The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the

Q: Wait, why do we need the mean of the second sample? Can't we just use the variance formula directly?

That is a great question! The combined variance formula relies on the difference between the means of the two groups, $(\bar{x}_1 - \bar{x}_2)^2$. If you don't know $\bar{x}_2$, you cannot calculate the 'external' spread, which is the variance contributed by the fact that the two groups are centered at different points. Without it, your calculation will be incomplete.

Q: What is that second term in the combined variance formula? It looks like an extra penalty.

It is not a penalty, but a correction factor! Think of it this way: if you combine two groups that have the same mean, the variance of the combined group is just the weighted average of their individual variances. But if they have different means, the combined group is more spread out because the data points are now distributed around a new, combined mean. That second term mathematically accounts for this additional spread.

Q: Is there a shortcut to avoid the long calculation?

While there is no 'magic' shortcut, you can simplify your life by working with fractions instead of decimals for as long as possible. Also, notice that the term $\frac{n_1n_2}{(n_1 + n_2)^2}$ is a constant that can be calculated early. Keeping your work organized and clean is the best 'shortcut' in JEE problems!

Question

The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation $\sqrt{13.44}$, then the standard deviation of the second sample is :

Options:

(A) 8

(B) 6

(C) 4

(D) 5

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two datasets. Sample one has $n_1 = 100$ items, a mean $\bar{x}_1 = 15$, and a standard deviation $\sigma_1 = 3$.

The combined dataset has a total size $n = 250$, a combined mean $\bar{x} = 15.6$, and a combined variance $\sigma^2 = 13.44$. Our goal is to determine the standard deviation $\sigma_2$ of the second sample.

### Finding the Size

Before we dive into the complex variance formulas, we need the basics. We know the total number of items is $n = 250$, and the first sample contributes $n_1 = 100$.

It is trivial but essential to calculate the size of the second sample:
$$n_2 = 250 - 100 = 150$$

### The Anchor of the Mean

Variance is a measure of spread, but spread is always measured relative to a center—the mean. If we do not know where the second sample is centered, we cannot calculate its variance.

We use the combined mean formula:
$$\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$$

Substituting our known values:
$$15.6 = \frac{100(15) + 150(\bar{x}_2)}{250}$$

With algebraic discipline, we multiply $15.6$ by $250$ to get $3900$. Subtracting the contribution of the first sample ($100 	imes 15 = 1500$), we are left with:
$$2400 = 150\bar{x}_2$$

Dividing by $150$, we find that $\bar{x}_2 = 16$. Our second sample is centered at $16$.

### The Heart of the Problem: Combined Variance

Now, we reach the most critical part of our journey. The combined variance formula is a beautiful piece of statistical architecture:

$$\sigma^2 = \frac{n_1\sigma_1^2 + n_2\sigma_2^2}{n_1 + n_2} + \frac{n_1n_2}{(n_1 + n_2)^2}(\bar{x}_1 - \bar{x}_2)^2$$

This formula tells us that the total variance is the sum of two distinct parts: the weighted average of the individual variances (the internal spread) and the variance of the means (the external spread). If you ignore that second term, you are missing the 'distance' between the two groups, which contributes significantly to the total variance.

### The Final Calculation

Let us plug in our values: $\sigma^2 = 13.44$, $\sigma_1^2 = 3^2 = 9$, $n_1 = 100$, $n_2 = 150$, $\bar{x}_1 = 15$, and $\bar{x}_2 = 16$. The equation becomes:

$$13.44 = \frac{100(9) + 150\sigma_2^2}{250} + \frac{100 	imes 150}{(250)^2}(15 - 16)^2$$

Simplifying the fractions, we get:
$$13.44 = \frac{900 + 150\sigma_2^2}{250} + \frac{15000}{62500}(-1)^2$$

This reduces to:
$$13.44 = 3.6 + 0.6\sigma_2^2 + 0.24$$

Combining the constants, we have $13.44 = 3.84 + 0.6\sigma_2^2$. Subtracting $3.84$ from $13.44$ gives us $9.6 = 0.6\sigma_2^2$.

Finally, dividing by $0.6$, we find $\sigma_2^2 = 16$. The standard deviation is the square root of the variance, so $\sigma_2 = \sqrt{16} = 4$.

We have solved the mystery. The **standard deviation of the second sample is 4**.

Animated Solution for Mathematics - Statistics: The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation $13.44$ , then the standard deviation of the second sample is :

Select Answer:

Visualized Solution

Problem Overview

Calculate Size of Sample $2$

Combined Mean Formula

Substitute Values for Mean

Solve for $\overset{x}{ˉ}_{2}$

Combined Variance Formula

Substitute Values into Variance Formula

Simplify the Equation

Solve for $σ_{2}^{2}$

Final Answer

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

Finding the Size

The Anchor of the Mean

The Heart of the Problem

The Final Calculation

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