Let the mean of the data
\begin{tabular}{|l|l|l|l|l|l|}
\hline$x$ & 1 & 3 & 5 & 7 & 9 \\
\hline Frequency $(f)$ & 4 & 24 & 28 & $\alpha$ & 8 \\
\hline
\end{tabular}
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 $\_\_\_\_$ .
