Sampling distribution?
952174
That is the central limit theorem.  Each distribution can be uniform, but the distributions themselves are normally distributed in their range, their value, their frequency.\nRead the attached link for more details.
928594
If you assume it's normal, then yeah, although I assume there is more data than you have given as you do not have the variance there, although you could treat this as a binomial distribution giving it mean 0.3, variance n0.3*0.7, although when calculating the confience interval you'll want to divide by n^2 for that variance.
307839
What do want to distribute? What specific items would you prefer.\nFind someone doing what you want to do and copy them.\nNo need to reinvent the wheel!
799156
Use Binomial distribution:\n\n(n nCR x) p^x (1-p)^(n-x)\n\nn nCR x = n!/(x!(n-x)!)\np=probabiliy (.10)\nn=patients (10)\nx= desired\n\na) P(X<=2) = P(X=0) + P(X=1) + P (X=2)\nb) 1- P (X<=2)... 1-prevoius answer
488365
Try again Paul C.  The population doesn't need to look normal.\n\nThis all comes down to the Central Limit Theorem.  Let's assume that I take many samples of size 1004 from the population.  Well, each sample will have a percentage that is different from the population percent, because samples are going to look different from the population.  Also, each sample is going to look different from each other, because they are different samples.  If we were to draw a graph that looked at the "behavior" of these many samples, we would notice the following:\n\n1.  If I were to average out all of these sample percents, I would get the population percent.  Hopefully you can see why this would happen.  Sample percents will be around the population percent, but just be off by a little bit.\n\n2.  If I were to look at the spread of the percents, the spread would be smaller as bigger samples were taken.  In other words, if I had taken samples of size 500, the sample percents would be more spread out than for samples of size 1004.  That is because if were take larger samples, then we have more information about the population, so we should expect to get closer to the population percent.\n\n3.  As I take larger and larger samples, the shape of the graph of sample percents would look more and more like a normal curve.  A normal curve is shaped like a bell, where the middle of the bell is were the population percent is.\n\nSo Paul C, it doesn't matter that the population doen't look normal; all we need is a sample that is large enough.\n\nNow the margin of error on polls is typically going to be\n\n2*sqrt(p(1-p)/n))*100%\n\nwhere p is the sample proportion (the sample percent divided by 100%) and n is the sample size.  It can be shown that this margin of error would be at its largest if p = 0.5, so the largest the margin of error would be in this case would be\n\n2*sqrt(0.5*0.5/1004)*100% = 3.2%\n\nThis is typically associated with 95% confidence.  This basically means that if I were again to take many, many samples of size 1004,  95% of the sample percents would be within a margin of error away from the true population percent.\n\nThis of course depends on the sample that is taken.  The online polls that you see would not be good estimates for the population percent.  Below I have a link to a story that MSNBC has on each of its online polls.  The title is "How 1,000 people can be more representative than 200,000."  I think that it is a very good explaination of this topic.
1022381
You can use a chi-square test for goodness of fit.  In the link below, O would be the observed number of times each card came up and E would be the expected.  (You would have to compute this 52 times and then add them up.).  So if you did this 500 times, the E for each card would be 500/52, which will be greater than 5, which is the typical way you determine if the sample size is large enough.  Under the assumption that the cards are at random, this would have a chi square distribution with 52-1 = 51 degrees of freedom.  You can do a test of significance using this test statistic and distribution.
905827
If a random variable has a standard deviation of s and you take N samples, then the standard deviation of the sample mean is s/sqrt(N).\n\nIn your case you assume that the standard deviation of the random variable is close to 50 because you have such a huge number of samples. Now if you turn around and only look at 30 samples then the standard deviation will be\n50/sqrt(30).
570786
First of all, if the population is normal, the distribution of sample means is normal regardless of sample size.  So I. is true.\n\nSecond, if the population is not normal, the distribution of sample means is approximately normal if the sample size is large.  So III. is also true.\n\nII. cannot be determined to be true unless we know what the population looks like.\n\nSo I would go with C.\n\nUPDATE: Apparently ghakh hasn't heard of the Central Limit Theorem.  You know, the "Fundamental Theorem of Statistics"?  In layman's terms, it states that if the sample size is large enough, then the sample means will be approximately normal if the sample size is large enough REGARDLESS OF WHAT THE POPULATION LOOKS LIKE!!!!!!  All you need is a finite variance, which for elementary statistics students is not something they need to be concerned with.\n\nMaybe you should take my statistics class, ghakh.  You might learn something!
