Let the mean and variance of 8 numbers x, y, 10, 12, 6, 12, 4, 8 be 9 and 9.25 respectively. If x

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

That is a fantastic question! While both formulas are mathematically equivalent, the definition $\frac{\sum (x_i - \bar{x})^2}{n}$ requires you to subtract the mean from every single data point, which is tedious and prone to arithmetic errors, especially with unknowns like $x$ and $y$. The computational formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ allows us to work with the sum of squares directly, which is much cleaner and faster for solving systems of equations.

Q: What if I get stuck solving the quadratic equation $t^2 - 20t + 91 = 0$?

Don't panic! If you cannot immediately see the factors, always fall back on the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a=1$, $b=-20$, and $c=91$. The discriminant $\sqrt{400 - 364} = \sqrt{36} = 6$ will guide you straight to the roots. You have the tools; just trust the process.

Q: Does the order of $x$ and $y$ matter in the final answer?

Absolutely! The problem explicitly states $x > y$. If you solve the quadratic and get $13$ and $7$, you must assign $x=13$ and $y=7$. If you swapped them, you would get $3(7) - 2(13) = 21 - 26 = -5$, which would be incorrect. Always check your constraints at the very end of the problem.

Question

Let the mean and variance of 8 numbers $x, y, 10, 12, 6, 12, 4, 8$ be 9 and 9.25 respectively. If $x > y$, then $3x - 2y$ is equal to ______.

Sigma Percentile · Accepted Answer

### The Detective Work of Statistics: Unlocking the Unknowns

Welcome, future engineer. Today, we are not just solving a statistics problem; we are embarking on a detective mission. We have a set of eight numbers: $x, y, 10, 12, 6, 12, 4, 8$.

We are given the mean and the variance, and we are tasked with finding the value of $3x - 2y$. It might look like a simple algebra problem, but it is a beautiful exercise in how we can use statistical properties to constrain and solve for unknown variables. Let us dive in.

### Phase 1: The Linear Constraint

Imagine you are standing before a locked door. The mean is your first key. The definition of the mean is the sum of all observations divided by the number of observations, expressed as:

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

We know the mean is $9$ and the total count $n$ is $8$. So, we write:

$$9 = \frac{x + y + 10 + 12 + 6 + 12 + 4 + 8}{8}$$

When we sum the known numbers, we get $52$. Thus, our equation becomes $x + y + 52 = 72$.

Subtracting $52$ from both sides, we arrive at our first vital piece of information: $x + y = 20$. This is our linear constraint. It tells us that $x$ and $y$ are tethered together; if one increases, the other must decrease to keep the mean constant.

### Phase 2: The Quadratic Constraint

Now, we need a second key to unlock the values of $x$ and $y$ individually. This is where variance comes in. Variance is a measure of spread, but algebraically, it is a goldmine.

We use the computational formula:

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

We are given $\sigma^2 = 9.25$ and $\bar{x} = 9$. Substituting these into our formula, we get:

$$9.25 = \frac{x^2 + y^2 + 10^2 + 12^2 + 6^2 + 12^2 + 4^2 + 8^2}{8} - 9^2$$

Notice how the variance formula naturally introduces the sum of squares, $x^2 + y^2$. This is the bridge we need. Calculating the squares of the known numbers: $100 + 144 + 36 + 144 + 16 + 64 = 504$.

Our equation simplifies to:

$$9.25 = \frac{x^2 + y^2 + 504}{8} - 81$$

Adding $81$ to both sides gives $90.25 = \frac{x^2 + y^2 + 504}{8}$. Multiplying by $8$, we find $722 = x^2 + y^2 + 504$.

Finally, subtracting $504$ yields $x^2 + y^2 = 218$. We now have two equations: $x + y = 20$ and $x^2 + y^2 = 218$.

### Phase 3: The Algebraic Bridge

We have the sum and the sum of squares. How do we find $x$ and $y$? We use the identity $(x + y)^2 = x^2 + y^2 + 2xy$.

This identity is the glue that connects linear sums to products. Substituting our known values:

$$(20)^2 = 218 + 2xy$$

This simplifies to $400 = 218 + 2xy$, which means $2xy = 182$, or $xy = 91$. Now we have the sum $S = 20$ and the product $P = 91$.

Any two numbers with a known sum and product are the roots of the quadratic equation $t^2 - St + P = 0$. Substituting our values, we get:

$$t^2 - 20t + 91 = 0$$

### Phase 4: The Final Reveal

Solving $t^2 - 20t + 91 = 0$ is straightforward. We look for two numbers that multiply to $91$ and add to $20$.

Those numbers are $13$ and $7$. So, $(t - 13)(t - 7) = 0$, giving us roots $13$ and $7$. We assume the condition $x > y$, which forces $x = 13$ and $y = 7$.

Finally, we calculate the objective:

$$3x - 2y = 3(13) - 2(7) = 39 - 14 = 25$$

We have successfully navigated the problem, turning statistical properties into a clear, definitive answer. Remember, in JEE Advanced, the math is never just about calculation; it is about recognizing the structure of the problem. You have done excellent work today.

Animated Solution for Mathematics - Statistics: Let the mean and variance of 8 numbers $x, y, 10, 12, 6, 12, 4, 8$ be 9 and 9.25 respectively. If $x > y$ , then $3 x - 2 y$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Problem Overview

Applying the Mean Formula

Finding the Sum $x + y$

The Variance Formula

Substituting into Variance

Summing the Squares

Finding $x^{2} + y^{2}$

Calculating the Product $x y$

Forming the Quadratic

Solving for $x$ and $y$

Applying the Constraint $x > y$

Final Calculation

Summary and Takeaway

The Sigma Insight: Variance and Standard Deviation

The Detective Work of Statistics

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Statistics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the mean and variance of 8 numbers x,y,10,12,6,12,4,8 be 9 and 9.25 respectively. If x>y, then 3x−2y is equal to ______.