std deviation/variance?
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.
553133
This is an example for standard deviation:\n\nSay you have these values and need to find the deviation for them\n5, 10, 15, 20, 25 \n\nMean = 15\n\nNow, calculate the difference of the values from the mean\n\n15-5 = 10\n15-10 = 5\n15-15 = 0\n20 - 15 = -5\n25 - 15 = -10\n\nNow, square the differences and add them\n\n10^2 + 5^2 + (-5)^2 + (-10)^2 = 100 + 25 + 25 + 100 = 250\n\nStandard deviation is the squared root of this sum divided by the mean\n\nStandard Deviation = (250/15)^(1/2) \n                   = 5 * (2/3)^(1/2)\n                   = 5 * 0.816\n                   = 4.082
47438
Take each of your values, and then subtract the mean from them.  E.g., 70 - 56 = 14, 60 - 56 = 4, etc.\n\nNow take each of these answers and square it.  Half of the starting numbers will be negative (eg 30 - 56 = -26) but once you square all of these numbers, they should all be positive.\n\nNow you'll have 7 positive numbers. Add them all together and divide the answer by 7.  Finally, take the square root of that value, and that's the standard deviation.\n\nSo 14, 4, -6, 4, 4, 9, and -26\nbecome 196, 16, 36, 16, 16, 81, and 676\nSummed together gives 1037\ndivided by 7 is ~148.14\nSquare-rooted is ~12.2\n\nThe standard deviation is 12.2.\n\nNOTE: the rt11guru answer that follows is based on a technique used in some larger sample sets where 1 is subtracted from the number of entries, ie the sum is divided by 7-1=6 rather than 7.  That is called the *sample standard deviation*, as shown in the link.
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.
967643
z score= (n-mean)/standard deviation\nstandard deviation = square root of variance\n\nz score = (50-24)/ 5\n            = 26/5\n            = 5.2\n\nThe z score of the data is 5.2, meaning n is 5.2 standard deviations from the mean.
64438
Let me see if I can answer your question in English. :-)\n\nI'm sure you have an equation in your notes or book that you can just start plugging stuff into.  It probably looks like this:\n\ns = square root of (X - X bar)^2/(n-1)\n\nBut if you want to know why it works, keep reading.\n\nYou have a list of numbers.  First, you need to find the average, or mean, of those numbers, by adding them up and dividing by how many there were.\n\nNow, if you subtract each number from the mean, you get a new list of numbers.  These are the deviation scores, and tell you how far above or below average each number was.  If you add them up, you will get zero, because some of them are positive and some negative.  That's not very interesting, so we square them to make them all positive.  Then we add up all those squared deviation scores.\n\nYou probably got a pretty large number, and not the one we want, yet.  We want the Standard Deviation, which is sort of like the average deviation.  So just like you divide by the number of scores to get the average, you need to divide the sum of the squared deviation scores by something.  Sometimes we divide it by the number of scores, like you would think, but for complicated reasons it works better if you divide by one less than that.  So if you started with 5 numbers, divide by 4 now.  (Be sure to check your notes or your book to see what your teacher said you should divide by.)\n\nWe're still one step away.  The number you should have now is called the Variance.  It is also a measure of how spread out your numbers were, but it's bigger than we want.  Remember how we squared the deviation scores earlier?  Now we need to undo that by taking the square root.  What you get is the Standard Deviation you wanted.
872957
Look at them carefully, find out the diff between each number and it should be obvious.\n\nEquations?
405752
Specifically: press STAT, then ENTER. This takes you to the lists. Put in a value and press enter, and continue until all your values are entered into the list. \nThen press STAT again. This time use the arrow key to move to CALC, then press 1-Var Stats. Then press 2nd then watever list you entered the data into. List one would be 2nd then 1 key, list two is 2nd and 2 key. L1, L2, L3 L4 L5 and L6 all have their own key. \n\nPress enter. The first number is x bar, or the mean. The SD, standard deviation is Sx. \n\nHope that helps
278272
You'll use Z-scores and the Z-table. Z-scores convert deviations to standard deviation units. \n\nIn this case, you're looking for the range of scores that makes up the middle 88% of the distribution - 44% on either side of the mean. \n\nThe z-score is 1.56 (look in the table for the z-score that cuts off 44% of the distribution.) This tells you that 44% of the scores fall between the mean and 1.56 standard deviation units above the mean - and 44% of all scores fall between the mean and 1.56 standard deviation units below the mean. \n\nNow, since each standard deviation unit in this particular distribution is 4ml (as per your question), 1.56 standard deviation units is equal to 1.56 x 4, which is equal to 6.24ml.\n\nBack to the problem - your range is 6.24 ml above and below the mean - which is 250ml - therefore the range of volumes is from 6.24ml less than 250 to 6.24ml more than 250 - or in other words - 243.76ml to 256.24 ml.\n\nHope that helped!
919154
It depends on whether you want the mean number of heads, the total number of heads, etc.\nIf you want the total, multiply the standard deviation by sqrt(n), where you have n tosses.\nIf you want the average, divide by sqrt(n).\n\n(Note, the formula talked about below is for the *margin* of error, not the standard error. And thats only for means.)
