If the mean and variance of the frequency distribution
$$x_i$$         2,    4,    6,    8,   10,   12,   14,   16
$$f_i$$         4,    4,    $$\alpha$$,  15,    8,    $$\beta$$,     4,     5
are 9 and 15.08 respectively, then the value of $$\alpha^2 + \beta^2 - \alpha\beta$$ is ______.

The frequency distribution below is known to have mean = 9 and variance = 15.08.

Using the condition for the mean, the total number of observations is $$N = 40 + \alpha + \beta$$ and the sum of the products is

$$\sum f_i x_i = 8 + 16 + 6\alpha + 120 + 80 + 12\beta + 56 + 80 = 360 + 6\alpha + 12\beta.$$

Since the mean equals 9,

$$\frac{360 + 6\alpha + 12\beta}{40 + \alpha + \beta} = 9$$ which leads to $$360 + 6\alpha + 12\beta = 360 + 9\alpha + 9\beta$$ and hence $$3\beta = 3\alpha \implies \alpha = \beta.$$

Next, applying the variance condition, we compute

$$\sum f_i x_i^2 = 16 + 64 + 36\alpha + 960 + 800 + 144\beta + 784 + 1280 = 3904 + 36\alpha + 144\beta.$$

With $$\alpha = \beta$$, this becomes $$\sum f_i x_i^2 = 3904 + 180\alpha$$ and $$N = 40 + 2\alpha.$$

Setting the variance equal to 15.08 gives

$$\frac{3904 + 180\alpha}{40 + 2\alpha} - 81 = 15.08$$ so $$\frac{3904 + 180\alpha}{40 + 2\alpha} = 96.08$$ and therefore $$3904 + 180\alpha = 96.08(40 + 2\alpha) = 3843.2 + 192.16\alpha,$$ leading to $$60.8 = 12.16\alpha \implies \alpha = 5.$$

It follows that $$\alpha = \beta = 5$$, and substituting these values gives

$$\alpha^2 + \beta^2 - \alpha\beta = 25 + 25 - 25 = 25.$$

The answer is $$25$$.