1. Given frequency table

the class intervals are: $0-6,\ 6-12,\ 12-18,\ 18-24,\ 24-30$ with frequencies: $a,\ b,\ 12,\ 9,\ 5$

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

2. Using the mean

Class marks are: $3,\ 9,\ 15,\ 21,\ 27$

Hence $\bar x=\frac{3a+9b+15\cdot 12+21\cdot 9+27\cdot 5}{a+b+26}$

Compute constants: $15\cdot 12=180,\quad 21\cdot 9=189,\quad 27\cdot 5=135$ so \bar x=\frac{3a+9b+180+189+135}{a+b+26}=rac{3a+9b+504}{a+b+26}

Given mean is $\frac{309}{22}$ thus $\frac{3a+9b+504}{a+b+26}=\frac{309}{22}$

Cross-multiplying: $22(3a+9b+504)=309(a+b+26)$ $66a+198b+11088=309a+309b+8034$ $3054=243a+111b$ Divide by 3: $1018=81a+37b \qquad ...(1)$

3. Using the median

Median formula for grouped data: $\text{Median}=l+\left(\frac{\frac N2-c_f}{f}\right)h$

Given median = 14.

Now 14 lies in the class $12-18$, so median class is $12-18$.

Thus:

• lower limit $l=12$
• class width $h=6$
• frequency of median class $f=12$
• cumulative frequency before median class $c_f=a+b$
• total frequency $N=a+b+26$

Apply formula: $14=12+\left(\frac{\frac{a+b+26}{2}-(a+b)}{12}\right)6$

So $2=\left(\frac{\frac{a+b+26}{2}-(a+b)}{12}\right)6$

Multiply both sides by 12: $24=6\left(\frac{a+b+26}{2}-(a+b)\right)$ $4=\frac{a+b+26}{2}-(a+b)$

Now simplify: $4=\frac{a+b+26-2a-2b}{2}=\frac{26-a-b}{2}$ $8=26-a-b$ $a+b=18 \qquad ...(2)$

4. Solve equations (1) and (2)

From (2): $b=18-a$

Substitute into (1): $81a+37(18-a)=1018$ $81a+666-37a=1018$ $44a=352$ $a=8$

Then $b=18-8=10$

5. Find $(a-b)^2$

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

Final Answer

$\boxed{4}$