878193
EDIT: This answer has been corrected.  See note below.\n\n\nIn the case of the dice, you have discrete values in the probability distribution.  It is not a "normal" (bell-curve) distribution.  But you can calculate the std dev. as follows:\n\nFor every possible value v of a throw i, compute P(vi).  In a large number of throws N, the expected number of times a value vi occurs, nvi = N*P(vi).  The mean value would then be m = (1/N)∑nv*v = (1/N)∑N*v*P(v) = ∑v*P(v).  The standard deviation would then be\n\nd^2 = [1/(N-1)]∑(xi - m)^2 =  [1/(N-1)]*{∑[N*vi*P(vi)]^2 - 2*N*m*∑v*P(vi) +N*m^2}\n\nd = √ [1/(N-1)]*{∑[N*vi*P(vi)]^2 - 2*N*m*∑vi*P(vi) +N*m^2}\n\n\nYou can't eliminate N (the total no of triails) but you can compute the result for a large N without actually rolling the dice.\n\nEDIT:  There is an error in the formula: the terms N and P(vi) should not be squared in the first term of the sum. The formula should be \n\nd = √ [1/(N-1)]*{∑N*P(vi)*vi^2 - 2*N*m*∑vi*P(vi) +N*m^2}\n\nThen, for large N, N-1≈1 and the N's cancel out to give\n\nd = √ [∑P(vi)*vi^2 - 2*m*∑vi*P(vi) +m^2]\n\nSo for all possible values of the dice (vi), compute the probabiity of that value P(vi), the mean value m = ∑vi*P(vi),  and sum over all value according to the above formula.\n\nPlugging the mean value into the formula gives the simplification:\n\nd = √ [∑P(vi)*vi^2 - 2*m^2 +m^2] = √ [∑P(vi)*vi^2 - m^2]\n\nFINAL:  Made final correction.  Above formula is now correct; the m^2 is not under the summation sign.\n\nCalculated for an ordinary pair of dice: m = 7, d = 2.415\nYou can see the calculation here http://img452.imageshack.us/img452/1734/dicestddevlk8.png
999718
what was the question?\n\n\nDo you want to know terms like mean, average, mode, standard deviation, etc.?
411069
http://en.wikipedia.org/wiki/STD%27s
697345
I get a different answer. Surely none of these answers can be right: in no case is the interval symmetric about the mean = 10.\n\nSample size n = 4\nSample mean xbar = 40/4 = 10\nPopulation mean = mu (unknown)\nSample variance s^2 = (0 + 4 + 1 + 1)/(4-1) = 2\n(Note: dividing here by n-1 = 4-1 is more accurate than by n because it makes s^2 an exactly unbiased estimator of the population variance sigma^2. This would matter little if n were large.)\n\nStd error = s/sqrt(n) = sqrt(2)/2 = 0.707\nSo xbar-mu is normally distributed with mean 0 and std dev 0.707 (as estimated from the sample; its exact but unknown value is sigma/sqrt(n) = sigma/2).\nSo the 95% confidence interval for mu is\n\n[10 - 1.96*0.707, 10 + 1.96*0.707] = [8.614,11.386].\n\n\nRemark 1: With such a small sample, you really have to take (n-1) = 3 instead of n = 4 in calculating the sample variance s^2. The formula for s^2 "mean of squares - square of mean" under-estimates sigma^2 by the factor n/(n-1) = 4/3 here. This is not inconsiderable. The unbiased estimate of s^2 from this sample is 2, and NOT 1.5 as in Wal C's solution.\nEven the value s^2 = 2 will lead to under-estimating the population std dev. sigma itself (because sqrt is a concave fctn, s is an under-estimating estimator of sigma even though s^2 is an unbiased estimator of sigma^2).\n\nRemark 2: Wal C's solution is even more seriously wrong. She should have used the std error, the std dev of the sample mean, which is sigma/sqrt(n). But she used the std dev. of a SINGLE trial, sigma. If that were correct, there would never be any point in increasing the sample size! The point is of course that the std error sigma/sqrt(n) goes down as n increases -- but you've got to use this fact, not ignore it!\n\nRemark 3: My solution above is also faulty. The sample size is so small that you must use the Student t-distribution and not the normal distribution. (xbar-mu)/(s/sqrt(n)) is not normally distributed; it has the Student t-distribution with 4-1=3 degrees of freedom. The 97.5 percentile is 3.18 (and not 1.96 as for the normal distr), and so the 95% confidence interval for mu is\n\n[10 - 3.18*0.707, 10 + 3.18*0.707] = [7.75, 12.25].
643078
Ok well, seems like u are from india... awaiting ur 10th standard results hun?( im an indian too but living overseas) Well, if im not wrong.. the 12th std results will be out on 5th june. So, erm.. the results of 10th std will be out a few days later... say... 20th June, maybe? Main thing is that.. until or unless the results are out, NO SITE wud reveal them to u.... u just have to wait....
308472
Find yourself a z-statistic table and look up critical value in the table.  Check your stats book on how to interpret the table.\nhttp://www.math2.org/math/stat/distributions/z-dist.htm
280200
If a random sample size of 25 corresponds to a standard deviation of 30, then you can expect a random sample of size 25N to roughly correspond to a standard deviation of 30 / sqrt(N).\n\nIf 99% of the area under the bell curve is between -Z and +Z standard deviations, then you want 30 Z / sqrt(N) to be less than or equal to 0.5.\n\nMaybe a little more because of the variation of the estimated standard deviation from the true standard deviation.\n\nAlso check that the theory and the measurements both suggest it is appropriate to use the normal or Gaussian distribution.