Animated Solution for Mathematics - Statistics: Consider the given data with frequency distribution $x_i$: 3, 8, 11, 10, 5, 4 $f_i$: 5, 2, 3, 2, 4, 4 Match each entry in List-I to the correct entries in List-II.

Welcome, future engineers. Today, we are not just solving a statistics problem; we are learning the art of systematic thinking. In the high-stakes environment of the JEE Advanced, statistics is often viewed as a 'calculation-heavy' topic, but it is truly the science of finding the center of gravity in a sea of numbers.

The Foundation of Order

Imagine you are given a bag of mixed-up numbers. Because the median is a positional measure, it cares about where a number sits in the hierarchy of values. Our first step is to sort the data.

We have $x_i$ values: $3,4,5,8,10,11$. Notice how the chaos transforms into a sequence; this is the first step of any great engineer: organizing the input.

We also calculate the total frequency $N=\sum f_i$. By summing $5+4+4+2+2+3$, we arrive at:

This $N$ is our denominator, our anchor for all subsequent calculations.

The Mean, Our Center of Gravity

Now, we seek the mean, denoted by $\bar{x}$. Think of the mean as the balance point of the distribution; if you were to place this distribution on a seesaw, the mean is where it would balance perfectly.

The formula is:

We create a new column for $f_i{x}_i$. We multiply each $x_i$ by its frequency $f_i$: $3\times 5=15$, $4\times 4=16$, and so on. Summing these products gives us:

Dividing by $N=20$, we get $\bar{x}=6$. It is elegant, it is precise, and it is our first match: (P) matches (3).

The Median, Our Positional Anchor

Next, we find the median. Since $N=20$ is even, there is no single middle number. We must look for the $10^{\text{th}}$ and $11^{\text{th}}$ observations.

We use the cumulative frequency ($cf$) to find them. The $cf$ tells us how many observations we have accumulated as we move up the sorted list. We see that up to $x_i=4$, we have $5+4=9$ observations.

The $10^{\text{th}}$ and $11^{\text{th}}$ observations must therefore fall into the next category, where $x_i=5$. Since both are $5$, their average is:

Thus, the median $M=5$. This is our second match: (Q) matches (2).

Measuring the Spread

Now, we tackle the Mean Deviation. Mean Deviation is simply the average distance of all points from a central value. The logic is identical for both mean and median:

For the Mean Deviation about the mean, we use $A=\bar{x}=6$. We calculate $|x_i-6|$ for each row, multiply by $f_i$, and sum them up. The sum is $54$. Dividing by $20$, we get $2.7$. This is our third match: (R) matches (4).

Finally, for the Mean Deviation about the median, we use $A=M=5$. We calculate $|x_i-5|$, multiply by $f_i$, and sum them up. The sum is $48$. Dividing by $20$, we get $2.4$. This is our final match: (S) matches (5).

The Beauty of Structure

Look at what we have achieved. By simply creating a table and following the definitions, we have navigated through mean, median, and deviations without breaking a sweat.

Statistics is not about memorizing formulas; it is about building a structure that allows the truth to emerge from the data. You have the tools and the method. Now, go forth and apply this systematic approach to every problem you face.