If the mean and variance of the following data: 6, 10, 7, 13, a, 12, b, 12 are 9 and 37/4 respective

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

Great question! While both formulas are mathematically equivalent, the first one is much more efficient for this specific problem. If you used $\frac{\sum (x_i - \bar{x})^2}{n}$, you would have to subtract the mean from every single term, including the unknowns $a$ and $b$, and then square those binomials. That creates a lot of extra algebra. The formula $\frac{\sum x_i^2}{n} - (\bar{x})^2$ allows you to isolate the sum of squares $a^2 + b^2$ much faster.

Q: Do I need to solve for $a$ and $b$ individually to get the answer?

Not at all! In fact, trying to solve for $a$ and $b$ individually might lead you into a trap. You would end up with a system of equations that could have multiple solutions or require solving a quadratic equation. Since the question only asks for $(a - b)^2$, you only need the values of $(a + b)$ and $(a^2 + b^2)$. Always look at what the question asks for before diving into unnecessary calculations.

Q: Why is the variance formula so sensitive to the mean?

Think of the mean as the 'center of gravity' of your data. The variance measures how far, on average, your data points are from that center. The term $(\bar{x})^2$ in the variance formula acts as a correction factor. When you calculate the raw sum of squares $\sum x_i^2$, you are measuring the distance of all points from zero. Subtracting $(\bar{x})^2$ shifts that reference point from zero to the mean, which is exactly what variance represents: the spread around the average.

Question

If the mean and variance of the following data: 6, 10, 7, 13, a, 12, b, 12 are 9 and $\frac{37}{4}$ respectively, then $(a - b)^2$ is equal to:

Options:

(A) 24

(B) 12

(C) 32

(D) 16

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to tackle a problem that perfectly illustrates the elegance of JEE mathematics. We are given eight numbers: $6, 10, 7, 13, a, 12, b, 12$.

We know the mean is $9$ and the variance is $\frac{37}{4}$. Our mission is to find $(a - b)^2$.

At first glance, you might feel the urge to find $a$ and $b$ individually, but let us pause and think like a strategist. We have two unknowns, so we need two independent pieces of information. The mean and the variance are exactly those two keys.

### Phase 1: The Balance Point

The mean is the heart of any dataset. Mathematically, the mean $\bar{x}$ is defined as:

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

With $n = 8$ and $\bar{x} = 9$, the sum of all our observations must be $8 	imes 9 = 72$.

Let us sum our knowns: $6 + 10 + 7 + 13 + 12 + 12 = 60$. So, our equation becomes $60 + a + b = 72$.

Instantly, we have our first vital link: $a + b = 12$. Keep this in your pocket; it is the foundation of our bridge.

### Phase 2: The Measure of Chaos

Now, let us talk about variance. Variance, $\sigma^2$, tells us how much our data deviates from the mean. The most efficient way to handle this is the computational formula:

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

We know $\sigma^2 = \frac{37}{4}$ and $\bar{x} = 9$. Substituting these, we get:

$$\frac{37}{4} = \frac{\sum x_i^2}{8} - 81$$

Let us isolate the sum of squares:

$$\frac{\sum x_i^2}{8} = \frac{37}{4} + 81$$

Calculating the right side, $81 = \frac{324}{4}$, so $\frac{37 + 324}{4} = \frac{361}{4}$. Now, multiply by $8$:

$$\sum x_i^2 = 8 	imes \frac{361}{4} = 2 	imes 361 = 722$$

This is the total sum of the squares of all our data points.

### Phase 3: The Algebraic Bridge

We know the total sum of squares is $722$. Let us break that down:

$$\sum x_i^2 = 6^2 + 10^2 + 7^2 + 13^2 + 12^2 + 12^2 + a^2 + b^2$$

Calculating the known squares: $36 + 100 + 49 + 169 + 144 + 144 = 642$. So, $642 + a^2 + b^2 = 722$.

Subtracting $642$ from $722$, we find $a^2 + b^2 = 80$.

Now, look at what we have: $a + b = 12$ and $a^2 + b^2 = 80$. We need $(a - b)^2$.

We know the identity $(a - b)^2 = a^2 + b^2 - 2ab$. We have $a^2 + b^2$, but we need $2ab$.

We can find $2ab$ using the identity $(a + b)^2 = a^2 + b^2 + 2ab$. Substituting our values:

$$12^2 = 80 + 2ab \Rightarrow 144 = 80 + 2ab \Rightarrow 2ab = 64$$

Finally, the moment of truth:

$$(a - b)^2 = a^2 + b^2 - 2ab = 80 - 64 = 16$$

And there it is! The answer is **$16$**. By focusing on the relationships between the variables rather than the variables themselves, we navigated the problem with precision and grace.

Animated Solution for Mathematics - Statistics: If the mean and variance of the following data: 6, 10, 7, 13, a, 12, b, 12 are 9 and $\frac{37}{4}$ respectively, then $(a - b)^{2}$ is equal to:

Select Answer:

Visualized Solution

Given Data

Mean Formula & Substitution

Solve for $a + b$

Variance Formula & Substitution

Calculate Sum of Squares

Solve for $a^{2} + b^{2}$

Algebraic Identity for $2 ab$

Final Calculation for $(a - b)^{2}$

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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