Standard deviation is the square root of the variance; so if the variance is 64, the std dev is 8.
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I believe you are interested in calculating the variance from a set of data related to salaries. Variance = square of the standard deviation, where: s= square root[sum (xi- mean)2/(n-1)] where mean of the set is the sum of all data divided by the number in the sample. X of i is a single data point (single salary). If instead of a sample of data, you have the entire population of size N, substitute N for n-1 in the above equation. You may find more information on the interpretation of variance, by searching wikipedia under variance and standard deviation. I note that an advantage of using the standard deviation rather than variance, is because the standard deviation will be in the same units as the mean.
You can't. You need an estimate of p (p-hat) q-hat = 1 - p-hat variance = square of std dev sample size n= p-hat * q-hat/variance yes you can- it would be the confidence interval X standard deviation / margin of error then square the whole thing
Standard deviation calculation is somewhat difficult.Please refer to the site below for more info
If the population standard deviation is sigma, then the estimate for the sample standard error for a sample of size n, is s = sigma*sqrt[n/(n-1)]
Standard deviations are measures of data distributions. Therefore, a single number cannot have meaningful standard deviation.