If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, x and y be 10 and 25 respectively, th

Q: Wait, why do we use the computational formula for variance instead of the definition $\frac{\sum(x_i - \bar{x})^2}{n}$?

You absolutely can use the definition! However, the computational formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ is a powerful shortcut. It saves you from calculating the deviation of every single data point from the mean, which is where most students make arithmetic errors. In JEE, speed and accuracy are your best friends.

Q: Do I need to solve for $x$ and $y$ individually to find their product?

Not at all! While you could solve the quadratic equation to find the values of $x$ and $y$, the question only asks for their product $xy$. Using the identity $(x + y)^2 = x^2 + y^2 + 2xy$ is much faster and less prone to calculation mistakes. Always look for the most direct path to the answer.

Q: What if the variance was zero? Does that change the approach?

If the variance were zero, it would mean all the numbers in the set are identical to the mean. In that case, $x$ and $y$ would both have to be 10. It's a great sanity check—variance is essentially a measure of how much the data deviates from the 'average' behavior. If there is no deviation, the variance must be zero.

Question

If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, $x$ and $y$ be 10 and 25 respectively, then $x \cdot y$ is equal to....

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are a data scientist tasked with reconstructing a missing piece of a puzzle. You have a set of eight numbers, but two of them, $x$ and $y$, are hidden.

You are given two vital clues: the mean, which tells you where the data is centered, and the variance, which tells you how much the data is spread out. This is a classic JEE Advanced problem that tests your ability to translate statistical concepts into algebraic relationships.

### Phase 1: The Mean as a Balancing Act

The mean, $\bar{x}$, is the fulcrum of your data. It is the point where, if you placed your numbers on a physical beam, it would perfectly balance.

We know that the mean of eight numbers is 10. The formula for the mean is:

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

With $n = 8$ and $\bar{x} = 10$, the sum of all eight numbers must be $8 	imes 10 = 80$. Our known numbers are 3, 7, 9, 12, 13, and 20.

Adding these up, we get $3 + 7 + 9 + 12 + 13 + 20 = 64$. Therefore, our first equation is $64 + x + y = 80$, which simplifies to:

$$x + y = 16$$

### Phase 2: The Variance as a Measure of Energy

Now, we turn to the variance, $\sigma^2 = 25$. While the definition $\frac{\sum(x_i - \bar{x})^2}{n}$ is intuitive, the computational formula is the secret weapon for efficiency:

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

Let's plug in our values: $25 = \frac{\sum x_i^2}{8} - (10)^2$. This simplifies to $25 = \frac{\sum x_i^2}{8} - 100$, or $125 = \frac{\sum x_i^2}{8}$.

Multiplying by 8, we find that the sum of the squares of all eight numbers is $1000$. Now, let's calculate the sum of the squares of our known numbers:

$$3^2 + 7^2 + 9^2 + 12^2 + 13^2 + 20^2 = 9 + 49 + 81 + 144 + 169 + 400 = 852$$

With this, we can find the sum of the squares of our unknowns: $852 + x^2 + y^2 = 1000$, which gives us:

$$x^2 + y^2 = 148$$

### Phase 3: The Algebraic Bridge

We have arrived at the final stage with two powerful equations: $x + y = 16$ and $x^2 + y^2 = 148$. We need to find the product $xy$.

We recall the algebraic identity:

$$(x + y)^2 = x^2 + y^2 + 2xy$$

By substituting our known sums, we get $(16)^2 = 148 + 2xy$. Calculating $16^2$ gives us 256.

So, $256 = 148 + 2xy$. Subtracting 148 from 256 leaves us with $108 = 2xy$.

Finally, dividing by 2, we find the result:

**$xy = 54$**

Animated Solution for Mathematics - Statistics: If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, $x$ and $y$ be 10 and 25 respectively, then $x \cdot y$ is equal to....

Enter Numerical Value:

Visualized Solution

Problem Setup

Formula for Mean

Substituting Values in Mean

Solving for $x + y$

Formula for Variance

Substituting Values in Variance

Calculating Sum of Squares

Simplifying the Variance Equation

Solving for $x^{2} + y^{2}$

Using Algebraic Identity

Substituting into Identity

Final Calculation for $x y$

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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