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Ask and answer stupid questions here! - Page 59
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Hryul
Austria2609 Posts
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d00p
711 Posts
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icystorage
Jollibee19350 Posts
On March 06 2014 17:01 d00p wrote: For the last couple of months peolpe have been saying "kappa" in the forums for whatever reason. What did I miss? its a twitch chat emoticon | ||
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d00p
711 Posts
Ugh, twitch chat. Didn't miss anything substantial then. Thanks. Edit: But what is it trying to emote? | ||
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TXRaunchy
United States131 Posts
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icystorage
Jollibee19350 Posts
On March 06 2014 17:08 d00p wrote: Ugh, twitch chat. Didn't miss anything substantial then. Thanks. Edit: But what is it trying to emote? its like troll face but kappa | ||
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ComaDose
Canada10357 Posts
http://www.teamliquid.net/forum/tl-community/423487-twitch-emotes-and-their-place-on-tl | ||
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Jonrock
Germany80 Posts
On March 06 2014 14:43 Najda wrote: A quick google turned up this, there may be other apps that do similar. http://m.cnet.com/news/wake-up-to-your-favorite-music-with-pandoras-new-alarm-clock/57614958?ds=1 Well, I'm not looking for an App. I'm actually looking for a physical alarm clock | ||
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Dark_Chill
Canada3353 Posts
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Simberto
Germany12004 Posts
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Dark_Chill
Canada3353 Posts
On March 07 2014 01:51 Simberto wrote: That is what a standard deviation is. Well, the standard deviation is the square root of the sum of the squares of each summand of that value, but that is just because it's easier to math with squares than having to make every negative number positive on it's own. That's what I thought at first, but then I noticed that the square root was around the whole function and not just the sum of squares. So the sample size was also affected by the square root which makes no sense to me. | ||
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o29
United States220 Posts
http://en.wikipedia.org/wiki/Average_absolute_deviation#Mean_absolute_deviation_.28MAD.29 Rather than: http://en.wikipedia.org/wiki/Standard_deviation "The standard deviation of a random variable, statistical population, data set, or probability distribution is the square root of its variance. It is algebraically simpler though in practice less robust than the average absolute deviation." | ||
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FiWiFaKi
Canada9859 Posts
On March 07 2014 01:51 Simberto wrote: That is what a standard deviation is. Well, the standard deviation is the square root of the sum of the squares of each summand of that value, but that is just because it's easier to math with squares than having to make every negative number positive on it's own. Wrong. You square it to put more emphasis on points that are really far away. You have to look at norms and their residuals in Numerical Analysis Methods, and have a pretty good understanding before you understand exactly which norms are useful, and when which apply. As to why a standard deviation is useful, let me ask you. You have 100 points of data randomly scattered, how do you specify how scattered the data is without telling me how far every single one of these 100 points is from the curve of best fit. Standard deviation in really simple terms will tell you: For a set of data, 68% of all data will be within one standard deviation, 95% will be within 2 standard deviations, and 99.7% of data will be within 3 standard deviations. Essentially normal distribution is given by: (1/(sigma*sqrt(2pi)))*exp(-(x-u)^2/(2sigma^2)) This is the "bell shaped curve people talk about, and x is given by your data set, u is your mean, and sigma is your standard deviation. To simplify the equation we substitute z=(x-u)/sigma. Now by simply integrating the normal distribution function, we can get the area under the curve and represent how our data is distributed. Sadly the term e^-z^2/2 is a term that cannot be integrated by classical mathematics as it doesn't exist, and hence we define the integral with the error function, 1/2*erf(z/sqrt(2)) use z-score tables instead. By having the normal distribution and by assuming the data is normally distributed (nature likes to be normally distributed), we can then find how precise our data is by having the 3 parameters: Mean, data average, and standard deviation. This is only one of many many distributions that exist, and as an engineer only a handful have been covered. However statistics and probability like this is definitely one of the most complex applied math degrees, with extremely brutal proofs and complex formulas that even the smartest engineers opt to just memorize the formulas opposed to learn their inner workings. | ||
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FiWiFaKi
Canada9859 Posts
On March 07 2014 02:54 Dark_Chill wrote: That's what I thought at first, but then I noticed that the square root was around the whole function and not just the sum of squares. So the sample size was also affected by the square root which makes no sense to me. Correct. It's not simply to get rid of the negative sign. Any self respecting mathematician would just use the sum of the absolute values. There is a reason (x-x_bar)^2 then square rooted is used instead of abs(x-x_bar), or (abs(x-x_bar))^3 then cubed rooted. The L1 norm, which is essentially a norm which removes the negative sign and sums all the solutions is bad, it yields more than one result dependent on how you draw your line of best fit, etc. Higher level statistics courses must be taken to understand it well, we were simply told which norms were good and briefly why... And it so happens the L2 norm, which is used in standard deviation happens to be pretty good for accuracy and for the simplicity of its residual. | ||
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corumjhaelen
France6884 Posts
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FiWiFaKi
Canada9859 Posts
![[image loading]](http://i.imgur.com/TdA4BKx.gif) I think the more interesting thing about standard deviation, especially for me while learning it, is why are you dividing by n-1 instead of n. (n being the sample size)... It's really cool, and how it's associated with degrees of freedom. If you are interested in the subject, I recommend you learn a little bit about norms... The wikipedia article is a brief overview, but you'll be able to pick a direction to follow research for anything you find interesting. http://en.wikipedia.org/wiki/Norm_(mathematics) On March 07 2014 03:30 corumjhaelen wrote: Central limit theorem is the reason why. +1 | ||
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Epishade
United States2267 Posts
A woman enters an elevator that's empty. As she pushes a floor button, the camera switches to her feet. Though in the scene before, nobody was there, there is now an extra pair of hanging legs hovering a few inches off the ground. The woman starts panicking, sensing something behind her but too terrified to turn around. She starts mashing the floor button again, and the dead guy/ghost behind her starts SLOWLY hovering closer and closer to her. When they're almost touching, the elevator opens and she rushes out, terrified. | ||
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Fharoc
Canada32 Posts
this the scene you're looking for? + Show Spoiler + | ||
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Sub40APM
6336 Posts
On March 06 2014 16:33 Epishade wrote: But surely, if you're the enemy, and somebody tells you they're James Bond, shouldn't that set off some red flags for you? Regardless of who actually assumes the name when they're promoted (if that's an actual theory), shouldn't his name alone be an indication for being a spy that you should be able to pick up on? Surely his enemies must have heard somewhere of the all great and powerful 007 James Bond from all of the times he's saved the world? well, a lot of times when he saves the world the world doesnt actually see it happen, just you the audience member does. So if you meet someone who is James Bond you have no idea who are you dealing with, some rookie who just killed his first two guys, or some super suave master spy who has a laser wrist watch that will have you and your henchmen all dead before anyone can even disable the safety on their guns... | ||
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Simberto
Germany12004 Posts
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