This problem requires us to find an unknown frequency, $x$, in a given frequency distribution, using the provided variance of the distribution. We will systematically apply the definitions and formulas for the mean and variance of a frequency distribution.

Key Concepts and Formulas

For a frequency distribution with class intervals, we use class marks ($x_i$) to represent each class, along with their corresponding frequencies ($f_i$).

1. Class Mark (Midpoint): The representative value for a class interval $(L_1 - L_2)$ is its midpoint: $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$

2. Mean ($\bar{x}$): The arithmetic average of the distribution, calculated as the sum of the products of each class mark and its frequency, divided by the total frequency: $\bar{x} = \frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i}$

3. Variance ($\sigma^2$): A measure of the spread of data points around the mean. It is defined as the average of the squared deviations from the mean: $\sigma^2 = \frac{\sum_{i=1}^{n} f_i (x_i - \bar{x})^2}{\sum_{i=1}^{n} f_i}$ An alternative computational formula is $\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - (\bar{x})^2$. For this problem, the direct definition involving $(x_i - \bar{x})^2$ is more straightforward due to the calculated mean.

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Step-by-Step Solution

Step 1: Data Preparation - Determine Class Marks ($x_i$) and Tabulate Data

First, we convert the given class intervals into their representative class marks ($x_i$). These, along with their frequencies ($f_i$), form our working data.

Step 2: Calculate the Mean ($\bar{x}$)

The mean is the central point of the distribution, from which we measure deviations for variance. We use the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.

1. Calculate $\sum f_i x_i$ (Sum of products of frequency and class mark):

   • For the first class: $f_1 x_1 = 2 \times 15 = 30$
   • For the second class: $f_2 x_2 = x \times 25 = 25x$
   • For the third class: $f_3 x_3 = 2 \times 35 = 70$
   • Sum: $\sum f_i x_i = 30 + 25x + 70 = 100 + 25x$

2. Calculate $\sum f_i$ (Total Frequency):

   • This is the sum of all frequencies: $\sum f_i = 2 + x + 2 = 4 + x$

3. Substitute these sums into the mean formula: $\bar{x} = \frac{100 + 25x}{4 + x}$ We can factor out 25 from the numerator: $\bar{x} = \frac{25(4 + x)}{4 + x}$ Since $x$ represents a frequency, it must be a non-negative value. Therefore, $4+x$ will always be positive and non-zero, allowing us to cancel the term $(4+x)$ from the numerator and denominator. $\bar{x} = 25$ Reasoning: The mean of this distribution is 25, regardless of the value of $x$. This is a significant simplification, occurring because the class marks (15, 25, 35) are symmetrically distributed around 25, and the frequencies of the extreme classes (10-20 and 30-40) are equal (both 2). The central class mark (25) is exactly the arithmetic mean of the extreme class marks.

Step 3: Calculate the Variance ($\sigma^2$) in terms of $x$

With the mean ($\bar{x} = 25$) determined, we can now calculate the variance using the formula $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$. We already know $\sum f_i = 4+x$.

1. Calculate $(x_i - \bar{x})^2$ for each class: These terms measure the squared deviation of each class mark from the mean.

   • For the first class: $(x_1 - \bar{x})^2 = (15 - 25)^2 = (-10)^2 = 100$
   • For the second class: $(x_2 - \bar{x})^2 = (25 - 25)^2 = (0)^2 = 0$
   • For the third class: $(x_3 - \bar{x})^2 = (35 - 25)^2 = (10)^2 = 100$

2. Calculate $f_i (x_i - \bar{x})^2$ for each class: Multiply each squared deviation by its corresponding frequency.

   • For the first class: $f_1 (x_1 - \bar{x})^2 = 2 \times 100 = 200$
   • For the second class: $f_2 (x_2 - \bar{x})^2 = x \times 0 = 0$
   • For the third class: $f_3 (x_3 - \bar{x})^2 = 2 \times 100 = 200$ Reasoning: The contribution of the middle class interval (20-30) to the sum of squared deviations is zero because its class mark ($x_2 = 25$) is exactly equal to the mean ($\bar{x} = 25$). This simplifies the numerator of the variance formula significantly.

3. Calculate $\sum f_i (x_i - \bar{x})^2$ (Sum of weighted squared deviations):

   • Sum: $\sum f_i (x_i - \bar{x})^2 = 200 + 0 + 200 = 400$

4. Substitute into the variance formula: $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} = \frac{400}{4 + x}$

**Step 4: Use the Given Variance to Solve for $x$}

We are given that the variance ($\sigma^2$) of the distribution is 50. We set our calculated variance expression equal to 50 and solve for $x$.

$50 = \frac{400}{4 + x}$

Now, solve this algebraic equation for $x$: Multiply both sides by $(4+x)$: $50 (4 + x) = 400$ Divide both sides by 50: $4 + x = \frac{400}{50}$ $4 + x = 8$ Subtract 4 from both sides: $x = 8 - 4$ $x = 4$ Since $x=4$ is a non-negative integer, it is a valid frequency.

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Common Mistakes & Tips

• Order of Operations: Always calculate the mean ($\bar{x}$) first, as it's a prerequisite for calculating variance.
• Class Marks are Key: Ensure you correctly calculate the midpoint for each class interval. Using interval limits directly in formulas is a common error.
• Symmetry Recognition: Recognizing symmetry in data can simplify calculations, as seen with the mean and the zero contribution of the middle class to variance.
• Squaring is Essential: A frequent mistake is forgetting to square the $(x_i - \bar{x})$ term in the variance formula. Remember, variance is based on squared deviations.

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Summary

To determine the unknown frequency $x$, we followed the standard procedure for analyzing frequency distributions. We first calculated the class marks and then the mean of the distribution, which, due to the symmetric nature of the given data, simplified to 25, independent of $x$. Next, we calculated the variance in terms of $x$ by summing the weighted squared deviations from the mean. Finally, by equating this expression to the given variance of 50, we solved the resulting algebraic equation to find $x$.

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