Standard deviation
Population standard deviation of grades of eight students

Suppose that the entire population of interest is eight students in a particular class. For a finite set of numbers, the population standard deviation is found by taking the square root of the average of the squared deviations of the values subtracted from their average value. The marks of a class of eight students (that is, a statistical population) are the following eight values: 2, 4, 4, 4, 5, 5, 7, 9. {\displaystyle 2,\ 4,\ 4,\ 4,\ 5,\ 5,\ 7,\ 9.}

These eight data points have the mean (average) of 5: μ = 2 + 4 + 4 + 4 + 5 + 5 + 7 + 9 8 = 40 8 = 5. {\displaystyle \mu ={\frac {2+4+4+4+5+5+7+9}{8}}={\frac {40}{8}}=5.}

First, calculate the deviations of each data point from the mean, and square the result of each: (2 − 5) 2 = (− 3) 2 = 9 (5 − 5) 2 = 0 2 = 0 (4 − 5) 2 = (− 1) 2 = 1 (5 − 5) 2 = 0 2 = 0 (4 − 5) 2 = (− 1) 2 = 1 (7 − 5) 2 = 2 2 = 4 (4 − 5) 2 = (− 1) 2 = 1 (9 − 5) 2 = 4 2 = 16. {\displaystyle {\begin{array}{lll}(2-5)^{2}=(-3)^{2}=9&&(5-5)^{2}=0^{2}=0\\(4-5)^{2}=(-1)^{2}=1&&(5-5)^{2}=0^{2}=0\\(4-5)^{2}=(-1)^{2}=1&&(7-5)^{2}=2^{2}=4\\(4-5)^{2}=(-1)^{2}=1&&(9-5)^{2}=4^{2}=16.\\\end{array}}}

The variance is the mean of these values: σ 2 = 9 + 1 + 1 + 1 + 0 + 0 + 4 + 16 8 = 32 8 = 4. {\displaystyle \sigma ^{2}={\frac {9+1+1+1+0+0+4+16}{8}}={\frac {32}{8}}=4.}

and the population standard deviation is equal to the square root of the variance: σ = 4 = 2. {\displaystyle \sigma ={\sqrt {4}}=2.}

This formula is valid only if the eight values with which we began form the complete population. If the values instead were a random sample drawn from some large parent population (for example, they were 8 students randomly and independently chosen from a class of 2 million), then one divides by 7 (which is n − 1) instead of 8 (which is n) in the denominator of the last formula, and the result is s = 32 / 7 ≈ 2.1. {\textstyle s={\sqrt {32/7}}\approx 2.1.} In that case, the result of the original formula would be called the sample standard deviation and denoted by s {\textstyle s} instead of σ. {\displaystyle \sigma.} Dividing by n − 1 {\textstyle n-1} rather than by n {\textstyle n} gives an unbiased estimate of the variance of the larger parent population. This is known as Bessel's correction. Roughly, the reason for it is that the formula for the sample variance relies on computing differences of observations from the sample mean, and the sample mean itself was constructed to be as close as possible to the observations, so just dividing by n would underestimate the variability.
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