Consider the following frequency distribution: Class: 0 - 6 6 - 12...

We know that, mean$= \dfrac{∑ {f}_{i}{x}_{i}}{∑ {f}_{i}}$

Given, mean$= \dfrac{309}{22}$

$⇒ \dfrac{3a + 9b + 180 + 189 + 135}{a + b + 26} = \dfrac{309}{22}$

$⇒ \dfrac{504 + 3a + 9b}{a + b + 26} = \dfrac{309}{22}$

$⇒ 66a + 198b + 11088 = 309a + 309b + 8034$

$⇒ 243a + 111b = 3054$

$⇒ 81a + 37b = 1018 . . . \left(1\right)$

Now, median$= 14,$ which lies in the interval $12 - 18,$ thus $12 - 18$ is the median class and we have $= l + \dfrac{\left(\dfrac{N}{2} - c\right) \times h}{f},$

Where, $l$ is the lower limit of the median class i.e. $l = 12, c$ is the cumulative frequency of the class above the median class i.e. $c = a + b, f$ is the frequency of the median class i.e. $f = 12$ and $h$ is the length of the interval i.e. $h = 6 .$

Thus, we have $12 + \left(\dfrac{\left(\dfrac{a + b + 26}{2}\right) - ( a + b )}{12}\right) \times 6 = 14$

$⇒ \dfrac{\left(\dfrac{a + b + 26 - 2a - 2b}{2}\right)}{2} = 14 - 12$

$⇒ \dfrac{26 - a - b}{4} = 2$

$⇒ 26 - a - b = 8$

$⇒ a + b = 18 . . . \left(2\right)$

On solving the equations \left(1\right)$and \left(2\right),$ we get $a = 8, b = 10 .$

$∴ {\left(a - b\right)}^{2} = {\left(8 - 10\right)}^{2} = 4 .$