Marks obtains by all the students of class 12 are presented in a freqency distribution with classes

Q: Wait, why is the cumulative frequency $cf$ equal to $18$? The problem says 'marks less than 12 is 18'.

That is a great observation! In grouped data, the cumulative frequency of a class is the sum of frequencies of all classes up to that point. If the number of students with marks less than $12$ is $18$, it means that the sum of frequencies for all classes before the median class ($12-18$) is exactly $18$. Thus, $cf = 18$ is correct.

Q: Can we use the mean formula instead of the median formula?

No, you cannot. The mean and median are distinct statistical measures. The problem specifically provides the median and the median class, which are parameters used exclusively in the median formula for grouped data. Using the mean formula would require information about the sum of all marks, which is not provided.

Q: How did you get the class width $h = 6$?

The class width $h$ is the difference between the upper limit and the lower limit of any class interval. For the median class $12-18$, the width is $18 - 12 = 6$. Since the problem states the classes have equal width, this $h=6$ applies to all classes.

Question

Marks obtains by all the students of class 12 are presented in a freqency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18, then the total number of students is

Options:

(A) 52

(B) 48

(C) 44

(D) 40

Sigma Percentile · Accepted Answer

### The Art of Reverse Engineering: Unlocking the Median

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a statistics problem; we are performing a forensic analysis of data.

Often, in the heat of an exam, we see a formula and think, 'I just need to plug in the numbers.' But the true beauty of mathematics lies in the ability to look at a formula and see it as a map—a map that can be traversed in either direction.

### Phase 1: Decoding the Median Class

Imagine you are looking at a frequency distribution table. You have a sea of numbers, but the problem gives you a lighthouse: the median is $14$.

Since $14$ sits comfortably between $12$ and $18$, we immediately know our 'median class' is $12-18$. This is our anchor.

The lower limit of this class, which we denote as $L$, is $12$. The frequency of this specific class, $f$, is $12$.

The cumulative frequency of all the classes that came before it is $cf = 18$. Finally, the width of our interval, $h$, is simply $18 - 12 = 6$. We have all our pieces on the table.

### Phase 2: The Formula as a Map

The formula for the median of grouped data is our most powerful tool:

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

Think of this not as a rigid equation, but as a relationship. It tells us how the median is constructed from the total number of students, $N$.

We are not looking for the median; we are looking for $N$. We are reverse-engineering the data.

### Phase 3: The Algebraic Journey

Now, let us substitute our values into the equation:

$$14 = 12 + \left( \frac{\frac{N}{2} - 18}{12} ight) 	imes 6$$

I know that seeing variables like $N$ inside a fraction can feel daunting, but let us take a breath and simplify. First, subtract $12$ from both sides.

This gives us:

$$2 = \left( \frac{\frac{N}{2} - 18}{12} ight) 	imes 6$$

Now, look at that fraction $\frac{6}{12}$. It simplifies beautifully to $\frac{1}{2}$. Our equation becomes:

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

Multiply both sides by $2$, and we get $4 = \frac{N}{2} - 18$. The path is clearing!

Add $18$ to both sides: $22 = \frac{N}{2}$. Finally, multiply by $2$, and we arrive at **$N = 44$**.

### Conclusion: The Elegance of Logic

We started with a set of parameters and a single unknown, and through the elegance of algebra, we uncovered the total number of students: **$44$**.

This problem teaches us that statistics is not just about calculating averages; it is about understanding the structure of data. Whenever you face a problem where a parameter is missing, do not panic.

Trust the formula, trust your ability to isolate the variable, and enjoy the process of discovery. You have mastered this concept today—keep that momentum going!

Select Answer:

Visualized Solution

Identifying the Median Class

Extracting Frequency Parameters

The Median Formula

Substituting the Values

Isolating the Fraction

Simplifying the Terms

Cross Multiplication

Solving for N

Final Conclusion

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

The Art of Reverse Engineering

Phase 1

Phase 2

Phase 3

Conclusion

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