If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, α, β, 60 where α β are 56 and 66.

Q: Wait, do I need to solve for $\alpha$ and $\beta$ individually? I tried setting up a quadratic equation but it got messy.

That is exactly the trap! You do not need to find $\alpha$ and $\beta$ individually. The question asks for $\alpha^2 + \beta^2$. By using the computational formula for variance, you can find the sum of their squares directly without ever knowing their individual values. This is a classic JEE strategy to save time.

Q: Why is the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ better than the definition $\frac{\sum (x_i - \bar{x})^2}{n}$?

Both are mathematically equivalent, but the computational formula is significantly faster in this scenario. Using the definition would require you to calculate the deviation of each point from the mean, which would involve $\alpha$ and $\beta$ in every single term, leading to a much more complex algebraic expression. The computational formula lets you work with the sum of squares directly.

Q: Is it possible that $\alpha$ and $\beta$ are not real numbers?

In the context of JEE statistics problems, we assume the data points are real numbers. Even if they were complex, the algebraic manipulation for $\alpha^2 + \beta^2$ remains valid as long as the mean and variance are defined as real numbers. Stick to the algebraic path, and the numbers will resolve themselves.

Question

If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, $\alpha, \beta, 60$ where $\alpha > \beta$ are 56 and 66.2 respectively, then $\alpha^2 + \beta^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are a detective at a crime scene, but instead of fingerprints, you are looking at a dataset. You have ten numbers, but two of them, $\alpha$ and $\beta$, are missing.

You are given the mean and the variance of the entire set. As an elite JEE aspirant, you know that the path of least resistance is often the path of greatest insight. We are not here to find the individuals; we are here to find the sum of their squares.

### The Mean's Secret

The mean is the heartbeat of any dataset. It is defined as:

$$\bar{x} = \frac{\sum x_i}{n}$$

We know the mean is $56$ and the number of observations $n$ is $10$. This gives us a total sum of $560$.

When we look at our data, we see eight known numbers: $65, 68, 58, 44, 48, 45, 60, 60$. Summing these up, we get $448$.

Thus, the equation:

$$56 = \frac{448 + \alpha + \beta}{10}$$

simplifies beautifully to $\alpha + \beta = 112$. This is a vital piece of the puzzle, but we still require the sum of squares.

### The Variance's Power

Now, we turn to the variance, $\sigma^2 = 66.2$. We use the computational formula:

$$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$

This formula is a gift. It allows us to jump straight to the sum of squares without needing to know the individual values of $\alpha$ and $\beta$.

Substituting our known values, we get:

$$66.2 = \frac{\sum x_i^2}{10} - (56)^2$$

Calculating $56^2$ gives us $3136$. So, $66.2 = \frac{\sum x_i^2}{10} - 3136$.

Adding $3136$ to both sides, we get $\frac{\sum x_i^2}{10} = 3202.2$. Multiplying by $10$, we find the total sum of squares:

$$\sum x_i^2 = 32022$$

### Final Calculation

We are now at the finish line. The total sum of squares is the sum of the squares of the eight known numbers plus $\alpha^2 + \beta^2$.

Let us calculate the sum of the squares of the knowns:

$$65^2 + 68^2 + 58^2 + 44^2 + 48^2 + 45^2 + 60^2 + 60^2$$

This yields:

$$4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 3600 + 3600 = 25678$$

Finally, we subtract this from our total sum of squares:

$$\alpha^2 + \beta^2 = 32022 - 25678 = 6344$$

The elegance of this solution lies in its efficiency. We never needed to know $\alpha$ or $\beta$.

Remember this: in the JEE, the most elegant solution is often the one that avoids unnecessary work. You have mastered the data, and the final result is **$6344$**.

Animated Solution for Mathematics - Statistics: If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, $α, β, 60$ where $α > β$ are 56 and 66.2 respectively, then $α^{2} + β^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Data

The Mean Formula

Substituting into Mean

Summing the Knowns

Finding $α + β$

The Variance Formula

Substituting into Variance

Squaring the Mean

Isolating the Sum of Squares

Sum of Known Squares

The Final Step for $α^{2} + β^{2}$

Conclusion and Summary

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Mean's Secret

The Variance's Power

Final Calculation

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