Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and α (0

Q: Wait, why can't I just take the average of the variances?

That is the most common trap in statistics! The variance measures the spread of data points. When you combine two groups, the total spread is not just the average of the internal spreads; it also includes the spread *between* the two group means. That is why the $d_i^2$ terms (the squared deviations of the means) are absolutely essential.

Q: Why do we have two values for $\alpha$ and how do I choose the right one?

The quadratic equation $\alpha^2 - 40\alpha + 300 = 0$ yields two roots: $10$ and $30$. If $\alpha = 30$, then the standard deviation of class B, which is $30 - \alpha$, becomes $0$. While mathematically possible, a standard deviation of $0$ implies every single student in class B scored exactly the same mark, which is statistically unrealistic in a real-world scenario. Thus, we reject $30$ and accept $10$.

Q: Is the deviation $d_i$ always positive?

No, $d_i$ is defined as the difference between the individual mean and the combined mean ($\bar{x}_i - \bar{x}$). It can be negative, as seen with class A ($40 - 50 = -10$). However, because we square this deviation in the variance formula ($d_i^2$), the sign doesn't matter for the final calculation. The square ensures the contribution to variance is always positive.

Question

Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and $\alpha (>0)$, and the mean and standard deviation of marks of class B of $n$ students be respectively 55 and $30 - \alpha$. If the mean and variance of the marks of the combined class of $100 + n$ students are respectively 50 and 350, then the sum of variances of classes A and B is:

Options:

(A) 500

(B) 650

(C) 450

(D) 900

Sigma Percentile · Accepted Answer

### The Heartbeat of Data: Understanding Combined Statistics

Welcome, future engineers! Today, we are diving into a problem that sits at the very core of statistical analysis. It is not just about crunching numbers; it is about understanding how data behaves when we merge different populations.

Imagine you are a data scientist analyzing the performance of two different classes. You have the individual metrics, but when you combine them, the landscape changes. This is the essence of the JEE Advanced statistics challenge.

### Phase 1: The Weighted Mean

We start with two classes, $A$ and $B$. Class $A$ has $n_1 = 100$ students with a mean $\bar{x}_1 = 40$ and standard deviation $\sigma_1 = \alpha$. Class $B$ has $n_2 = n$ students with a mean $\bar{x}_2 = 55$ and standard deviation $\sigma_2 = 30 - \alpha$.

When we merge these groups, the combined mean $\bar{x}$ becomes $50$. The formula for the combined mean is a weighted average:

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

Substituting our values, we get:

$$50 = \frac{100(40) + n(55)}{100 + n}$$

Solving this linear equation is our first victory: $5000 + 50n = 4000 + 55n$, which simplifies to $5n = 1000$, giving us $n = 200$. We have successfully identified the size of class $B$.

### Phase 2: The Geometry of Variance

Now, we tackle the variance. This is where most students stumble. The combined variance $\sigma^2$ is not a simple average. It is governed by the formula:

$$\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$$

Here, $d_1$ and $d_2$ represent the 'distance' of each class mean from the combined mean. Think of this as the 'Parallel Axis Theorem' of statistics.

For class $A$, $d_1 = 40 - 50 = -10$. For class $B$, $d_2 = 55 - 50 = 5$. These distances are the physical 'gaps' that contribute to the total variance of the combined group.

### Phase 3: The Quadratic Dance

With $n=200$ and the combined variance $\sigma^2 = 350$, we substitute everything into our formula:

$$350 = \frac{100(\alpha^2 + (-10)^2) + 200((30 - \alpha)^2 + 5^2)}{100 + 200}$$

Multiplying by $300$ and simplifying, we get:

$$105000 = 100(\alpha^2 + 100) + 200((30 - \alpha)^2 + 25)$$

Dividing by $100$ yields $1050 = (\alpha^2 + 100) + 2((30 - \alpha)^2 + 25)$. Expanding the square term $(30 - \alpha)^2$ gives us $900 + \alpha^2 - 60\alpha$.

The equation becomes $1050 = \alpha^2 + 100 + 2(900 + \alpha^2 - 60\alpha + 25)$. After careful algebraic expansion and grouping, we arrive at $3\alpha^2 - 120\alpha + 900 = 0$, which reduces to the elegant quadratic:

$$\alpha^2 - 40\alpha + 300 = 0$$

### Phase 4: The Final Decision

Factoring our quadratic, we find $(\alpha - 10)(\alpha - 30) = 0$, leading to $\alpha = 10$ or $\alpha = 30$. We must now apply a physical reality check.

If $\alpha = 30$, the standard deviation of class $B$ becomes $30 - 30 = 0$, implying zero variance—a scenario that is statistically improbable for a class of $200$ students. Thus, we reject $30$ and accept $\alpha = 10$.

The question asks for the sum of variances $\sigma_1^2 + \sigma_2^2 = \alpha^2 + (30 - \alpha)^2$. With $\alpha = 10$, this is:

$$10^2 + (30 - 10)^2 = 100 + 400 = 500$$

We have conquered the problem! The final answer is **500**.

Select Answer:

Visualized Solution

Data Extraction

Combined Mean Formula

Finding Student Count $n$

Deviation Calculation

Combined Variance Formula

Substituting Known Values

Simplifying the Equation

Forming the Quadratic Equation

Solving for $α$

Final Variance Sum Calculation

Eliminating the Extraneous Root

The Sigma Insight: Variance and Standard Deviation

The Heartbeat of Data

Phase 1

Phase 2

Phase 3

Phase 4

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