What is Standard Deviation? How it is calculated? What are the formulas of Standard Deviation?
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
107133
You have all the information you need.\n\nThe Mean is the same as the average. It's just another term for a simple average. So it's $7500\n\nThe standard error of the mean is the standard deviation divided by the square root of the number of samples. So it's just $650 divided by 10, or 65.
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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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.
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Finance:\nStandard deviation might be used to characterize how much spread to expect in a group of investments.  High standard deviation implies high uncertainty.\n\nManufacturing:\nLikewise for manufactured parts.  High standard deviation here means highly variable parts.  Manufacturers was low variability, so that parts fit together well.\n\nMedicine:\nA drug might be tested and shown to reduce blood pressure.  If the amount of reduction has a high standard deviation, though, the drug may not be reliable.
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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).
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it is the average amount that each sample (or reading number) deviates from the mean (average). It is a measure of how widely the data is scattered away from the mean.\n\nThis is highly simplified, you might want to look on wikipedia for a better explaination. (it involves square roots and other nasty maths)
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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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There are actually many different types of "mean." The most common "mean"  is the arithmetic mean, also called the average in everyday language.\n\nCalculating the arithmetic mean is simple: Add all the values in the sample together and divide by the total number of values in the sample.\n\nThe arithmetic mean is often used to describe data. The problem is that it doesn't always necessarily give the correct picture.  Sometimes it is better to use the median, which is the number found exactly in the middle of the set of values, or the mode, which is the number most frequently reported. It depends on the data involved and the question you are asking about the data.\n\nWhat this all comes down to is: the mean is a measure of central tendency (as are the median and the mode).\n\nThe other two most common means are the geometric mean and the harmonic mean. If you are interested in leanring more, visit Wikipedia.
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Standard deviation tells you the spread of the data, for example the data may be very close together or very spread out.\n\nConfidence limits tell you how accurate a particular model or hypothesis test is\n\nQ-test tells you about extreme values\n\nPolled standard deviation is the standard deviation of the sample of data, not necessarily the same as the actual standard deviation
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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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Standard deviation is often called standard error.  Say you take a sample of 100 people to know their averga ages.  You calculate the standard deviation by first knowing what the average is.  Let´s say the average age is 35 years old.  By definition, the standard deviation would be how many years are the people on your sample older or younger than the average. It could be about 3 years.  That means that most people in your sample has ages between 32 (35-3) and 38 (35+3) years.\nby now you should have the formula, and some example.  Notice that if you draw a graph made by columns with your data, the standard deviation measures exactly a proportion of the width of a column in your graph.
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You are correct. The definition of mean is the sum of the data divided by the sample size. It is a mathematical average of a set of numbers or measurements. To find a mean you first add up all the numbers that you have. Then you divide that figure by how many numbers you have. Keep up the good work.\nEx. 234+410+355+288=1,287\n      1,287/4=321.75 or 322
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The standard deviation is the square root of the variance. Now, the variance is the "average squared difference (distance) of each observation from the mean".\n \nLets think about it. You want a measure of how dispersed the data is, right? Well, suppose you have heights for your data and your trying to compare two groups. For example, you want to know if the height´s of New Yorkers vary more or less than the heights of Floridians. You are NOT trying to see which are taller.\n \nSo you can calculate the average height of each group and then see haw far away from this average each individual height is and then calculate the average distance from the average height. If everyone had the same height, the average distance would be ZERO, reflecting that the population´s height is very CONCENTRATED around its average.\n \nBut why do you use the squared difference? Because if you were to calculate the average of "x - average(x)" you would always get zero, so by squaring and taking the square, you always get a positive number (the distance is always positive).\n \nGot it?
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2 standard deviations above average (100).\n\nThat means 130 or 142 depending on the IQ test(s) being used.\n\nI joined a few years ago, mostly to prove I could, but found the whole thing pretty dull.
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The mean is simply the average of the data.  Add them all up and divide by 15.\n\n69.54
149503
Right, so feed the numbers into your calculator and press the SD button.\n\nOr enter it into Excel and use teh standard deviation function.