The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observa

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

Both are mathematically identical! The first one is often preferred in competitive exams because it simplifies the calculation by avoiding the subtraction of the mean from each individual term. It's a shortcut that saves time and reduces the chance of arithmetic errors.

Q: Can I just solve for $x_4$ and $x_5$ using the quadratic equation?

Absolutely! You could form a quadratic equation $t^2 - (x_4+x_5)t + x_4x_5 = 0$. However, that would require finding the product $x_4x_5$ first. While valid, using the identity $(x_4 - x_5)^2 = 2(x_4^2 + x_5^2) - (x_4 + x_5)^2$ is much more elegant and direct. In JEE, speed and elegance are your best friends.

Q: What if the value inside the square root is negative?

That would be a red flag! Variance is a measure of spread, and it is always non-negative. If you ever find yourself with a negative value inside a square root for a variance-related problem, pause and re-check your calculations. It's a sign that a calculation error occurred earlier in the process.

Question

The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4; then the absolute value of the difference of the other two observations, is :

Options:

(A) $1$

(B) $3$

(C) $7$

(D) $5$

Sigma Percentile · Accepted Answer

### The Heartbeat of Data: Decoding Mean and Variance

Welcome, fellow explorer of the mathematical universe! Today, we are going to dive into a problem that might seem like a simple exercise in statistics, but it is actually a beautiful dance of algebra and data analysis.

Imagine you are standing before a set of five numbers. Three of them are known: $3, 4, 4$. The other two, let's call them $x_4$ and $x_5$, are shrouded in mystery.

We are given the mean and the variance, and our mission is to uncover the absolute difference between these two unknown values. Let's embark on this journey together.

### Phase 1: The Balance Point (The Mean)

First, let's talk about the mean. The mean, $\bar{x}$, is the center of gravity of your data. It is the point where, if you were to balance these numbers on a scale, everything would be perfectly level.

We know the mean is $4$ and the total number of observations $n$ is $5$. The formula for the mean is:

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

Substituting our values, we get:

$$4 = \frac{3 + 4 + 4 + x_4 + x_5}{5}$$

Multiplying both sides by $5$, we find that the sum of all observations is $20$. Thus, $11 + x_4 + x_5 = 20$, which leads us to our first crucial piece of information:

$$x_4 + x_5 = 9$$

We have successfully captured the sum of our unknowns!

### Phase 2: The Measure of Chaos (The Variance)

Now, let's turn our attention to the variance, $\sigma^2$. If the mean is the center, the variance is the measure of how much the data spreads out from that center. It is the 'chaos' factor.

We are given $\sigma^2 = 5.20$. The formula we will use is the computational form of variance:

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

This formula is a powerful tool in your JEE arsenal. Let's plug in our values:

$$5.20 = \frac{3^2 + 4^2 + 4^2 + x_4^2 + x_5^2}{5} - 4^2$$

Simplifying this, we get:

$$5.20 = \frac{9 + 16 + 16 + x_4^2 + x_5^2}{5} - 16$$

Adding $16$ to both sides gives us:

$$21.20 = \frac{41 + x_4^2 + x_5^2}{5}$$

Multiplying by $5$, we find $106 = 41 + x_4^2 + x_5^2$, which means:

$$x_4^2 + x_5^2 = 65$$

We now have the sum of the squares of our unknowns!

### Phase 3: The Algebraic Bridge

We are so close! We have $x_4 + x_5 = 9$ and $x_4^2 + x_5^2 = 65$. We need to find $|x_4 - x_5|$.

Instead of solving for $x_4$ and $x_5$ individually, which could lead to messy quadratic equations, let's use a brilliant algebraic identity:

$$(x_4 - x_5)^2 = 2(x_4^2 + x_5^2) - (x_4 + x_5)^2$$

This identity is the shortcut to the truth. Let's substitute our values:

$$(x_4 - x_5)^2 = 2(65) - (9)^2$$

This gives us:

$$(x_4 - x_5)^2 = 130 - 81 = 49$$

Finally, taking the square root, we find:

$$|x_4 - x_5| = 7$$

### The Final Victory

Look at that! With just a few steps of logical deduction and the power of algebraic identities, we have unmasked the difference between our two unknown observations.

The beauty of this problem lies not in the arithmetic, but in the way the mean and variance act as constraints, guiding us directly to the solution.

Remember, in JEE Advanced, it is rarely about brute force; it is about finding the most elegant path. You have mastered this path today. The final answer is **7**.

Animated Solution for Mathematics - Statistics: The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4; then the absolute value of the difference of the other two observations, is :

Select Answer:

Visualized Solution

Given Data for $n = 5$

The Mean Formula $\overset{x}{ˉ}$

Setting up the Mean Equation

Finding the Sum $x_{4} + x_{5}$

The Variance Formula $σ^{2}$

Setting up the Variance Equation

Finding the Sum of Squares $x_{4}^{2} + x_{5}^{2}$

The Algebraic Identity

Calculating the Squared Difference

Final Absolute Difference $∣ x_{4} - x_{5} ∣$

The Sigma Insight: Measures of Dispersion

The Heartbeat of Data

Phase 1

Phase 2

Phase 3

The Final Victory

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