What is the variance of the following numbers?: 63, 54, 62, 59, 52,

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Answer 1 · 2015-12-27T20:51:14

$σ^{2} = 18.8$ $sigma^2 = 18.8$

$m e a n = \frac{63 + 54 + 62 + 59 + 52}{5}$ $mean = (63 + 54 + 62 + 59 + 52)/5$
$m e a n = 58$ $mean = 58$
$n = 5$ $n = 5$

63
$x - m e a n = 63 - 58 = 5$ $x - mean = 63 - 58 = 5$
${(x - m e a n)}^{2} = 5^{2} = 25$ $(x - mean)^2 = 5^2 = 25$

54
$x - m e a n = 54 - 58 = - 4$ $x - mean = 54 - 58 = -4$
${(x - m e a n)}^{2} = {(- 4)}^{2} = 16$ $(x - mean)^2 = (-4)^2 = 16$

62
$x - m e a n = 62 - 58 = 4$ $x - mean = 62 - 58 = 4$
${(x - m e a n)}^{2} = 4^{2} = 16$ $(x - mean)^2 = 4^2 = 16$

59
$x - m e a n = 59 - 58 = 1$ $x - mean = 59 - 58 = 1$
${(x - m e a n)}^{2} = 1^{2} = 1$ $(x - mean)^2 = 1^2 = 1$

52
$x - m e a n = 52 - 58 = - 6$ $x - mean = 52 - 58 = -6$
${(x - m e a n)}^{2} = {(- 6)}^{2} = 36$ $(x - mean)^2 = (-6)^2 = 36$

$Sigma(x - mean)^2 = 25 + 16 + 16 + 1 + 36 = 94$

$sigma^2 = (Sigma(x - mean)^2)/n = 94/5 = 18.8$