Consider a set of 3n numbers having variance 4. In this set, the mean of first 2n numbers is 6 and t

Q: Wait, I thought adding a constant to every term doesn't change the variance. Why does it change here?

That is a brilliant observation! You are absolutely right: if you add a constant $c$ to *every* single number in a set, the variance remains unchanged because the relative distances between the numbers don't change. However, in this problem, we are adding $+1$ to one group and $-1$ to another. We are effectively pulling the two groups apart, which increases the 'spread' relative to the new mean. That is why the variance changes!

Q: Do I really need to calculate the sum of squares? It feels like a lot of algebra.

In the heat of a JEE Advanced exam, algebra is your best friend. While there are properties of variance involving covariance that could solve this, they are often more prone to sign errors. The method of calculating $\sum x_i^2$ is robust, systematic, and foolproof. Trust the process—the algebra will simplify beautifully if you keep your terms organized.

Q: Why is the final answer asking for $9k$? Is there a trick I missed?

No trick at all! Often, JEE examiners ask for a multiple of the answer (like $9k$ or $100k$) simply to ensure the final result is an integer. It prevents ambiguity in the answer key. Don't let the $9k$ distract you; focus entirely on finding the value of $k$ first. Once you have $k$, the final step is just a simple multiplication.

Question

Consider a set of $3n$ numbers having variance 4. In this set, the mean of first $2n$ numbers is 6 and the mean of the remaining $n$ numbers is 3. A new set is constructed by adding 1 into each of first $2n$ numbers, and subtracting 1 from each of the remaining $n$ numbers. If the variance of the new set is $k$, then $9k$ is equal to ______.

Sigma Percentile · Accepted Answer

### The Symphony of Statistics: Understanding Data Shifts

Welcome, my dear student. Today, we are going to dive into a problem that, at first glance, looks like a tedious exercise in algebra. But I want you to pause and look deeper.

Statistics is not just about crunching numbers; it is the art of understanding how a collection of points behaves in space. When we talk about variance, we are talking about the 'spread'—the chaotic energy of a dataset. When we transform that data, we are essentially changing the geometry of that spread.

Let us embark on this journey together.

### Phase 1: The Original State

Imagine you have $3n$ numbers scattered on a number line. We have two groups: a larger group of $2n$ numbers with a mean of $6$, and a smaller group of $n$ numbers with a mean of $3$.

Before we do anything else, we must find the center of gravity of this entire system—the combined mean, $\bar{x}$. The formula is our compass:

$$\bar{x} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$$

Substituting our values, we get:

$$\bar{x} = \frac{(2n 	imes 6) + (n 	imes 3)}{3n} = \frac{12n + 3n}{3n} = \frac{15n}{3n} = 5$$

So, our original mean is $5$. This is the anchor point for our entire calculation. Now, we are given that the variance $\sigma^2$ is $4$.

Recall the fundamental definition of variance:

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

This equation is the most powerful tool in your statistical arsenal. It links the raw data (the sum of squares) to the variance and the mean. Let us rearrange it to find the sum of squares of our original numbers:

$$4 = \frac{\sum x_i^2}{3n} - (5)^2$$

$$4 = \frac{\sum x_i^2}{3n} - 25$$

$$\frac{\sum x_i^2}{3n} = 29 \Rightarrow \sum x_i^2 = 87n$$

This value, $87n$, is the 'energy' of our original distribution. Keep it safe; we will need it soon.

### Phase 2: The Transformation

Now, the problem introduces a transformation. We are adding $1$ to each of the first $2n$ numbers (let's call them $a_i$) and subtracting $1$ from each of the remaining $n$ numbers (let's call them $b_j$).

Imagine the number line. The first group is shifting to the right, and the second group is shifting to the left. They are moving away from each other.

First, let's find the new mean, $\bar{x}_{new}$. The sum of the new numbers is:

$$\sum (a_i + 1) + \sum (b_j - 1) = (\sum a_i + 2n) + (\sum b_j - n)$$

Since $\sum a_i = 12n$ and $\sum b_j = 3n$, the new sum is:

$$12n + 2n + 3n - n = 16n$$

Dividing by the total count $3n$, we get the new mean:

$$\bar{x}_{new} = \frac{16n}{3n} = \frac{16}{3}$$

### Phase 3: The Final Calculation

Now, we need the new sum of squares, $S_{new} = \sum (a_i + 1)^2 + \sum (b_j - 1)^2$. Let's expand these squares carefully:

$$S_{new} = \sum (a_i^2 + 2a_i + 1) + \sum (b_j^2 - 2b_j + 1)$$

Grouping the terms, we get:

$$S_{new} = (\sum a_i^2 + \sum b_j^2) + 2\sum a_i - 2\sum b_j + (2n + n)$$

We already know $\sum a_i^2 + \sum b_j^2 = 87n$. We also know $\sum a_i = 12n$ and $\sum b_j = 3n$. Substituting these in:

$$S_{new} = 87n + 2(12n) - 2(3n) + 3n$$

$$S_{new} = 87n + 24n - 6n + 3n = 108n$$

We are almost there! The new variance $k$ is:

$$k = \frac{S_{new}}{3n} - (\bar{x}_{new})^2$$

$$k = \frac{108n}{3n} - \left(\frac{16}{3}ight)^2$$

$$k = 36 - \frac{256}{9}$$

$$k = \frac{324 - 256}{9} = \frac{68}{9}$$

Finally, the problem asks for $9k$. Multiplying our result by $9$, we get **$68$**.

See how the complexity melted away? By systematically tracking the sum of squares and the mean, we navigated the transformation without getting lost in the chaos. Keep this methodical approach in your heart, and no JEE problem will ever be too daunting.

Enter Numerical Value:

Visualized Solution

Understanding the Data Groups

Calculating the Combined Mean

The Mean of the Original Set

Applying the Variance Formula

Finding the Sum of Squares

The Data Transformation

Calculating the New Mean

Evaluating the New Mean

Expanding the New Sum of Squares

Grouping the Terms

Substituting Known Values

Calculating the New Variance $k$

Final Result for $9 k$

The Sigma Insight: Variance and Standard Deviation

The Symphony of Statistics

Phase 1

Phase 2

Phase 3

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