If the variance of the following frequency distribution: Class : 10-20, 20-30, 30-40 Frequency : 2,

Q: Wait, why did the mean turn out to be exactly 25? Is that a coincidence?

It is not a coincidence! Because the distribution is perfectly symmetric around the central class (20-30) with frequencies 2, $x$, and 2, the mean must lie at the center of the distribution. In symmetric distributions, the mean, median, and mode often coincide. Recognizing this symmetry is a powerful tool to verify your calculations.

Q: Can I use the Step Deviation method to solve this? It looks like the numbers are getting huge.

Absolutely! The Step Deviation method is excellent for reducing calculation load. You could define $d_i = \frac{x_i - A}{h}$, where $A=25$ and $h=10$. This would make your $d_i$ values -1, 0, and 1. It simplifies the arithmetic significantly, though for this specific problem, the direct method is also quite manageable.

Q: What happens if I get a negative value for $x$? Does that mean I made a mistake?

Yes, absolutely. Frequency represents a count of occurrences, so it must always be a non-negative real number (and in this context, an integer). If you calculate a negative $x$, re-check your arithmetic, especially the expansion of the squares or the cross-multiplication step.

Question

If the variance of the following frequency distribution:
Class : 10-20, 20-30, 30-40
Frequency : 2, $x$, 2
is 50, then $x$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In the world of JEE Advanced, statistics is about finding the hidden order within chaos. We are analyzing a frequency distribution with three classes: $10-20$, $20-30$, and $30-40$.

The midpoints, or class marks $x_i$, serve as the anchors of our analysis. They are calculated as $x_i = \frac{	ext{Lower Limit} + 	ext{Upper Limit}}{2}$.

Thus, we find the class marks to be:
$x_1 = 15$, $x_2 = 25$, and $x_3 = 35$.

The corresponding frequencies are $2$, $x$, and $2$. This symmetry indicates that the distribution is perfectly balanced around the central class, $20-30$.

### The Mean - A Moment of Elegance

To find the variance, we first determine the mean, $\bar{x}$. The formula is defined as:
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$

Substituting our known values, we get:
$$\bar{x} = \frac{2(15) + x(25) + 2(35)}{4 + x}$$

Simplifying the numerator, we obtain $30 + 25x + 70$, which results in $100 + 25x$. Factoring out $25$, we see:
$$\bar{x} = \frac{25(4 + x)}{4 + x}$$

The $(4+x)$ terms cancel out beautifully, leaving us with the elegant result:
$$\bar{x} = 25$$

### The Variance - The Heart of the Problem

We are given the variance $\sigma^2 = 50$. The standard formula for variance is:
$$\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - (\bar{x})^2$$

First, we calculate the sum of squares, $\sum f_i x_i^2$:
$$\sum f_i x_i^2 = 2(15^2) + x(25^2) + 2(35^2)$$

Performing the arithmetic:
$$2(225) + x(625) + 2(1225) = 450 + 625x + 2450 = 2900 + 625x$$

Now, we substitute these values into the variance equation:
$$50 = \frac{2900 + 625x}{4 + x} - (25)^2$$

Since $25^2 = 625$, the equation becomes:
$$50 = \frac{2900 + 625x}{4 + x} - 625$$

### The Final Resolution

We add $625$ to both sides to isolate the fractional term:
$$675 = \frac{2900 + 625x}{4 + x}$$

Cross-multiplying yields:
$$675(4 + x) = 2900 + 625x$$

Expanding the left side gives:
$$2700 + 675x = 2900 + 625x$$

Subtracting $625x$ from both sides and $2700$ from both sides, we arrive at:
$$50x = 200$$

Solving for $x$, we find:
**$x = 4$**

The mystery is solved; the missing frequency is **4**. In these problems, the algebra is merely the vehicle, while the real skill lies in the setup and the confidence to trust the process.

Animated Solution for Mathematics - Statistics: If the variance of the following frequency distribution: Class : 10-20, 20-30, 30-40 Frequency : 2, $x$ , 2 is 50, then $x$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Data

Finding Midpoints ( $x_{i}$ )

Calculating Total Frequency

Calculating the Mean ( $\overset{x}{ˉ}$ ) - Setup

Calculating the Mean ( $\overset{x}{ˉ}$ ) - Simplification

The Variance Formula

Calculating $\sum f_{i} x_{i}^{2}$

Setting up the Variance Equation

Simplifying the Equation

Solving for x - Cross Multiplication

Solving for x - Final Result

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Mean - A Moment of Elegance

The Variance - The Heart of the Problem

The Final Resolution

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