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# The variance of 6, 8, 10, 12, 14 is

$\begin {array} {1 1} (A)\;1 & \quad (B)\;8 \\ (C)\;12 & \quad (D)\;16 \end {array}$

Mean $= \overline x = \large\frac{6+8+10+12+14}{5}$
$= \large\frac{50}{5}$
$= 10$
$\sigma^2=\large\frac{ \Sigma (x-M)^2}{n}$
$\sigma^2=8$
Ans : (B)