If the mean deviation about the mean is m and | vedclass

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(D) First,we calculate the mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4(1) + 24(3) + 28(5) + 16(7) + 8(9)}{4 + 24 + 28 + 16 + 8} = \frac{4

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$6$

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(D) First,we calculate the mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4(1) + 24(3) + 28(5) + 16(7) + 8(9)}{4 + 24 + 28 + 16 + 8} = \frac{4 + 72 + 140 + 112 + 72}{80} = \frac{400}{80} = 5$. Next,the mean deviation about the mean $m = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{4|1-5| + 24|3-5| + 28|5-5| + 16|7-5| + 8|9-5|}{80} = \frac{4(4) + 24(2) + 28(0) + 16(2) + 8(4)}{80} = \frac{16 + 48 + 0 + 32 + 32}{80} = \frac{128}{80} = \frac{8}{5} = 1.6$. The variance $\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - (\bar{x})^2 = \frac{4(1)^2 + 24(3)^2 + 28(5)^2 + 16(7)^2 + 8(9)^2}{80} - 5^2 = \frac{4 + 216 + 700 + 784 + 648}{80} - 25 = \frac{2352}{80} - 25 = 29.4 - 25 = 4.4 = \frac{22}{5}$. Finally,$m + \sigma^2 = \frac{8}{5} + \frac{22}{5} = \frac{30}{5} = 6$.

$x$	$1, 3, 5, 7, 9$
$f$	$4, 24, 28, 16, 8$

If the mean deviation about the mean is $m$ and variance is $\sigma^2$ for the following data,then $m + \sigma^2 =$

$x$ $1, 3, 5, 7, 9$

$f$ $4, 24, 28, 16, 8$

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If the mean deviation about the mean is $m$ and variance is $\sigma^2$ for the following data,then $m + \sigma^2 =$ $x$ $1, 3, 5, 7, 9$ $f$ $4, 24, 28, 16, 8$