Let the mean of the data

$x$	1	3	5	7	9
Frequency ($f$)	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 __________

$$ \begin{aligned} & 5=\bar{x}=\frac{\sum x_i f_i}{\sum f_i}=\frac{4+72+140+7 \alpha+72}{64+\alpha} \\\\ & \Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Rightarrow \alpha=16 \end{aligned} $$ <br/><br/>$$ \begin{aligned} \sum f_i & =80 \\\\ \text { M.D } & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\ & =\frac{4+4+24 \times 2+0+16 \times 2+8 \times 4}{80} \\\\ & =\frac{8}{5} \end{aligned} $$ <br/><br/>$$ \begin{aligned} \sigma^2 & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\ & =\frac{4+16+24 \times 4+0+16 \times 4+8 \times 16}{80}=\frac{22}{5} \end{aligned} $$ <br/><br/>So, $\frac{3 \alpha}{m+\sigma^2}=\frac{3 \times 16}{\frac{8}{5}+\frac{22}{5}}=8$