Consider the following frequency distribution:

If mean = $$\frac{309}{22}$$ and median = 14, then the value $$(a-b)^2$$ is equal to ___.

The frequency distribution is: Class 0-6 (frequency $$a$$), 6-12 (frequency $$b$$), 12-18 (frequency 12), 18-24 (frequency 9), 24-30 (frequency 5).

The total frequency is $$N = a + b + 12 + 9 + 5 = a + b + 26$$.

The class midpoints are 3, 9, 15, 21, 27. Using the mean formula: $$\text{Mean} = \frac{3a + 9b + 15(12) + 21(9) + 27(5)}{N} = \frac{3a + 9b + 180 + 189 + 135}{a + b + 26} = \frac{3a + 9b + 504}{a + b + 26}$$.

Setting this equal to $$\frac{309}{22}$$ and cross-multiplying: $$22(3a + 9b + 504) = 309(a + b + 26)$$.

Expanding: $$66a + 198b + 11088 = 309a + 309b + 8034$$.

Simplifying: $$243a + 111b = 3054$$. Dividing by 3: $$81a + 37b = 1018$$. $$-(1)$$

For the median, the cumulative frequencies are: up to 6 is $$a$$, up to 12 is $$a + b$$, up to 18 is $$a + b + 12$$, up to 24 is $$a + b + 21$$, up to 30 is $$a + b + 26$$.

Since the median is 14, it falls in the class 12-18 (with frequency 12). Using the median formula: $$\text{Median} = L + \frac{\frac{N}{2} - F}{f} \times h$$, where $$L = 12$$, $$F = a + b$$ (cumulative frequency before the median class), $$f = 12$$, and $$h = 6$$.

Substituting: $$14 = 12 + \frac{\frac{N}{2} - (a+b)}{12} \times 6$$.

So $$2 = \frac{N/2 - (a+b)}{2}$$, which gives $$N/2 - (a+b) = 4$$. $$-(2)$$

Since $$N = a + b + 26$$, we get $$\frac{a + b + 26}{2} - (a + b) = 4$$, so $$a + b + 26 - 2(a+b) = 8$$, giving $$a + b = 18$$. $$-(3)$$

Substituting $$b = 18 - a$$ into equation $$(1)$$: $$81a + 37(18 - a) = 1018$$.

$$81a + 666 - 37a = 1018$$, so $$44a = 352$$, giving $$a = 8$$.

From $$(3)$$: $$b = 18 - 8 = 10$$.

Therefore $$(a - b)^2 = (8 - 10)^2 = (-2)^2 = 4$$.