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The variance of 1, 2, 3, 4, 5, 6 is

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$\frac{35}{12}$

$- \frac{3}{5}$

$\frac{1}{3}$

$\frac{2}{9}$

answer is D.

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Detailed Solution

Mean = $\frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{7}{2}$

$c^{2} + \frac{49}{4} = \frac{1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}$

$σ^{2} = \frac{91}{6} - \frac{49}{4} = \frac{35}{12}$

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