177395
The easiest way to generate random number from a given distribution is to generate a uniform (0,1) random number; then compute the inverse cummulative probability function from that number:\nLet u be a uniform random number\nthen, x=Pinv(u,l) distributes Poisson if Pinv stands for the inverse function of the cummulative poisson distribution.\nIt's not relevant what language you use for programming
1152776
Depending on your computer type you can download or purchase programs that will allow you to 'chop up' samples as you please.  I've heard on of one, I just have to tell you the name.  It's for PC users by the way.  It's called Mr. Free Free.  I couldn't believe it when I saw it.  Funniest name for software I've seen.
896151
To answer your question, i need to give u an example.\nConsider all real integers from 1 to 10. Let this be our population. \nNow i draw a sample, let this sample be x\nx ={ 3, 4, 5, 6}\nto find population mean, divide the sum of (all integers from 1 to 10) by 10. This will give you 5.4\nnow our population mean is = 5.4\nConsider the sample x now.\nexpectation is the mean of a sample.\nE(X) = (3 + 4 + 5 + 6)/4 = 4.5\nNow, if you take a larger sample (y)\ny = { 2, 3, 4, 5, 6, 7, 8, 9}\nE(Y) = (2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)/8 = 5.5\ncompare this with the population mean, which is 5.4.\nTherefore, if you take a sample of N distinct numbers, where N <10,\n"the expectation of the sample will converge to the population mean as N approaches 10." - this is how the population mean and the expectation of a sample are related.
23153
Assuming you mean some random numbers, try http://www.random.org
408817
X = sample mean, U = theoretical mean, S = standard deviation, and N = number of data...\n\ni assume u know how to calculate X and S from above. \n\n.95 = P(1.96< (X-U)/(S/sqrt N) < 1.96)...solve for U..U will be in between to values, which means that you are 95% confident that U will lie in that range.
1087310
In statistics, a sampling frame is the source from which a sample is drawn. It is a list of all those within a population who can be sampled, and may include individuals, households or institutions
236313
Wouldn't the probability than 5 of the 38 thought it was exciting be\n\n38C5(.79^5)(.21^33) = 6.63x10^-18 : basically impossible?  Even if you added in the probabilities of 4, 3, 2, 1, and 0, it's still tiny.\n\nSo I would infer that either the 79% is wrong or the sample was not independently chosen.
1166976
As to part one, I can not recommend any acceptable, known test for such data (perhaps what I've written below might work).\n\nAs to part two, this is a horrid approximation since the midrange as the mean is a major assumption (even more far-reaching than assuming something about the populations' probability distributions).\n\nFor part 3 and also part 1:\n\nThe first problem you run into is that you do not know anything about the nature of the population distribution (normal, uniform, etc).  Thus, a nonparametric test would be required.  I know of no nonparametric tests for any of the statistics you have (minimum, maximum or range).\n\nIf it is possible to obtain the actual values for each of the samples in each set, the Kruskal-Wallis test should suffice.  It is nonparametric, so there are no problems with the requirement of assumptions about population distributions.\n\nFurther, its job is to determine if two samples are far enough apart such that the null hypothesis that the population distributions are the same can be rejected.  This would give you what you desire.\n\nAs far as your problem is concerned, the only statistics you have are minimum, maximum and range.   Perhaps one could derive a test to test the minima or maxima of each set, but one still runs into the problem that there can be no parameter placed on the population distribution.\n\nIf you can make an assumption about the distributions of the populations of interest (normal or whatnot), then it would be possible to derive such a test.  This would be a tedious task, however, as it would require thorough knowledge of order statistics and distribution functions.\n\nOne would need to use a test statistic based of off the difference in the two statistics that you use (I would use the maxima since they are further apart than the minima and probability distributions for ranges are uncharted territory unless you're a Ph.D.).  Thus, your test statistic is based off of the difference between the maxima (Y(n) - Z(n) where the (n) are subscripts, which is 2 in this case).\n\nTo perform this test, you would need to derive the probability distribution of the difference of the two maxima and use that to find the probability of the difference being two or greater.  This is quite a task for anyone and requires a large amount of knowledge in mathematical statistics.  I believe I could perform such a task, but it would likely take some time to complete.\n\nRemember though that you must make a critical assumption about the population before you can even do this.  I would recommend using the assumption of uniformity if you are unsure of the populations' distributions.  This is because the derived test would then be least likely to give a type I error, where you reject the data improperly, because the test is less likely to reject.
487348
The mean of the average of 100 randomly selected travelers would just be the same as the population mean, which is 190.\n\nThe SD of the avereage of 100 randomly selected travelers would be the SD of the population divided by the square root of the sample size, which is 20/sqrt(100) = 2.