Let M denote the median of the following frequency distribution. Class: 0-4, 4-8, 8-12, 12-16, 16-20

Q: Why do we use the cumulative frequency of the *preceding* class ($C$) instead of the median class itself?

Great question! The formula is essentially an interpolation. We are starting at the lower limit $L$ and adding a fraction of the class width to reach the median position. The term $\frac{N}{2} - C$ represents how many steps *into* the median class we need to go to reach the exact middle. If we used the cumulative frequency of the median class, we would be overshooting!

Q: Is it possible for the median to be outside the median class?

Absolutely not! By definition, the median must lie within the median class. If your calculation gives you a value outside the range of your median class, it is a clear signal that you have made a calculation error, likely in the substitution of $L$, $C$, $f$, or $h$.

Q: Why do we need to multiply by $20$ at the end? Is this just to make the problem harder?

In competitive exams like JEE, this is a common technique to test your attention to detail. It is not just about finding the median; it is about following the instructions to the very end. Always re-read the final line of the question before bubbling your answer!

Question

Let $M$ denote the median of the following frequency distribution. $$\begin{array}{|c|c|c|c|c|c|} \hline 	ext{Class} & 0 - 4 & 4 - 8 & 8 - 12 & 12 - 16 & 16 - 20 \ \hline 	ext{Frequency} & 3 & 9 & 10 & 8 & 6 \ \hline \end{array}$$ Then $20M$ is equal to :

Options:

(A) 416

(B) 104

(C) 52

(D) 208

Sigma Percentile · Accepted Answer

### The Hidden Order in Data: Unlocking the Median

Welcome, future engineers! Today, we are going to peel back the layers of a statistical distribution. Statistics is not just about crunching numbers; it is about finding the heartbeat of a dataset.

When we are given a grouped frequency distribution, we are essentially looking at a crowd of data points organized into neat little rooms, and our goal is to find the person standing exactly in the middle. This is the **median**.

### Phase 1: The Cumulative Frequency (The Running Total)

To find the median, we cannot simply look at the middle value because the data is grouped. We need to track how the frequencies accumulate. Think of this as a running total.

We create a new column for Cumulative Frequency ($C$). For the first class, $0-4$, the frequency is $3$, so the cumulative frequency is $3$.

For the next class, $4-8$, we add the frequency $9$ to our previous total, giving us $3 + 9 = 12$. We continue this process: $12 + 10 = 22$, $22 + 8 = 30$, and finally $30 + 6 = 36$. This $36$ is our total number of observations, $N$.

### Phase 2: The Median Class (The Search)

Now that we have our map, where is the middle? The median divides the data into two equal halves.

Since $N = 36$, the median position is:

$$ \frac{N}{2} = \frac{36}{2} = 18 $$

We need to find the eighteenth observation. Looking at our cumulative frequency column, we see that the first class where the cumulative frequency is at least $18$ is the class $8-12$, which has a cumulative frequency of $22$.

This is our **Median Class**. This is the neighborhood where our answer lives.

### Phase 3: The Formula (The Interpolation)

Now, we use the median formula for grouped data:

$$ M = L + \left( \frac{\frac{N}{2} - C}{f} ight) 	imes h $$

Let us extract our variables:
*   $L$ (lower limit of the median class) $= 8$
*   $C$ (cumulative frequency of the class preceding the median class) $= 12$
*   $f$ (frequency of the median class) $= 10$
*   $h$ (class width) $= 12 - 8 = 4$

Now, we substitute these into our formula:

$$ M = 8 + \left( \frac{18 - 12}{10} ight) 	imes 4 $$

### Phase 4: The Final Calculation (The Precision)

Let us simplify this step-by-step. The numerator inside the bracket is $18 - 12 = 6$.

So, we have:

$$ M = 8 + \left( \frac{6}{10} ight) 	imes 4 $$

This simplifies to:

$$ M = 8 + 0.6 	imes 4 = 8 + 2.4 = 10.4 $$

We have found the median $M = 10.4$. But wait! The question asks for $20M$.

This is the final test of our focus. We calculate:

$$ 20 	imes 10.4 = 208 $$

And there it is! The final answer is **208**. The elegance of statistics lies in this precision. Keep practicing, stay focused, and remember that every calculation is a step closer to your goal.

Animated Solution for Mathematics - Statistics: Let $M$ denote the median of the following frequency distribution. $Class Frequency 0 - 4 3 4 - 8 9 8 - 12 10 12 - 16 8 16 - 20 6$ Then $20 M$ is equal to :

Select Answer:

Visualized Solution

Problem Overview

Cumulative Frequency ( $C$ )

Calculating $C F$ Values

Finding $\frac{N}{2}$

Identifying the Median Class

The Median Formula

Extracting Formula Variables

Substituting Values

Simplifying the Expression

Calculating the Median $M$

Calculating $20 M$

The Sigma Insight: Measures of Central Tendency (Mean, Median, Mode)

The Hidden Order in Data

Phase 1

Phase 2

Phase 3

Phase 4

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