how do you figure standard deviation?
114212
First you have to find the mean (average) of the set of numbers. Standard deviation measures how spread out the values in a data set are. To find standard deviation, take the difference of each number in the data set and the mean, square them individually and then add them together. Then divide that by the number of values. And then take the square root of all of that. It's really hard to write that equation in just plain text, but the link shows you what it looks like.\n\nIt's easier to see with an example.\nFind the standard deviation of 45.7, 43.8, 42.6, and 48.1.\n\nThe mean is the sum of all the numbers divided by the amount.\n(45.7+43.8+42.6+48.1)/4= 180.2/4= 45.05\n\nNow find standard deviation, sd:\nsd^2= [(45.7-45.05)^2 + (43.8-45.05)^2 + (42.6-45.05)^2 + (48.1-45.05)^2]/4\n\nsd^2= [(.65)^2 + (-1.25)^2 + (-2.45)^2 + (3.05)^2]/4\n\nsd^2= (.4425 + 1.5625 + 6.0025 + 9.3025)/4\n\nsd^2= 17.29/4=4.3225\n\nsd=sqrt(4.3226)= 2.08
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Range: 19\nStdev: 6.27\n\nRange is just the max value minus the min value.  To learn more about calculating standard deviation, visit http://en.wikipedia.org/wiki/Standard_deviation\n\nSo, I saw someone got a different standard deviation than me, so I checked it again on Excel and I still got 6.27.  Instead of dividing by 7, I divided by 6 (n-1).\n\nIf these data are a sample of a larger population, then you divide by 6 (n-1) to estimate the stdev of the population, so the stdev is 6.27.  If you just want to know the stdev of those numbers, then you divide by 7 (n) so the stdev is 5.81.
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standard deviation is one of measure of variability use in statistics which has the same unit as the given data and is equivalent ot the square root of the variance.\n\nIt measures how all the data varies. It indicates how the data in a distribution scatter or spread.\n\nThe lower the standard deviation, the closer are all the data around the mean.\nThat is, the data are not scattered and they are less heterogenuous, less variable.\n\nThe greater its value, the more the spreadness of the data around the mean.\n\nFor example,\nGiven:\nA: 45,46,40,43,48,46\nB:10,45,28,100,158,468\n\nThe standard deviation of the data A is 2.804757863\nThe standard deviation of the data B is 171.9400089\n\n]since the statndard deviation of the data A is less that in B,\nthe data in A are closer to each other, compare to the data B.
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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.
588275
Do you have area tables for the standard normal distribution? If so, look in the table for 0.2514 and read of the z-value (approx 0.68). That means 25.14% is 0.68 standard deviations from the mean. If 30 is 0.68 standard deviations from the mean, then 30/0.68 = 1 standard deviation. Do the sum then go and get your A+ grade.\n\nIf you haven't studied the standard normal distribution, then your teacher is asking too much too soon.
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1 standard deviation range is\n\n74-13=61\nto\n74+13=87\n\nSo any score between 61-87 is 1 standard deviation of the mean 74. So a 62 is in that range.
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By calculationg it's mean. Add all the values of observations and then divide by the number of observations.\nThat's all!
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Gosh.   I have no idea how to do it in excel.... \n\nbut for a normal distribution with an error function that always returns 1 (no error)  then the percentage of the population between -n standard deviations and +n standard deviations is n/(sqrt(2) \n\nSo, if you want there to be 90% of the population between -n and +n so that 5% is below -n and 5% is above +n then \n\n0.9 = n/sqrt(2)\n0.9 * sqrt(2) = n\n\n1.27279221 = n \n\nso,  now we take 1.27279221 time the standard deviation of 0.87 and then subtract it from 10.42 \n\nUnderweight = 10.42 - (1.27279921 * 0.87)\n\n= 9.31266469\n\nand then you do the same thing for 80% (except you add) to get the weight at 10% over....
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Standard deviation tells us how far from the mean the data points tend to be.\n\nFor example, if the mean is 20 and SD is 5. Then all the numbers used to calculate the average (20) can be 5 standard deviation away from it. So the numbers can be from anywhere of 15-25.
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Approximately 95% of the data is within 2 standard deviations of the mean.\nCheck out this website\nhttp://en.wikipedia.org/wiki/Confidence_interval\nand look under the section\n"Confidence Intervals in Measurement"
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Standard error is standard deviation divided by n^1/2. In other words, it estimates the standard deviation of the sample mean _based on the population mean_.
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Your going to compare your mean with the population mean and see if your class really better than the average of the population.\nHo: u = xbar\nHa: u < xbar\n\nalpha = 0.10\none tailed test (of course)\ntherefore: z(table) = -1.28 (i'm not sure)\n\nuse z test since you know the population standard deviation.\n\nz = (x - u)/o/sqrt of n\n\nthat's it.\n\ngood luck
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Find the midpoint of each interval.\nFor 1-3, the midpoint is (3+1)/2 = 4/2 = 2\nFor 4-6, the midpoint is (6+4)/2 = 10/2 = 5\nFor 7-9, the midpoint is (9+7)/2 = 16/2 = 8\n2(1) + 5(6) + 8(3)\n= 2+30+24\n= 56\nn = number of data points = 1+6+3 = 10\nvariance = [1/(10-1)][((2^2)(1) +(5^2)(6) +(8^2)(3)) - (56^2)/10]\n= (1/9)((4+150+192) - 3136/10)\n= (1/9)(346 - 313.6)\n= (1/9)(32.4)\n= 3.6\nstandard deviation = sqrt(variance)\n= sqrt(3.6) \n= 1.9
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Standard deviation of a random variable X is defined as: \nό=√E((X-E(X))2) = √E(X2)-(E(X))2\n(the 2's of course are squares)
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The standard deviation tells you how far away something is from the average.  You can do a standard deviation on any set of data, however most people are familiar with the Normal Distribution.
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Angle of deviation, d = i1 + i2 - A\nA= r1 + r2\nIn minimum deviation, µ = {sin(A+D)/2}/(sinA/2)
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Variance = (∑x²)/n - (∑x/n)² \n= (4 + 25 + 64)/3 - [(2 + 5 + 8)/3]²\n= 31 - 25\n= 6\nThus standard deviation = √(variance)\n= √6