What is the variance of {7, 3, -1, 1, -3, 4, -2}?

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What is the variance of {7, 3, -1, 1, -3, 4, -2}?

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Answer 1 · 2016-03-14T22:47:58

Variance $σ^{2} = \frac{542}{49} = 11.0612$ $sigma^2=542/49=11.0612$

Explanation:

Solve the mean $¯ x$ $barx$ first

$¯ x = \frac{7 + 3 + (- 1) + 1 + (- 3) + 4 + (- 2)}{7} = \frac{9}{7}$ $barx=(7+3+(-1)+1+(-3)+4+(-2))/7=9/7$

Solve Variance $σ^{2}$ $sigma^2$

$σ^{2} = \frac{{(7 - \frac{9}{7})}^{2} + {(3 - \frac{9}{7})}^{2} + {(- 1 - \frac{9}{7})}^{2} + {(1 - \frac{9}{7})}^{2} + {(- 3 - \frac{9}{7})}^{2} + {(4 - \frac{9}{7})}^{2} + {(- 2 - \frac{9}{7})}^{2}}{7}$ $sigma^2=((7-9/7)^2+(3-9/7)^2+(-1-9/7)^2+(1-9/7)^2+(-3-9/7)^2+(4-9/7)^2+(-2-9/7)^2)/7$

$σ^{2} = \frac{542}{49} = 11.0612$ $sigma^2=542/49=11.0612$

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What is the variance of {7, 3, -1, 1, -3, 4, -2}?

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