We are given the discrete data:

$$\begin{array}{c|ccccc} x & 1 & 3 & 5 & 7 & 9 \\ \hline f & 4 & 24 & 28 & \alpha & 8 \end{array}$$

and its mean is $5$.

We need to find:

$$\frac{3\alpha}{m+\sigma^2}$$

where $m$ is the mean deviation about the mean and $\sigma^2$ is the variance.

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1. Use the given mean to find $\alpha$

Mean of a frequency distribution is:

$$\bar{x}=\frac{\sum fx}{\sum f}$$

Since mean is $5$,

$$\frac{4(1)+24(3)+28(5)+\alpha(7)+8(9)}{4+24+28+\alpha+8}=5$$

Now simplify numerator:

$$4+72+140+7\alpha+72=288+7\alpha$$

Denominator:

$$4+24+28+\alpha+8=64+\alpha$$

So,

$$\frac{288+7\alpha}{64+\alpha}=5$$ $$288+7\alpha=320+5\alpha$$ $$2\alpha=32$$ $$\alpha=16$$

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2. Find mean deviation about the mean, $m$

Mean is $5$, so

$$m=\frac{\sum f|x-5|}{\sum f}$$

First compute $|x-5|$:

• For $x=1$, $|1-5|=4$
• For $x=3$, $|3-5|=2$
• For $x=5$, $|5-5|=0$
• For $x=7$, $|7-5|=2$
• For $x=9$, $|9-5|=4$

Now multiply by frequencies:

$$\sum f|x-5|=4\cdot 4+24\cdot 2+28\cdot 0+16\cdot 2+8\cdot 4$$ $$=16+48+0+32+32=128$$

Total frequency:

$$N=4+24+28+16+8=80$$

Thus,

$$m=\frac{128}{80}=\frac{8}{5}$$

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3. Find the variance, $\sigma^2$

Variance about the mean is:

$$\sigma^2=\frac{\sum f(x-5)^2}{\sum f}$$

Now compute $(x-5)^2$:

• For $x=1$, $(1-5)^2=16$
• For $x=3$, $(3-5)^2=4$
• For $x=5$, $(5-5)^2=0$
• For $x=7$, $(7-5)^2=4$
• For $x=9$, $(9-5)^2=16$

Therefore,

$$\sum f(x-5)^2=4\cdot 16+24\cdot 4+28\cdot 0+16\cdot 4+8\cdot 16$$ $$=64+96+0+64+128=352$$

So,

$$\sigma^2=\frac{352}{80}=\frac{22}{5}$$

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4. Compute the required value

$$m+\sigma^2=\frac{8}{5}+\frac{22}{5}=\frac{30}{5}=6$$

Also,

$$3\alpha=3\cdot 16=48$$

Hence,

$$\frac{3\alpha}{m+\sigma^2}=\frac{48}{6}=8$$

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5. Final answer

$$\boxed{8}$$

This matches the stored correct answer.