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(A) The total frequency $N = a + b + 12 + 9 + 5 = a + b + 26$.The mean is given by $\frac{\sum f_i x_i}{N} = \frac{a(3) + b(9) + 12(15) + 9(21) + 5(27

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$5$

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(A) The total frequency $N = a + b + 12 + 9 + 5 = a + b + 26$.The mean is given by $\frac{\sum f_i x_i}{N} = \frac{a(3) + b(9) + 12(15) + 9(21) + 5(27)}{a + b + 26} = \frac{3a + 9b + 180 + 189 + 135}{a + b + 26} = \frac{3a + 9b + 504}{a + b + 26} = \frac{309}{22}$.Cross-multiplying: $22(3a + 9b + 504) = 309(a + b + 26)$.$66a + 198b + 11088 = 309a + 309b + 8034$.$243a + 111b = 3054$. Dividing by $3$: $81a + 37b = 1018$ (Equation $1$).For the median,since $	ext{median} = 14$,the median class is $12-18$. Here $l = 12, f = 12, cf = a + b, h = 6$.$	ext{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} ight) 	imes h = 12 + \left( \frac{\frac{a+b+26}{2} - (a+b)}{12} ight) 	imes 6 = 14$.$12 + \frac{\frac{a+b+26 - 2a - 2b}{2}}{2} = 14 \Rightarrow \frac{26 - a - b}{4} = 2$.$26 - a - b = 8 \Rightarrow a + b = 18$ (Equation $2$).From $b = 18 - a$,substitute into Equation $1$: $81a + 37(18 - a) = 1018$.$81a + 666 - 37a = 1018 \Rightarrow 44a = 352 \Rightarrow a = 8$.Then $b = 18 - 8 = 10$.Therefore,$(a - b)^{2} = (8 - 10)^{2} = (-2)^{2} = 4$. (Note: The provided options seem to contain a typo; the calculated value is $4$).

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

Consider the following frequency distribution:

Class $0-6$ $6-12$ $12-18$ $18-24$ $24-30$

Frequency $a$ $b$ $12$ $9$ $5$

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

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Consider the following frequency distribution: Class $0-6$ $6-12$ $12-18$ $18-24$ $24-30$ Frequency $a$ $b$ $12$ $9$ $5$ If $\text{mean} = \frac{309}{22}$ and $\text{median} = 14$,then the value of $(a-b)^{2}$ is equal to $.....$