Consider the following frequency distribution :

Class :	0-6	6-12	12-18	18-24	24-30
Frequency :	$a$	$b$	12	9	5

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

<table class="tg"> <thead> <tr> <th class="tg-0lax">Class</th> <th class="tg-0lax">Frequency</th> <th class="tg-0lax">${x_i}$</th> <th class="tg-0lax">${f_i}{x_i}$</th> </tr> </thead> <tbody> <tr> <td class="tg-0lax">0-6</td> <td class="tg-0lax">a</td> <td class="tg-0lax">3</td> <td class="tg-0lax">3a</td> </tr> <tr> <td class="tg-0lax">6-12</td> <td class="tg-0lax">b</td> <td class="tg-0lax">9</td> <td class="tg-0lax">9b</td> </tr> <tr> <td class="tg-0lax">12-18</td> <td class="tg-0lax">12</td> <td class="tg-0lax">15</td> <td class="tg-0lax">180</td> </tr> <tr> <td class="tg-0lax">18-24</td> <td class="tg-0lax">9</td> <td class="tg-0lax">21</td> <td class="tg-0lax">189</td> </tr> <tr> <td class="tg-0lax">24-30</td> <td class="tg-0lax">5</td> <td class="tg-0lax">27</td> <td class="tg-0lax">135</td> </tr> <tr> <td class="tg-0lax"></td> <td class="tg-0lax">$N = (26 + a + b)$</td> <td class="tg-0lax"></td> <td class="tg-0lax">$(504 + 3a + 9b)$</td> </tr> </tbody> </table><br><br> Mean = ${{3a + 9b + 180 + 189 + 135} \over {a + b + 26}} = {{309} \over {22}}$<br><br>$\Rightarrow 66a + 198b + 11088 = 309a + 309b + 8034$<br><br>$\Rightarrow 243a + 111b = 3054$<br><br>$\Rightarrow 81a + 37b = 1018$ $\to$ (1)<br><br>Now, Median $= 12 + {{{{a + b + c} \over 2} - (a + b)} \over {12}} \times 6 = 14$<br><br>$\Rightarrow {{13} \over 2} - \left( {{{a + b} \over 4}} \right) = 2$<br><br>$\Rightarrow {{a + b} \over 4} = {9 \over 2}$<br><br>$\Rightarrow a + b = 18$ $\to$ (2)<br><br>From equation (1) $ (2)<br><br>a = 8, b = 10<br><br>$\therefore$ $${(a - b)^2} = {(8 - 10)^2}$ = 4