Let the mean of the data

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Question

$5=\bar{x}=\frac{\sum x_i f_i}{\sum f_i}=\frac{4+72+140+7 \alpha+72}{64+\alpha}$

$\Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Ri

Vedclass · Accepted Answer

$8$

Vedclass · Answer

$5=\bar{x}=\frac{\sum x_i f_i}{\sum f_i}=\frac{4+72+140+7 \alpha+72}{64+\alpha}$

$\Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Rightarrow \alpha=16$

$	ext { M.. }(\bar{x})=\frac{\sum f _{ i }\left| x _{ i }-\overline{ x }ight|}{\sum f _{ i }} 	ext { where } \sum f _{ i }=64+16=80$

$	ext { M.D. }(\bar{x})=\frac{4 	imes 4+24 	imes 2+28 	imes 0+16 	imes 2+8 	imes 4}{80}$

$=\frac{8}{5}$

$	ext { Variance }=\frac{\sum f _{ i }\left( x _{ i }-\overline{ x }ight)^2}{\sum f _{ i }}$

$=\frac{4 	imes 16+24 	imes 4+0+16 	imes 4+8 	imes 16}{80}=\frac{352}{80}$

$	herefore \frac{3 \alpha}{m+\sigma^2}=\frac{3 	imes 16}{\frac{128}{80}+\frac{352}{80}}=8$

$X$	$1$	$3$	$5$	$7$	$9$
$(f)$	$4$	$24$	$28$	$\alpha$	$8$

$\mathrm{x}$	$\mathrm{x}_{1}=2$	$\mathrm{x}_{2}=6$	$\mathrm{x}_{3}=8$	$\mathrm{x}_{4}=9$
$\mathrm{f}$	$4$	$4$	$\alpha$	$\beta$

Let the mean of the data

$X$ $1$ $3$ $5$ $7$ $9$

$(f)$ $4$ $24$ $28$ $\alpha$ $8$

be $5.$ If $m$ and $\sigma^2$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 \alpha}{m+\sigma^2}$ is equal to $..........$.

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