[HonestSean]

I knew my students when I cared. After the first year, I didn't.

I was freinds with my undergrad professors. My graduate school professors I mostly didn't know (except some of the old school ones).

Jon Miller

[Broorbbub]

no, you have it messed up again. heres what I get.

standard deviation sample = squarroot(Var(X_bar)/n)
in this case, var(X_bar) is 2.89/1000 = 0.00289.
the standard deviation of the sample is squarrout(0.00289) = 0.054, which is close to zero.

[welihiedginly]

anything else you have to say to me before I go to bed?

[NeroASERCH]

I need to remind myself to bump this tomorrow so that people with a clue (other than myself and Kuci) can enjoy it.

[avarberickibe]

Sorry. Didn't mean to cut you out of that.

[joOEMcheapSOFTWARE]

I just love these math threads.

[uC4F0NVL]

Originally posted by KrazyHorse
They spoil the kids at Hopkins. We're combination graders, tutors and assistant professors. Same here.

I don't know some of my professors, either...

[exchpaypaleg]

Originally posted by self biased
what if i'm only looking for certain values? for example, what if i'm looking for a 3+ on a d6 on 15 dice? odds say that ten of those fifteen will be a three or better. dropping some dice now i get: 2, 3, 1, 1, 2, 2, 1, 4, 6, 2, 4, 6, 6, 6, and 1. more than half my values are less than my target of 3+.

is there any way of shortcutting this? more often than not, we're looking for 3+ or 4+ on anywhere from four to thirty-six dice. 15 choose 10 divided by 2^15 for 10 4+'s. You'll need a calculator.

In general, if you're making n rolls and you want to know the probability of getting k of some result (with a probability p of getting that result on any single roll, e.g. p=1/2 for a 4+), use the formula (n choose k) * p^k * (1-p)^k.

edit: (n choose k) = n! / ((n-k)! * k!), where ! = factorial.

[invasuant]

Originally posted by Kuciwalker

That said, Standard Deviation doesn't approach zero ever [unless it is of a set of identical items] iirc. It's not supposed to be a measure of the total variance of the set's mean from the true mean, but rather a predictor of how far from the mean any given single item in the set is likely to be.

So close, and yet so far. The standard deviation of the set's mean does approach zero. Yes, that would be the variance of the set's mean from the true mean... the standard deviation itself does not approach zero [unless the set is identical items].

[glamourcitys]

kuci: i'll digest your post after i've had some sleep.
All: thanks for the help and/or backbiting.

[Avoireeideree]

I'm trying to imagine how the world would be if LoA was right.

My head hurts.

[Quiniacab]

Great reading material

Much better than physics threads with KH's participation, I can't really understand those

[BostonDoctorTTT]

My stat exam this afternoon was easy

[Aaron757]

Originally posted by Kuciwalker
Btw, you should use Google for your calculator. Then, if you want to find the odds of getting 10 3+'s in 15 rolls, just type in "(15 choose 10) * (2/3)^10 * (1 - 2/3)^5". No need to do the factorial stuff. i think a point is being missed that i'm not making clear enough.

i know how to figure out the odds of rolling a particular number or higher on d6. 3+ is 2/3 (or near 67%) on 1d6. the odds say that on fifteen dice, ten will be 3+ and five will not. if i roll 15d6 and get say twelve results at 3+ (which is 4/5 or 80%), is this within a standard deviation? i realize i may not be using the proper terminology for this. all i know is that i'd kill for a set of dice that always rolled a straight statistical average.

[Hankie]

You mean the difference between N and N-1?

[7UENf0w7]

Originally posted by Lawrence of Arabia
so easy that you havnt learned the difference between sample and population variance eh? Neither the sample nor population variance approaches zero as the number of samples grows to infinity.

[juidizHusw]

Originally posted by self biased


i think a point is being missed that i'm not making clear enough.

i know how to figure out the odds of rolling a particular number or higher on d6. 3+ is 2/3 (or near 67%) on 1d6. the odds say that on fifteen dice, ten will be 3+ and five will not. if i roll 15d6 and get say twelve results at 3+ (which is 4/5 or 80%), is this within a standard deviation? i realize i may not be using the proper terminology for this. all i know is that i'd kill for a set of dice that always rolled a straight statistical average. Then you use the formula for standard devation I gave: sqrt(n * p * (1 - p)). In the case of 15 dice at 3+, that's sqrt(15 * 2/3 * 1/3) = sqrt(10/3) ~= 1.8. So you're likely to get 8-12 successes.