Math question? How can the measures of variation be used in the real world?
929935
variations matter greatly in interpreting data.  Something might seem like it is returning significantly different results than the control but stat evaluation may show that the variations are minimal at best
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.
667767
You can refer to this blog to calculate comparative metrics : \n\nhttp://ns-2.blogspot.com/2006/03/how-to-calculate-comparative-metrics.html\n\nGood luck!
156694
Then the coefficient of variation is .333
400910
"...the variance of a random variable is a measure of its statistical dispersion, indicating how far from the expected value its values typically are."\nhttp://en.wikipedia.org/wiki/Variance\n\nHere are some descriptions of how to calculate it:\nhttp://davidmlane.com/hyperstat/A16252.html\n\nCheers!
861788
There is no accurate way to do this because too many variables are involved. You can do like I do though and run a series of tests and then average them.\n\nhttp://www.toast.net/
920953
1) X/25\n2) X/25\n\nExplanation:\n1) If you divide a set of values by a constant, then both the mean and the standard deviation are divided by that constant.\nSince your question is about the variance, which is the SQUARE of the standard deviation, the variance is divided by the SQUARE of the constant (5, in this case).\n\n2) When you add a constant to a set of values, the mean is increased by the amount of that constant, but the standard deviation is unchanged. This is true because the group of values has all moved up by a constant amount, but their relation to each other is unchanged, and their variation from each other is unchanged. Since the standard deviation is unchanged, the variance (the square of the standard deviation) is also unchanged.
1030781
By reading your question, it appears that you are first and foremost seeking to "define" mathmatically what it is that you are attempting to do to reach your results-- presumably so that you can study that area of statistics further and apply it to your experiment.\n\nWith that in mind, what you are dealing with in your experiment is called "Event Probability."  In Event Probability, you are dealing with concepts of "conditional probability", "frequency", and "relative frequency."  For instance, what you are doing is very similar to the flow charts that appear in many family's medical books with a "Decision Tree" that allows you to answer "Yes" and "No" questions along a Flow Chart to reach an ultimate probability that you have a particular illness.  In other words, you are creating an algorithm by measuring the "relative frequency" of events and compound events and drawing a conclusion from that information.  People who perform these experiments are called "frequentists."  For all intents and purposes, this is what actuaries do at insurance companies and financial firms.  I'd read about those areas of statistics first which should prove very helpful in providing you with a structure for reaching the most helpful and accurate conclusions from your experiment.
242388
Example 1: If y varies inversely as x, and y = 6 when x = , write an equation describing this inverse variation.\n\nk = (6) = 8\nxy = 8 or y =  \nExample 2: If y varies inversely as x, and the constant of variation is k = , what is y when x = 10?\n\nxy = \n10y = \ny = × = × =  \nk is constant. Thus, given any two points (x1, y1) and (x2, y2) which satisfy the inverse variation, x1y1 = k and x2y2 = k. Consequently, x1y1 = x2y2 for any two points that satisfy the inverse variation. \nExample 3: If y varies inversely as x, and y = 10 when x = 6, then what is y when x = 15?\n\nx1y1 = x2y2\n6(10) = 15y\n60 = 15y\ny = 4\nThus, when x = 6, y = 4. \n\n\ndo u get it now
736479
Broad question! Focusing it would help us answer better.\n\nReally basic stats would do things like tally and sum, average, find the median of the subjects you are testing.\n\nOnce you have those descriptive statistics, you can do things like compare outcomes between groups. Suppose you have two samples of people. One group has a mean life expectancy X. The other has mean life expectancy Y. To determine whether X and Y are significantly different from each other, you could do a Student t-test to determine whether the difference in values of X and Y are truly significant, or differ only by chance.\n\nthat's just one example. there are tons more!
372526
some calculators can do this\n\nmathematica software will do it
149140
Don't know Levey Jennings, but in general the variance is the sum of the squared deviations from the mean divided by the number of samples (for large data sets) or the number of samples minus 1 (for smaller data sets).  The standard deviation (SD) is then the square root of the variance.\n\nSo, if "m" is the mean and each sample is x1..n, the variance is:\n\n[(m-x1)^2 + (m-x2)^2 + (m-x3)^2 ... (m-xn)^2]/n\n\nThen, take the square root of the variance to get the standard deviation.
1161445
The equation for direct variation is y = kx.\nk is the constant of variation.\n\n#1.  -10 = k(2)    Solve for k\n           -5 = k\nSo, the equation for direct variation is y = -5x -- just substitute the k back into the equation.\n\n#2.  7 = k(3)   Solve for k\n       7/3 = k\n     y = (7/3)x
972191
The two most common methods of normalizing data are:\n\n1. Subtract the mean, then divide by the standard deviation (or variance)\n\nor...\n\n2. Subtract the minimum, then divide by the range.\n\nThe first method above yields a distribution with a mean of zero and a standard deviation (or variance) of 1.\n\nThe second method yields a distribution which ranges from 0 to 1.\n\nOf course, other methods are possible, such as subtracting the median then dividing by the inter-quartile range.
165255
Take a look at the error function, and modify it so that you are dealing with distributions in 2 dimensions, using polar coordinates.\n\nFor example, the counterpart of the error function in 2 dimensions, in polar coordinates can actually be integrated, to produce the comulative distribution function\n\nF(R,s)=1-e^[-(R^2)/(2*s^2)]\n\nwhere s is the standard deviation, and R is the radius of a circle centered on the origin. \n\nAs R--> infinitity, then F(R,s)----> 1\nF(0,s)=0\n\nSuppose that the target has an edge R0 such that\n\nF(R0,s)=.5 for a given value of s. \n\nThen, you have to find the value of S for which F(R0,S)=0.9.
1143920
I don't know enough about your specific topic but you can express variables in science as ratios or functions of other variables.\nIn order to be specific about how these variables effect and relate to each other you must then be specific about the values of the invariables, or constants as they are more commonly called.\n\nSo .. you should be able to relate and analyse variants in terms of the invariants but without being specific as to the nature of the invariants.\n\nI think that would work.
402618
I have no idea about the math, but I'm thinking the assumption is that the data are relatively correlated linearly.  There's a lot of math behind the wiki link.