This problem requires us to determine two unknown frequencies, $a$ and $b$, in a grouped frequency distribution. We are provided with the mean and median of this distribution. Once $a$ and $b$ are found, we need to calculate the value of $(a-b)^2$. To achieve this, we will systematically apply the formulas for the mean and median of grouped data, which will lead to a system of two linear equations in $a$ and $b$.

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1. Key Concepts and Formulas

To solve this problem, we will use the standard formulas for the mean and median of a grouped frequency distribution:

• Mean for Grouped Data ($\bar{X}$): The mean is calculated as the sum of the products of each class's frequency ($f_i$) and its class mark ($x_i$), divided by the total frequency ($\sum f_i$). $\bar{X} = \frac{\sum f_i x_i}{\sum f_i}$ Here, $x_i$ is the midpoint of the $i$-th class interval.

• Median for Grouped Data: The median of a grouped frequency distribution is found using the formula: $\text{Median} = L + \left( \frac{\frac{N}{2} - C_f}{f} \right) \times h$ where:

  • $L$ = lower limit of the median class.
  • $N$ = total frequency ($\sum f_i$).
  • $C_f$ = cumulative frequency of the class preceding the median class.
  • $f$ = frequency of the median class.
  • $h$ = class size (width of the median class interval).

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2. Step-by-Step Solution

Step 1: Organize the Data and Calculate Necessary Components To efficiently apply both the mean and median formulas, we first organize the given data in a table and calculate the class marks ($x_i$), products of frequency and class mark ($f_i x_i$), and cumulative frequencies ($C_f$). This step is crucial for minimizing errors and having all required values readily available.

From this table, we derive the following sums:

• Total frequency ($N$): $N = a + b + 12 + 9 + 5 = a + b + 26$
• Sum of products ($\sum f_i x_i$): $\sum f_i x_i = 3a + 9b + 180 + 189 + 135 = 3a + 9b + 504$

Step 2: Formulate the First Equation using the Mean We use the given mean value ($\bar{X} = \frac{309}{22}$) and the calculated sums to establish our first algebraic relationship (a linear equation) between $a$ and $b$.

Substitute the values into the mean formula: $\bar{X} = \frac{\sum f_i x_i}{N}$ $\frac{309}{22} = \frac{3a + 9b + 504}{a + b + 26}$ Now, we simplify this equation by cross-multiplication: $309(a + b + 26) = 22(3a + 9b + 504)$ Distribute the terms on both sides: $309a + 309b + 8034 = 66a + 198b + 11088$ Group the terms with $a$ and $b$ on one side and constants on the other: $(309a - 66a) + (309b - 198b) = 11088 - 8034$ $243a + 111b = 3054$ All coefficients in this equation are divisible by 3. Dividing the entire equation by 3 simplifies it: $\frac{243a}{3} + \frac{111b}{3} = \frac{3054}{3}$ $\implies 81a + 37b = 1018 \quad \text{(Equation 1)}$

Step 3: Formulate the Second Equation using the Median The given median value (14) provides a second independent linear equation involving $a$ and $b$. This is crucial for solving for two unknown variables.

First, we identify the median class. Since the median is 14, it falls within the class interval 12-18. This is our median class.

Now, we identify the parameters required for the median formula from our table and the median class:

• $L$ (Lower limit of median class): $L = 12$
• $h$ (Class size): $h = 18 - 12 = 6$
• $f$ (Frequency of median class): $f = 12$ (frequency of the 12-18 class)
• $C_f$ (Cumulative frequency of the class preceding the median class): The class preceding 12-18 is 6-12. Its cumulative frequency is $a+b$. So, $C_f = a+b$.
  • Important Note: Always use the cumulative frequency of the preceding class, not the median class itself.
• $N$ (Total frequency): $N = a+b+26$. Therefore, $\frac{N}{2} = \frac{a+b+26}{2}$.

Substitute these values into the median formula: $\text{Median} = L + \left( \frac{\frac{N}{2} - C_f}{f} \right) \times h$ $14 = 12 + \left( \frac{\frac{a+b+26}{2} - (a+b)}{12} \right) \times 6$ Simplify the equation step-by-step: Subtract 12 from both sides: $14 - 12 = \left( \frac{\frac{a+b+26}{2} - (a+b)}{12} \right) \times 6$ $2 = \left( \frac{\frac{a+b+26 - 2(a+b)}{2}}{12} \right) \times 6$ $2 = \left( \frac{a+b+26 - 2a - 2b}{24} \right) \times 6$ $2 = \frac{-a - b + 26}{4}$ Multiply both sides by 4: $2 \times 4 = -a - b + 26$ $8 = -a - b + 26$ Rearrange the terms to isolate $a$ and $b$: $a + b = 26 - 8$ $\implies a + b = 18 \quad \text{(Equation 2)}$

Step 4: Solve the System of Linear Equations We now have a system of two linear equations with two variables ($a$ and $b$):

1. $81a + 37b = 1018$
2. $a + b = 18$

To find the specific values of $a$ and $b$, we solve this system. From Equation 2, we can easily express $b$ in terms of $a$: $b = 18 - a$ Substitute this expression for $b$ into Equation 1: $81a + 37(18 - a) = 1018$ $81a + (37 \times 18) - 37a = 1018$ $81a + 666 - 37a = 1018$ Combine the terms with $a$: $(81a - 37a) = 1018 - 666$ $44a = 352$ $a = \frac{352}{44}$ $a = 8$ Now, substitute the value of $a=8$ back into Equation 2 to find $b$: $b = 18 - a$ $b = 18 - 8$ $b = 10$ So, the unknown frequencies are $a=8$ and $b=10$.

Step 5: Calculate the Final Value The problem asks for the value of $(a-b)^2$. Substitute the values of $a=8$ and $b=10$: $(a-b)^2 = (8 - 10)^2$ $= (-2)^2$ $= 4$

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3. Common Mistakes & Tips

• Cumulative Frequency for Median: A common error is to use the cumulative frequency of the median class itself ($a+b+12$) instead of the class preceding it ($a+b$). Always be careful with $C_f$.
• Algebraic Errors: The most frequent mistakes involve distributing terms incorrectly or errors in basic arithmetic during the simplification of equations. Double-check all calculations, especially when dealing with large numbers or multiple steps.
• Identifying Median Class: Ensure the median class is correctly identified by locating which class interval contains the given median value.

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4. Summary

By systematically applying the formulas for the mean and median of grouped data, we derived two linear equations: $81a + 37b = 1018$ (from the mean) and $a + b = 18$ (from the median). Solving this system of equations yielded the unknown frequencies $a=8$ and $b=10$. Finally, we calculated the required value $(a-b)^2 = (8-10)^2 = (-2)^2 = 4$.

The final answer is $\boxed{4}$.