The mean and S.D. of 1, 2, 3, 4, 5, 6 are | vedclass

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(A) The mean $\bar{x} = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = \frac{7}{2}$.The variance $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$.F

Vedclass · Accepted Answer

$\frac{7}{2}, \sqrt{\frac{35}{12}}$

Vedclass · Answer

(A) The mean $\bar{x} = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = \frac{7}{2}$.The variance $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$.First,calculate $\sum x_i^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91$.Then,$\sigma^2 = \frac{91}{6} - (\frac{7}{2})^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12}$.The standard deviation $S.D. = \sigma = \sqrt{\frac{35}{12}}$.

$x_i$	$60$	$61$	$62$	$63$	$64$	$65$	$66$	$67$	$68$
$f_i$	$2$	$1$	$12$	$29$	$25$	$12$	$10$	$4$	$5$

The mean and $S.D.$ of $1, 2, 3, 4, 5, 6$ are

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