saying things doesn't make them so. Like it or not, a binomial distribution has a very specific definition; which Bungee Jumper was unaware of until I provided the Hyperstats link and the textbook quote. You are required to classify your results into only two categories to use a binomial distribution; instead of using eleven categories as Bungee Jumper was doing in his so-called "binomial distribution." 867864[/snapback] we posted links and i clearly stated why you were wrong. But one again: 1. you choose to ignore it, because it proves you wrong 2. you dont comprehend the situation enough to reasonably discuss it We are right, you are wrong. But keep on arguing. It fun lighting you up like a Christmas tree.
Bungee Jumper Posted December 15, 2006 Posted December 15, 2006 1. you choose to ignore it, because it proves you wrong2. you dont comprehend the situation enough to reasonably discuss it 867879[/snapback] Hey, look! A binomial distribution!
Orton's Arm Posted December 15, 2006 Author Posted December 15, 2006 we posted links and i clearly stated why you were wrong. But one again: 1. you choose to ignore it, because it proves you wrong 2. you dont comprehend the situation enough to reasonably discuss it We are right, you are wrong. But keep on arguing. It fun lighting you up like a Christmas tree. 867879[/snapback] Too bad you didn't understand the links which you posted. A binomial distribution must have only two categories--"success" and "failure" if you will. Bungee Jumper was using eleven categories in his dice rolling example. Even you should have the mental capacity to see the difference between two categories and eleven categories.
Chilly Posted December 15, 2006 Posted December 15, 2006 Too bad you didn't understand the links which you posted. A binomial distribution must have only two categories--"success" and "failure" if you will. Bungee Jumper was using eleven categories in his dice rolling example. Even you should have the mental capacity to see the difference between two categories and eleven categories. 867882[/snapback] O RLY? Verbal example: a pair of dice, roll an 11, the next roll will more likely be less than 11 than not, because there's 33 possible outcomes of being less than 11, but only three of being greater than or equal to 11...and thus, the value regresses toward the mean, because the difference between 11 and 7 (the mean value of two dice) is relatively high. You'll note two things: 1) this is a verbal example of the mathematically CORRECT definition of regression to the mean, and 2) THERE IS NO ERROR INVOLVED, it is strictly a function of the binomial probability distribution. I see two outcomes: 1.) Less than eleven 2.) Greater than or equal to 11 Wheres the rest?
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