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Variance is the squared deviation from the mean. (X bar - X data)^2

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Q: Is variance based on deviations from the mean?
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When computing the sample variance the sum of squared deviations about the mean is used for what reason?

You want some measure of how the observations are spread about the mean. If you used the deviations their sum would be zero which would provide no useful information. You could use absolute deviations instead. The sum of squared deviations turns out to have some useful statistical properties including a relatively simple way of calculating it. For example, the Gaussian (or Normal) distribution is completely defined by its mean and variance.


Why do you square the deviations to get the variance and then take the reverse action of taking the square root of the variance to return the variance to sigma?

because of two things- a) both positive and negative deviations mean something about the general variability of the data to the analyst, if you added them they'd cancel out, but squaring them results in positive numbers that add up. b) a few larger deviations are much more significant than the many little ones, and squaring them gives them more weight. Sigma, the square root of the variance, is a good pointer to how far away from the mean you are likely to be if you choose a datum at random. the probability of being such a number of sigmas away is easily looked up.


Is the variance of a group of scores the same as the sum of the squared deviations or the average of the squared deviations?

Given a set of n scores, the variance is sum of the squared deviation divided by n or n-1. We divide by n for the population and n-1 for the sample.


Why are variances squared?

The variances are squared so that all deviations above and below the mean become positive values. Taking the square root of the variance then gives a measure of the differences from the mean: the standard deviaton. Squaring the deviations also makes the bigger differences stand out. Look at 100 squared vs 10 squared.


How many standard deviations is 16.50 from the mean?

How many standard deviations is 16.50 from the mean?

Related questions

The sum of the deviations about the mean always equals what?

The sum of total deviations about the mean is the total variance. * * * * * No it is not - that is the sum of their SQUARES. The sum of the deviations is always zero.


Can the standard deviation or variance be negative?

No, a standard deviation or variance does not have a negative sign. The reason for this is that the deviations from the mean are squared in the formula. Deviations are squared to get rid of signs. In Absolute mean deviation, sum of the deviations is taken ignoring the signs, but there is no justification for doing so. (deviations are not squared here)


When computing the sample variance the sum of squared deviations about the mean is used for what reason?

You want some measure of how the observations are spread about the mean. If you used the deviations their sum would be zero which would provide no useful information. You could use absolute deviations instead. The sum of squared deviations turns out to have some useful statistical properties including a relatively simple way of calculating it. For example, the Gaussian (or Normal) distribution is completely defined by its mean and variance.


What is the variance of of the scores 3 4 8 and 9?

sum of scores: 24 mean of scores : 24/4 = 6 squared deviations from the mean: 9, 4,4,9 sum of these: 26 sample variance: 26/4 = 6.5


How do you find the variance 415 10 8 33 25 36 66 63 50 64 49 40?

First mean is calculated.Then calculate deviations from the mean.Then the deviations are squared.Then the squared deviations are summed up.Finally this sum is divided by number of items for which the variance is being calculated. For a population, it is by the number of values, in this case 12. If it is a sample, then we divide by one less, which is 11,For these figures, the variance for the population is 11069.24306. If it is a sample, it is 12075.53788 as the result.


Why do you square the deviations to get the variance and then take the reverse action of taking the square root of the variance to return the variance to sigma?

because of two things- a) both positive and negative deviations mean something about the general variability of the data to the analyst, if you added them they'd cancel out, but squaring them results in positive numbers that add up. b) a few larger deviations are much more significant than the many little ones, and squaring them gives them more weight. Sigma, the square root of the variance, is a good pointer to how far away from the mean you are likely to be if you choose a datum at random. the probability of being such a number of sigmas away is easily looked up.


Why do you divide by instead of when calculating the sample variance?

Usually the sum of squared deviations from the mean is divided by n-1, where n is the number of observations in the sample.


Is the variance of a group of scores the same as the sum of the squared deviations or the average of the squared deviations?

Given a set of n scores, the variance is sum of the squared deviation divided by n or n-1. We divide by n for the population and n-1 for the sample.


What is the sum of the deviations from the mean?

The sum of standard deviations from the mean is the error.


Why are variances squared?

The variances are squared so that all deviations above and below the mean become positive values. Taking the square root of the variance then gives a measure of the differences from the mean: the standard deviaton. Squaring the deviations also makes the bigger differences stand out. Look at 100 squared vs 10 squared.


How many standard deviations is 16.50 from the mean?

How many standard deviations is 16.50 from the mean?


Difference between standard deviation and mean devition?

The mean deviation (also called the mean absolute deviation) is the mean of the absolute deviations of a set of data about the data's mean. The standard deviation sigma of a probability distribution is defined as the square root of the variance sigma^2,