If the mean and variance of the frequency distribution xi: 2, 4, 6, 8, 10, 12, 14, 16; fi: 4, 4, α,

Q: Why do we use the formula $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2$ instead of the definition $\frac{\sum f_i (x_i - \bar{x})^2}{N}$?

Great question! While both are mathematically equivalent, the computational formula $\frac{\sum f_i x_i^2}{N} - (\bar{x})^2$ is significantly faster in exams. It avoids calculating $(x_i - \bar{x})^2$ for every single term, which is prone to arithmetic errors. In JEE, efficiency is as important as accuracy.

Q: I got $\alpha = \beta$ from the mean, but what if the mean equation was more complex?

If the mean equation didn't simplify so cleanly, you would still have a system of two linear equations with two variables ($\alpha$ and $\beta$). You would solve for one in terms of the other and substitute it into the variance equation. Always look for the simplest path first!

Q: Why is the variance calculation so decimal-heavy? Is there a trick?

The JEE examiners love to test your patience and precision. There isn't always a 'trick' to avoid decimals. The key is to keep your fractions as long as possible and only convert to decimals at the very last step if necessary. Here, the numbers were designed to cancel out, but you must trust your algebra!

Question

If the mean and variance of the frequency distribution $$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x_i & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 \ \hline f_i & 4 & 4 & \alpha & 15 & 8 & \beta & 4 & 5 \ \hline \end{array}$$ are 9 and 15.08, respectively, then the value of $\alpha^2 + \beta^2 - \alpha\beta$ is _____.

Sigma Percentile · Accepted Answer

### The Dance of Data: Unlocking the Mystery of Missing Frequencies

Welcome, students! Today, we embark on a journey through the heart of statistics. We are presented with a frequency distribution table, a classic structure in JEE Advanced problems.

But there is a twist: two of our frequencies, $\alpha$ and $\beta$, are shrouded in mystery. We are given the mean and the variance, and our mission is to find the value of $\alpha^2 + \beta^2 - \alpha\beta$.

This isn't just about plugging numbers into formulas; it is about understanding the balance of data.

### Phase 1: The Foundation of Total Frequency

Before we can calculate anything, we must define our universe. The total frequency $N$ is the sum of all individual frequencies.

Looking at our table, we have $N = 4 + 4 + \alpha + 15 + 8 + \beta + 4 + 5$. Simplifying this, we get:

$$N = 40 + \alpha + \beta$$

This $N$ is the denominator for both our mean and variance. It is the anchor of our entire calculation.

### Phase 2: The Mean as a Balancing Act

The mean $\bar{x}$ is the center of gravity of our distribution. We are given $\bar{x} = 9$. The formula for the mean is $\bar{x} = \frac{\sum f_i x_i}{N}$.

Let us calculate the numerator, $\sum f_i x_i$. By multiplying each $x_i$ by its corresponding $f_i$, we get:

$$(2 	imes 4) + (4 	imes 4) + (6 	imes \alpha) + (8 	imes 15) + (10 	imes 8) + (12 	imes \beta) + (14 	imes 4) + (16 	imes 5)$$

Summing these, we find $\sum f_i x_i = 360 + 6\alpha + 12\beta$. Now, we set up our equation:

$$\frac{360 + 6\alpha + 12\beta}{40 + \alpha + \beta} = 9$$

Cross-multiplying gives us $360 + 6\alpha + 12\beta = 9(40 + \alpha + \beta)$, which simplifies to $360 + 6\alpha + 12\beta = 360 + 9\alpha + 9\beta$. The $360$ cancels out, leaving $3\beta = 3\alpha$, or simply $\alpha = \beta$.

This is a beautiful, elegant simplification!

### Phase 3: The Variance as a Measure of Spread

Now, we tackle the variance $\sigma^2 = 15.08$. The computational formula is $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2$.

We need $\sum f_i x_i^2$. Squaring each $x_i$ and multiplying by its frequency, we get:

$$4(2^2) + 4(4^2) + \alpha(6^2) + 15(8^2) + 8(10^2) + \beta(12^2) + 4(14^2) + 5(16^2)$$

This simplifies to $\sum f_i x_i^2 = 3904 + 36\alpha + 144\beta$. Since we know $\alpha = \beta$, we substitute $\beta$ with $\alpha$. Our sum becomes $3904 + 180\alpha$, and our total frequency $N$ becomes $40 + 2\alpha$.

### Phase 4: The Final Resolution

We plug these into our variance equation:

$$15.08 = \frac{3904 + 180\alpha}{40 + 2\alpha} - 9^2$$

Adding $81$ to $15.08$, we get $96.08 = \frac{3904 + 180\alpha}{40 + 2\alpha}$. Cross-multiplying, we get $96.08(40 + 2\alpha) = 3904 + 180\alpha$.

Expanding this, $3843.2 + 192.16\alpha = 3904 + 180\alpha$. Rearranging, $12.16\alpha = 60.8$, which leads us to $\alpha = 5$.

Since $\alpha = \beta$, then $\beta = 5$. Finally, we evaluate:

$$\alpha^2 + \beta^2 - \alpha\beta = 5^2 + 5^2 - (5 	imes 5) = 25 + 25 - 25 = 25$$

We have arrived at our destination. The complexity of the decimals was merely a test of your resolve. You have mastered the data! The final answer is **25**.

Animated Solution for Mathematics - Statistics: If the mean and variance of the frequency distribution $x_{i} f_{i} 2444 6 α 815108 12 β 144165$ are 9 and 15.08, respectively, then the value of $α^{2} + β^{2} - α β$ is _____.

Enter Numerical Value:

Visualized Solution

Understanding the Problem

Total Frequency $N$

Mean Equation Setup

Finding Relation $α = β$

Variance Equation Setup

Reducing to One Variable

Solving for $α$

Final Values of $α$ and $β$

Final Calculation

The Sigma Insight: Variance and Standard Deviation

The Dance of Data

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

Understanding the Problem

Total Frequency .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

Mean Equation Setup

Finding Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=β