If the mean and variance of six observations 7, 10, 11, 15, a, b are 10 and 20/3, respectively, then

Q: Wait, why do we reject the negative root here?

Actually, we are looking for the absolute difference $|a - b|$. Since $(a - b)^2$ is always non-negative for real numbers, the square root will always be positive. If you ever get a negative value for a squared term, it's a sign to re-check your arithmetic! We aren't rejecting a root; we are simply taking the magnitude of the difference.

Q: Why can't we just use energy conservation?

I love the enthusiasm! While this isn't a mechanics problem, you are thinking correctly about 'constraints.' In statistics, the Mean and Variance act as our conservation laws. They constrain the system, limiting the possible values of $a$ and $b$ just like energy conservation limits the motion of a particle.

Q: Is there a faster way to find $a$ and $b$?

You could solve the quadratic equation $x^2 - (a+b)x + ab = 0$, but that is overkill. The identity $(a - b)^2 = 2(a^2 + b^2) - (a + b)^2$ is the 'JEE shortcut.' It allows you to bypass solving for $a$ and $b$ individually, saving you precious minutes during the exam.

Question

If the mean and variance of six observations 7, 10, 11, 15, a, b are 10 and $\frac{20}{3}$, respectively, then the value of $|a - b|$ is equal to :

Options:

(A) 9

(B) 11

(C) 7

(D) 1

Sigma Percentile · Accepted Answer

### The Mystery of the Missing Variables

My dear student, welcome to the world of statistics, where numbers are not just cold, hard digits—they are stories waiting to be told. Today, we are faced with a classic JEE Advanced challenge.

We have six observations: $7, 10, 11, 15, a,$ and $b$. Two of these are shrouded in mystery. We are given the Mean and the Variance, and our mission is to find the absolute difference between these two unknown values, $|a - b|$.

Many students see this and immediately panic, trying to solve for $a$ and $b$ individually. But hold on! In the JEE, the path of least resistance is often the path of greatest elegance. Let us walk through this together, step by step.

### Phase 1: The Balancing Act (The Mean)

Imagine the Mean, $\bar{x} = 10$, as the center of gravity for our data set. The formula for the mean is the bedrock of statistics:

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

We have six observations, so $n = 6$. When we plug in our values, we get:

$$10 = \frac{7 + 10 + 11 + 15 + a + b}{6}$$

By multiplying both sides by $6$, we find the total sum of our observations:

$$60 = 43 + a + b$$

Subtracting $43$ from $60$, we arrive at our first vital clue: $a + b = 17$. This is our first equation. It tells us that no matter what $a$ and $b$ are, their sum is locked at $17$. Keep this in your pocket; we will need it soon.

### Phase 2: The Spread (The Variance)

Now, let us tackle the Variance, $\sigma^2 = \frac{20}{3}$. Variance is a measure of how 'spread out' our data is. While the definition formula involves deviations from the mean, the computational formula is far more powerful:

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

This formula is a lifesaver. It prevents us from having to calculate $(x_i - \bar{x})^2$ for every single term. Let us substitute our knowns:

$$\frac{20}{3} = \frac{7^2 + 10^2 + 11^2 + 15^2 + a^2 + b^2}{6} - 10^2$$

Calculating the squares is straightforward: $49, 100, 121,$ and $225$. Their sum is $495$. Now, watch the algebra unfold:

$$\frac{20}{3} + 100 = \frac{495 + a^2 + b^2}{6}$$

$$\frac{320}{3} = \frac{495 + a^2 + b^2}{6}$$

When we multiply both sides by $6$, the $3$ in the denominator cancels out beautifully, leaving us with a factor of $2$ multiplied by $320$, which is $640$. Thus:

$$640 = 495 + a^2 + b^2$$

$$a^2 + b^2 = 145$$

### Phase 3: The Algebraic Elegance

We are now at the finish line. We have $a + b = 17$ and $a^2 + b^2 = 145$. Most students would now try to solve for $a$ and $b$ using a quadratic equation. But why do that when we can use the beauty of algebraic identities?

We want $|a - b|$. Consider this identity:

$$(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)$$

This is a powerful tool in your arsenal. It connects the sum, the difference, and the sum of squares. Let us substitute our values:

$$(17)^2 + (a - b)^2 = 2(145)$$

$$289 + (a - b)^2 = 290$$

$$(a - b)^2 = 290 - 289$$

$$(a - b)^2 = 1$$

Taking the square root of both sides, we find that **$|a - b| = 1$**.

### Conclusion

And there it is! The distance between our two mysterious points is exactly $1$. We didn't need to find $a$ or $b$ individually. We didn't need to solve a complex quadratic.

We simply used the properties of the data to guide us to the answer. Remember, in JEE Advanced, it is not just about calculating; it is about choosing the most elegant path. Keep practicing, keep visualizing, and keep falling in love with the logic of mathematics!

Animated Solution for Mathematics - Statistics: If the mean and variance of six observations 7, 10, 11, 15, a, b are 10 and $\frac{20}{3}$ , respectively, then the value of $∣ a - b ∣$ is equal to :

Select Answer:

Visualized Solution

Understanding the Data

The Mean Formula

Substituting into the Mean

Finding $a + b$

The Variance Formula

Substituting into Variance

Summing the Squares

Simplifying the Equation

Finding $a^{2} + b^{2}$

The Algebraic Identity

Substituting into the Identity

Calculating $∣ a - b ∣$

The Sigma Insight: Variance and Standard Deviation

The Mystery of the Missing Variables

Phase 1

Phase 2

Phase 3

Conclusion

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