1. Key Concepts and Formulas

To calculate the variance of a frequency distribution, especially for grouped data, we rely on the following fundamental statistical formulas:

• Class Midpoint ($x_i$): For each class interval, the midpoint is the representative value. It is calculated as the average of the lower and upper limits of the class. $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$
• Mean ($\bar{x}$): The mean of a frequency distribution is the weighted average of the class midpoints, where the weights are the frequencies. $\bar{x} = \frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i}$ Here, $f_i$ is the frequency of the $i$-th class and $x_i$ is its midpoint.
• Variance ($\sigma^2$): Variance measures the dispersion or spread of the data points around the mean. For a frequency distribution, a computationally efficient formula is: $\sigma^2 = \frac{\sum_{i=1}^{n} f_i x_i^2}{\sum_{i=1}^{n} f_i} - (\bar{x})^2$

2. Step-by-Step Solution

Step 1: Calculate Class Midpoints ($x_i$) We begin by determining the midpoint for each given class interval. This is crucial because for grouped data, calculations assume observations are concentrated at these midpoints.

Step 2: Determine the Unknown Frequency ($x$) using the Given Mean We are provided that the mean ($\bar{x}$) of the distribution is 28. We will use the mean formula to find the unknown frequency $x$. This is a necessary step, as all frequencies must be known to calculate the variance.

First, let's calculate the sum of frequencies ($\sum f_i$) and the sum of products of frequency and midpoint ($\sum f_i x_i$):

• Sum of Frequencies: $\sum f_i = 2 + 3 + x + 5 + 4 = 14 + x$

• Sum of Products ($f_i x_i$): $\sum f_i x_i = (2 \times 5) + (3 \times 15) + (x \times 25) + (5 \times 35) + (4 \times 45)$ $\sum f_i x_i = 10 + 45 + 25x + 175 + 180$ $\sum f_i x_i = 410 + 25x$

Now, substitute these sums and the given mean ($\bar{x} = 28$) into the mean formula: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ $28 = \frac{410 + 25x}{14 + x}$ To solve for $x$, multiply both sides by $(14+x)$: $28(14 + x) = 410 + 25x$ $392 + 28x = 410 + 25x$ Group terms with $x$ on one side and constants on the other: $28x - 25x = 410 - 392$ $3x = 18$ $x = 6$ Thus, the unknown frequency is 6.

Step 3: Calculate Variance ($\sigma^2$) With all frequencies now known (including $x=6$), we can proceed to calculate the variance using the formula $\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - (\bar{x})^2$. To do this, we need to calculate $\sum f_i x_i^2$.

Let's update our table with $x=6$ and add columns for $f_i x_i$ and $f_i x_i^2$:

Important Check: We can verify our mean calculation with the complete data: $\bar{x} = \frac{560}{20} = 28$. This matches the given mean, confirming our value of $x=6$ is correct.

Now, substitute the values into the variance formula: $\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - (\bar{x})^2$ $\sigma^2 = \frac{18700}{20} - (28)^2$ First, perform the division: $\sigma^2 = 935 - (28)^2$ Next, calculate the square of the mean: $28^2 = 784$ Finally, subtract this from the first term: $\sigma^2 = 935 - 784$ $\sigma^2 = 151$

3. Common Mistakes & Tips

• Midpoint Calculation: A frequent error is miscalculating class midpoints. Always ensure they are the average of the class limits.
• Algebraic Precision: Be meticulous when solving for the unknown frequency $x$. Small arithmetic or algebraic errors can lead to an incorrect value of $x$, which then propagates through the variance calculation.
• Summation Errors: Double-check all summations ($\sum f_i$, $\sum f_i x_i$, $\sum f_i x_i^2$). A calculator should be used carefully, or sums should be re-verified.
• Variance Formula: Ensure you use the correct variance formula. A common mistake is forgetting to subtract $(\bar{x})^2$ or omitting the square in $x_i^2$.

4. Summary

This problem required a systematic application of statistical formulas for grouped frequency distributions. We first calculated the midpoints for each class. Then, using the given mean, we solved for the unknown frequency $x$. Finally, with all frequencies determined, we used the appropriate variance formula to calculate the variance of the distribution. The calculations consistently lead to a variance of 151.

The final answer is $\boxed{151}$.