What is the variance of {-2, 5, 18, -8, -10, 14, -12, 4}?

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What is the variance of {-2, 5, 18, -8, -10, 14, -12, 4}?

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Answer 1 · 2016-03-15T11:00:27

Variance $σ^{2} = \frac{6903}{64} = 107.8593$ $sigma^2=6903/64=107.8593$

Explanation:

compute the arithmetic mean $μ$ $mu$ first

$n = 8$ $n=8$

$μ = \frac{- 2 + 5 + 18 + (- 8) + (- 10) + 14 + (- 12) + 4}{8}$ $mu=(-2+5+18+(-8)+(-10)+14+(-12)+4)/8$

$μ = \frac{- 32 + 41}{8}$ $mu=(-32+41)/8$

$μ = \frac{9}{8}$ $mu=9/8$

compute the variance $σ^{2}$ $sigma^2$ using the variance formula for population

$σ^{2} = \frac{\sum {(x - μ)}^{2}}{n}$ $sigma^2=(sum(x-mu)^2)/n$

$σ^{2} = \frac{{(- 2 - \frac{9}{8})}^{2} + {(5 - \frac{9}{8})}^{2} + {(18 - \frac{9}{8})}^{2} + {(- 8 - \frac{9}{8})}^{2} + {(- 10 - \frac{9}{8})}^{2} + {(14 - \frac{9}{8})}^{2} + {(- 12 - \frac{9}{8})}^{2} + {(4 - \frac{9}{8})}^{2}}{8}$ $sigma^2=((-2-9/8)^2+(5-9/8)^2+(18-9/8)^2+(-8-9/8)^2+(-10-9/8)^2+(14-9/8)^2+(-12-9/8)^2+(4-9/8)^2)/8$

$σ^{2} = \frac{6903}{64}$ $sigma^2=6903/64$

$σ^{2} = 107.8593$ $sigma^2=107.8593$

God bless ....I hope the explanation is useful.

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What is the variance of {-2, 5, 18, -8, -10, 14, -12, 4}?

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