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

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

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Answer 1 · 2016-06-24T11:29:55

The population variance of the data set is

$σ^{2} = 35$ $sigma^2=35$

Explanation:

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 $n / (n - 1)$ $n//(n-1)$

The formula for the population variance is

$σ^{2} = \frac{1}{N} N \sum i = 1 {(x_{i} - μ)}_{i}^{2}$ $sigma^2=1/N sum_(i=1)^N(x_i-mu)^2$

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

$μ = \frac{1}{N} N \sum i = 1 x_{i}$ $mu = 1/N sum_(i=1)^N x_i$

In our population the mean is

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

Now we can proceed with the variance calculation:

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

$σ^{2} = 35$ $sigma^2 = 35$

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

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