both in general and in the particular case where the variables are normal and the limit mean and variance are finite
135577
The central limit theorem still applies.\n\nWhen variables are correlated a principal component analysis will find the independent variables for which the CLT applies.
304375
mean is the average of the data, mode is the data that repeats the most.
421571
The roots of this problem lie in the fundamental problem addressed by kalman filters. You want to read up about kalman filters to fully understand the solution.\n\nHere is the approach...\n\ngiven two estimates of a value that is normally distributed, you assume that a better estimate exists with value E = wE1 + (1-w)E2.  The variance of this new estimate is given by S^2  = w^2.S1^2 + (1-w)^2.S2^2.  You need to differentiate the second equation, look for a minima and plug that value of the weight into the first equation to get your optimal estimate.  Hope this helps.
424339
The central limit theorem states that given a distribution with a mean μ and variance σ², the sampling distribution of the mean approaches a normal distribution with a mean (μ) and a variance σ²/N as N, the sample size, increases. \n\nThe amazing and counter-intuitive thing about the central limit theorem is that no matter what the shape of the original distribution, the sampling distribution of the mean approaches a normal distribution. \n\nFurthermore, for most distributions, a normal distribution is approached very quickly as N increases. Keep in mind that N is the sample size for each mean and not the number of samples. Remember in a sampling distribution the number of samples is assumed to be infinite.\n\nDEGREE OF FREEDOM:\nThat measure of variability which merely expresses the number of options available within a variable or space. \n\nDegrees of freedom are the equivalent of currency in statistics - you earn a degree of freedom for every data point you collect, and you spend a degree of freedom for each parameter you estimate.
363586
Yes, the mathematical definition of risk is exactly that: larger variability is the return. However, the variation is symmetrical around the mean. The probabilty of having a larger return than the mean is just as big as the probability of having a smaller return than the mean.
218186
It takes only one example to prove the statement to be false, and I provided one:\n\n(Now, while you may never naturally encounter the set of numbers I used, the definitions of mean and median fit, so the statement is false.)
655845
I would say variance, but I'm no statistics expert.
44655
(10-8) * 20 = 40. The other data is just for figuring out how much the options worth.
298485
Answer is 10.2% of the sample(25) have a mean score within 3 points of the population mean score of 220.\n\nSD=10 --> 34% (normal distribution)\n   3 from mean = 3/10 x 34% = 10.2%.
170100
The problem is incompletely or improperly described.\n\nEach member of the population has some characteristic value.  The mean is the average of this value taken over the entire population, which, assuming each member is an equal contributor, is M = sum(values)/N, where N is the population.  If each member has the same value, say 515, then mean would be 515 and the standard deviation would be zero.  If each even-numbered member has value 1030, and each odd-number member has value 0, the mean is 515 and the standard deviation is (approximately) 515.\n\nSo, you have provided enough information to solve the problem satisfactorily.
448301
The standardized and studentized version of x differ in such a way that the former requires parameters (that is, population mean and population variance) while the latter uses statistic (that is, sample mean and sample variance).\n\n\nstandardized\n\nz test = (x - mu)(sqrt of n)/sigma\nz test = (108 - 100)(sqrt of 4)/16\nz test = 1\n\nstudentized\n\nt test = (x - muo)(sqrt of n)/s\nt test = (108 - 100)(sqrt of 4)/12\nt test = 1.3333\n\n\nlast\nBetween the two curves with different degress of freedom, the one with the greater value approximate the standard noramal curve more closely.\nIn fact, if the degrees of freedom approach infinity, the resulting curve will be the standard normal curve.\n\nTherefore the one with degress of freedom of 20 approximate the standard normal curve more closely than with the value of 12.
146516
Raw series occur most frequently in statistics, economics and physics (at a low level, this is where the simple pendulum is simplified, and perturbation theory is applied).\nIn statistics, the average of something is a sum, so if that something has an infinite sample space, as is the norm for any interesting random variables, you're dealing with a series.
388555
You need the standard deviation of the wieght of samples of 50 people.\n\nThe mean weight of samples of 50 people is 150 * 50, or 7500 pounds. Four tons is 8000 pounds.\n\nWith such a large sample as 50, you would expect that your sample means will each be quite close to 150. In fact, the standard deviation of the means of your samples is\n\n35 / sqrt(50)  = 35 / 5sqrt(2) = 7 / sqrt(2) = 7sqrt(2) / 2\n\nor around 5 pounds. That means that 2/3 of the time, the weight of 50 people will lie between 7495 and 7505 pounds, and 95% of the time, their weight will lie between 7490 and 7510 pounds.\n\nThe design difference between 8000 pounds and 7500 pounds represents 100 standard deviations of the sample means. That's off the charts. There is essentially "zero" probability that a sample as large as 50 will be so far from the mean as that.\n\nModulo does some calculations, but upon finding the result "enormous" abandons reason and uses some other (wrong) method that gives a more "reasonable" result. He was almost right the first time. The problem supposes an enormous design margin.
368117
That sounds familiar... Think I'm working on it now! But I don't remember q. It is n in my book. In which case, q (if I'm thinking right) is the number of samples. Then again what's up with the ^'s? If np was > 10, then it meant that the distribution was approx. normal. You might be ahead of me right now...
48554
Yes. What you're talking about is the adjustment of the MEDIAN (definition below). Because "good" and "bad" are relative ("good" is simply BETTER than what "isn't good") but "normal" is what is average, your expectation for what normal days are will rise when you experience a lot of good days. When you encounter what USED to be a "normal" day, it is now "not as good as the current normal day" and therefore bad. Confused? Good.
397848
I can tell that nobody wants to just do this problem.  There's nothing interesting about it.  It's a typical problem that a stat student does to show understanding.  Get out your textbook and study up how to solve it.
579380
it's c\n\neven functions have the property that evaluating the function at their negative and positive values for the variable produces the same result.