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 _____.

Given the frequency table with $$x$$: 1, 3, 5, 7, 9 and frequencies $$f$$: 4, 24, 28, $$\alpha$$, 8, with mean = 5.

Finding $$\alpha$$:

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

$$\frac{288 + 7\alpha}{64 + \alpha} = 5$$

$$288 + 7\alpha = 320 + 5\alpha \implies \alpha = 16$$

Total $$N = 80$$.

Variance $$\sigma^2$$:

$$\sigma^2 = \frac{1}{N}\sum f_i(x_i - \bar{x})^2 = \frac{4(16) + 24(4) + 28(0) + 16(4) + 8(16)}{80}$$

$$= \frac{64 + 96 + 0 + 64 + 128}{80} = \frac{352}{80} = 4.4$$

Mean deviation $$m$$:

$$m = \frac{1}{N}\sum f_i|x_i - \bar{x}| = \frac{4(4) + 24(2) + 28(0) + 16(2) + 8(4)}{80}$$

$$= \frac{16 + 48 + 0 + 32 + 32}{80} = \frac{128}{80} = 1.6$$

Result:

$$\frac{3\alpha}{m + \sigma^2} = \frac{3 \times 16}{1.6 + 4.4} = \frac{48}{6} = 8$$

The answer is $$8$$.