Animated Solution for Mathematics - Statistics: The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and ${\sigma }^2$ respectively. If the variance of all the 30 numbers in the two sets is 13, then ${\sigma }^2$ is equal to

Imagine you are standing in a vast field, looking at two distinct clouds of fireflies. Each cloud has its own center—its mean—and each cloud has its own level of chaos—its variance.

In the world of JEE Advanced statistics, we are often asked to merge these clouds and understand the chaos of the new, combined swarm. This is not just about crunching numbers; it is about understanding the geometry of spread.

When we merge two sets of data, the new variance is not simply the average of the old ones. If you think that, you have fallen into the most classic trap in the syllabus! Let us embark on a journey to uncover why.

The Combined Mean

Before we can talk about the spread, we must find the center of our new, combined universe. We have two sets, each with $n_1=15$ and $n_2=15$ numbers.

The means are ${\bar{x}}_1=12$ and ${\bar{x}}_2=14$. Because the sizes of the sets are identical, the combined mean ${\bar{x}}_c$ is simply the arithmetic average of the two individual means:

This is our anchor. It is the point around which all thirty numbers now revolve.

Now, we must ask: how far is each group's center from our new anchor? This is where the concept of deviation comes in.

For the first set, the deviation is $d_1={\bar{x}}_1-{\bar{x}}_c=12-13=-1$. For the second set, it is $d_2={\bar{x}}_2-{\bar{x}}_c=14-13=1$.

These values, $d_1$ and $d_2$, are the 'distances' that create the extra variance when we merge the sets. Even if the individual variances were zero, the mere fact that the groups are centered at different points would create a non-zero variance for the combined set. This is the physical soul of the combined variance formula.

We are now ready to wield the master formula:

This equation is a beautiful balance. It takes the internal variance of each group (${\sigma }_1^2$ and ${\sigma }_2^2$), adds the square of the shift ($d_1^2$ and $d_2^2$), and then averages them out.

We know the combined variance ${\sigma }_c^2=13$. Let us plug in our values:

Watch how the algebra simplifies with grace. The numerator becomes $15(14+1)+15({\sigma }^2+1)$, which is $15(15)+15({\sigma }^2+1)$.

We can factor out the $15$ to get $15(15+{\sigma }^2+1)=15(16+{\sigma }^2)$. Now, divide by the denominator of $30$.

Since $\frac{15}{30}=\frac{1}{2}$, our equation collapses into:

Multiplying by $2$ gives $26=16+{\sigma }^2$. Finally, subtracting $16$ from $26$ leaves us with ${\sigma }^2=10$.

The mystery is solved! The unknown variance is $10$. Remember, in the exam hall, do not fear the complexity of the formula. Visualize the data, respect the deviations, and the math will always guide you home.