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2022-10-05

What is the variance of {-4, 5, 8 ,-1, 0 ,4 ,-12, 4}?

Quinn Alvarez

Beginner2022-10-06Added 13 answers

First, let's assume that this is the entire population of values. Therefore we are looking for the population variance . If these numbers were a set of samples from a larger population, we would be looking for the sample variance which differs from the population variance by a factor of n/(n-1)

The formula for the population variance is

${\sigma}^{2}=\frac{1}{N}\sum _{i=1}^{N}({x}_{i}-\mu {)}^{2}$

where $\mu $ is the population mean, which can be calculated from

$\mu =\frac{1}{N}\sum _{i=1}^{N}{x}_{i}$

In our population the mean is

$\mu =\frac{-4+5+8-1+0+4-12+4}{8}=\frac{4}{8}=\frac{1}{2}$

Now we can proceed with the variance calculation:

${\sigma}^{2}$

$\frac{(-4-\frac{1}{2}{)}^{2}+(5-\frac{1}{2}{)}^{2}+(8-\frac{1}{2}{)}^{2}+(-1-\frac{1}{2}{)}^{2}+(0-\frac{1}{2}{)}^{2}+(4-\frac{1}{2}{)}^{2}+(-12-\frac{1}{2}{)}^{2}+(4-\frac{1}{2}{)}^{2}}{8}$

${\sigma}^{2}=35$

The formula for the population variance is

${\sigma}^{2}=\frac{1}{N}\sum _{i=1}^{N}({x}_{i}-\mu {)}^{2}$

where $\mu $ is the population mean, which can be calculated from

$\mu =\frac{1}{N}\sum _{i=1}^{N}{x}_{i}$

In our population the mean is

$\mu =\frac{-4+5+8-1+0+4-12+4}{8}=\frac{4}{8}=\frac{1}{2}$

Now we can proceed with the variance calculation:

${\sigma}^{2}$

$\frac{(-4-\frac{1}{2}{)}^{2}+(5-\frac{1}{2}{)}^{2}+(8-\frac{1}{2}{)}^{2}+(-1-\frac{1}{2}{)}^{2}+(0-\frac{1}{2}{)}^{2}+(4-\frac{1}{2}{)}^{2}+(-12-\frac{1}{2}{)}^{2}+(4-\frac{1}{2}{)}^{2}}{8}$

${\sigma}^{2}=35$

Read carefully and choose only one option

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