Let the mean and variance of four numbers 3, 7, x and y (x y) be 5 and 10 respectively. Then the m

Q: Wait, why do we use the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - \mu^2$ instead of $\frac{\sum (x_i - \mu)^2}{n}$?

Both formulas are mathematically equivalent! However, in competitive exams like JEE, the formula $\sigma^2 = \frac{\sum x_i^2}{n} - \mu^2$ is often faster because it avoids calculating the deviation of each term from the mean, which can be messy if the mean is not an integer. It directly uses the sum of squares, which is exactly what we need to solve for $x^2+y^2$.

Q: Why is the condition $x > y$ given in the problem?

Great observation! When we solve the system $x+y=10$ and $xy=9$, we get two possible pairs for $(x, y)$: $(9, 1)$ or $(1, 9)$. Without the condition $x > y$, we wouldn't know which value corresponds to $x$ and which to $y$. This condition ensures a unique solution for the final calculation.

Q: Can I use the property of variance under linear transformation to solve this faster?

While you can use properties like $Var(ax+b) = a^2 Var(x)$, it is actually more complex here because the new set of numbers is not a simple linear transformation of the original set. The new numbers are $3+2x, 7+2y, x+y, x-y$. Since these are mixed combinations of $x$ and $y$, the algebraic approach we used is the most reliable and straightforward method.

Question

Let the mean and variance of four numbers $3, 7, x$ and $y (x > y)$ be 5 and 10 respectively. Then the mean of four numbers $3 + 2x, 7 + 2y, x + y$ and $x - y$ is .

Sigma Percentile · Accepted Answer

### The Mystery of the Missing Numbers

Imagine you are standing at the edge of a mathematical landscape, looking at a set of four numbers: $3, 7, x,$ and $y$. You are told that their mean is $5$ and their variance is $10$.

In the world of JEE Advanced, statistics is not just about plugging numbers into formulas; it is about decoding the hidden relationships between variables. Let us embark on this journey to find $x$ and $y$ and solve the final mystery.

### Phase 1: Decoding the Mean

We start with the most fundamental tool in our statistical arsenal: the mean. The mean, denoted by $\mu$, is the center of gravity of our data. It is defined as the sum of all observations divided by the total number of observations.

We are given four numbers, so our equation is:

$$\mu = \frac{3 + 7 + x + y}{4} = 5$$

Multiplying both sides by $4$, we get $3 + 7 + x + y = 20$. Simplifying this, we find our first crucial relationship: $x + y = 10$.

This is our first anchor point. We have two variables, $x$ and $y$, and we have successfully reduced the complexity of the mean into a simple linear equation.

### Phase 2: The Variance Trap

Now, we turn to the variance, $\sigma^2 = 10$. The variance measures the spread of the data. A powerful way to express variance is the mean of the squares minus the square of the mean:

$$\sigma^2 = \frac{\sum x_i^2}{n} - \mu^2$$

Substituting our known values, we get:

$$10 = \frac{3^2 + 7^2 + x^2 + y^2}{4} - 5^2$$

This is where many students stumble. Let us be precise. $5^2$ is $25$. Adding $25$ to both sides, we get:

$$35 = \frac{9 + 49 + x^2 + y^2}{4}$$

Simplifying the numerator, $9 + 49 = 58$. So, $35 = \frac{58 + x^2 + y^2}{4}$. Cross-multiplying gives us $140 = 58 + x^2 + y^2$, which simplifies to $x^2 + y^2 = 82$.

We now have two equations: $x + y = 10$ and $x^2 + y^2 = 82$.

### Phase 3: The Algebraic Bridge

We have the sum and the sum of squares. To find the individual values, we need the product $xy$. This is where the classic algebraic identity shines:

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

Substituting our values, $10^2 = 82 + 2xy$, which means $100 = 82 + 2xy$. Subtracting $82$ from $100$ leaves us with $18 = 2xy$, so $xy = 9$.

Now, we have a system: $x + y = 10$ and $xy = 9$. These are the roots of the quadratic equation $t^2 - 10t + 9 = 0$.

Factoring this, we get $(t - 9)(t - 1) = 0$. Thus, the numbers are $9$ and $1$. Given the condition $x > y$, we conclude that $x = 9$ and $y = 1$.

### Phase 4: The Final Transformation

With $x = 9$ and $y = 1$, we can now construct the new set of numbers requested by the problem: $3 + 2x, 7 + 2y, x + y,$ and $x - y$. Substituting our values:

*   $3 + 2(9) = 3 + 18 = 21$
*   $7 + 2(1) = 7 + 2 = 9$
*   $9 + 1 = 10$
*   $9 - 1 = 8$

The new set is $21, 9, 10, 8$. The final step is to find the mean of this new set:

$$	ext{New Mean} = \frac{21 + 9 + 10 + 8}{4} = \frac{48}{4} = 12$$

And there we have it! The mean of the new set is **$12$**. By systematically breaking down the statistical parameters, we have navigated through the algebra to arrive at the solution.

Animated Solution for Mathematics - Statistics: Let the mean and variance of four numbers $3, 7, x$ and $y (x > y)$ be 5 and 10 respectively. Then the mean of four numbers $3 + 2 x, 7 + 2 y, x + y$ and $x - y$ is .

Enter Numerical Value:

Visualized Solution

Given Data

Mean Formula

Solve for $x + y$

Variance Formula

Simplify Variance

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

Algebraic Identity

Calculate $x y$

Find $x$ and $y$

New Set of Numbers

Final Mean Calculation

The Sigma Insight: Measures of Dispersion

The Mystery of the Missing Numbers

Phase 1

Phase 2

Phase 3

Phase 4

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