JEE Main 2025 — Statistics Question with Solution

The median for grouped data is given by:

$$\text{Median}=L+\left(\frac{\frac{n}{2}-CF}{f}\right)\times h$$

where

$$L$$ is the lower limit (or boundary) of the median class.

$$CF$$ is the cumulative frequency of all classes preceding the median class.

$$f$$ is the frequency of the median class.

$$h$$ is the class width.

$$n$$ is the total number of students.

Given:

Median $$=14$$

Median class interval is $$12-18$$, so $$L=12$$ and the class width $$h=18-12=6$$.

Frequency of median class $$f=12$$.

Cumulative frequency below the median class $$=18$$ (i.e., $$CF=18$$).

Plugging these into the formula:

$$14=12+\left(\frac{\frac{n}{2}-18}{12}\right)\times 6$$

Step 1: Subtract 12 from both sides:

$$2=\left(\frac{\frac{n}{2}-18}{12}\right)\times 6$$

Step 2: Simplify the multiplication factor:

$$\left(\frac{6}{12}\right)=\frac{1}{2}$$

So the equation becomes:

$$2=\frac{1}{2}\left(\frac{n}{2}-18\right)$$

Step 3: Multiply both sides by 2 to remove the fraction:

$$4=\frac{n}{2}-18$$

Step 4: Solve for $$\frac{n}{2}$$:

$$\frac{n}{2}=4+18=22$$

Step 5: Multiply both sides by 2 to find $$n$$:

$$n=44$$

Thus, the total number of students is $$44$$.