Let X = {11, 12, 13, ..., 40, 41} and Y = {61, 62, 63, ..., 90, 91} be the two sets of observations.

Q: Why is the combined mean just the average of the two means? Can we always do this?

Great observation! You can only do this when the number of elements in both sets is equal. Since $n_1 = n_2 = 31$, the combined mean is indeed the simple arithmetic mean of the two individual means. If the sample sizes were different, you would need to use the weighted average formula $\mu = \frac{n_1\bar{x} + n_2\bar{y}}{n_1 + n_2}$.

Q: Why does the linear term vanish in the expansion of $(k - 25)^2$?

This is the beauty of symmetry! When we define $k$ as the deviation from the mean, $k$ ranges from $-15$ to $15$. Because the set of values for $k$ is symmetric around zero, the sum of all $k$ values is exactly zero. This is a powerful shortcut that saves you from tedious arithmetic.

Q: Is there a faster way to calculate the sum of squares?

Absolutely. Instead of summing every single number, use the standard formula for the sum of the first $n$ squares: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$. In our case, we calculate this for $n=15$ and multiply by 2 (since we have both positive and negative $k$ values).

Question

Let $X = \{11, 12, 13, \dots, 40, 41\}$ and $Y = \{61, 62, 63, \dots, 90, 91\}$ be the two sets of observations. If $\bar{x}$ and $\bar{y}$ are their respective means and $\sigma^2$ is the variance of all the observations in $X \cup Y$, then $|\bar{x} + \bar{y} - \sigma^2|$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a statistics problem; we are uncovering the hidden symmetry within a set of numbers.

When you look at the sets $X = \{11, 12, \dots, 41\}$ and $Y = \{61, 62, \dots, 91\}$, what do you see? These are arithmetic progressions, perfectly spaced, marching in lockstep with a common difference of $d = 1$. This structure is our greatest ally.

### Phase 1: Finding the Center of Gravity

Before we can talk about variance—which is essentially a measure of how 'spread out' our data is—we must find the center of gravity, the mean. For an arithmetic progression, the mean is simply the average of the first and last terms.

For Set $X$, the first term is $11$ and the last is $41$. The mean $\bar{x}$ is:
$$\bar{x} = \frac{11 + 41}{2} = 26$$

For Set $Y$, the first term is $61$ and the last is $91$. The mean $\bar{y}$ is:
$$\bar{y} = \frac{61 + 91}{2} = 76$$

Since both sets contain exactly $31$ elements, the combined mean $\mu$ of the union $X \cup Y$ is the midpoint of these two means:
$$\mu = \frac{26 + 76}{2} = 51$$

We have found our anchor point. Now, we can measure how far every single number in our combined set deviates from this value of $51$.

### Phase 2: The Art of the Shift

Here is where most students get bogged down in arithmetic. Instead of calculating $(11-51)^2 + (12-51)^2 + \dots$ directly, let us use the 'shift' method.

Imagine we define every element $x$ in Set $X$ as $x = 26 + k$, where $k$ ranges from $-15$ to $15$. When we look at the deviation $(x - 51)$, it becomes $(26 + k - 51)$, which simplifies to $(k - 25)$.

Now, look at the sum of squared deviations for Set $X$:
$$\sum_{x \in X} (x - 51)^2 = \sum_{k=-15}^{15} (k - 25)^2$$

Expanding this, we get $\sum (k^2 - 50k + 625)$. Because $k$ is symmetric around zero, the term $\sum 50k$ vanishes completely! We are left with $\sum k^2 + 31(625)$.

### Phase 3: The Symmetry Miracle

We apply the exact same logic to Set $Y$. We define $y = 76 + k$, where $k$ again ranges from $-15$ to $15$. The deviation $(y - 51)$ becomes $(76 + k - 51)$, which is $(k + 25)$.

Expanding this, we get $\sum (k^2 + 50k + 625)$. Again, the linear term $\sum 50k$ vanishes. The sum for Set $Y$ is identical to the sum for Set $X$: $\sum k^2 + 31(625)$.

This symmetry is beautiful. It tells us that both sets are distributed identically around their respective means, and their combined variance is just a reflection of this shared structure.

### Phase 4: The Final Calculation

Now, we bring it all together. The total sum of squares is $2 	imes [\sum_{k=-15}^{15} k^2 + 31(625)]$.

Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ for $n=15$, we find:
$$\sum_{k=1}^{15} k^2 = \frac{15 	imes 16 	imes 31}{6} = 1240$$

Since we have both positive and negative $k$ values, the sum from $-15$ to $15$ is $2 	imes 1240 = 2480$. Adding this to $31 	imes 625 = 19375$, we get $21855$. Multiplying by $2$ gives us a total sum of $43710$.

Finally, the variance $\sigma^2$ is the total sum divided by the total number of observations, $N = 62$:
$$\sigma^2 = \frac{43710}{62} = 705$$

Our target expression is $|\bar{x} + \bar{y} - \sigma^2|$. Substituting our values:
$$|26 + 76 - 705| = |102 - 705| = |-603| = 603$$

We have arrived at the destination. The final answer is **603**. Remember: in JEE Advanced, the math is rarely about how fast you can calculate; it is about how clearly you can see the structure of the problem.

Enter Numerical Value:

Visualized Solution

Visualizing the Data Sets

Mean of Set $X$

Mean of Set $Y$

Combined Mean $μ$

The Variance Formula

Sum of Squared Deviations for $X$

Expanding the Sum for $X$

Sum of Squared Deviations for $Y$

Expanding the Sum for $Y$

Total Sum of Squares

Calculating Variance $σ^{2}$

The Final Calculation

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

Visualizing the Data Sets

Mean of Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

Mean of Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y

Combined Mean .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ

The Variance Formula

Sum of Squared Deviations for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

Expanding the Sum for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

Sum of Squared Deviations for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y

Expanding the Sum for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y