JEE Main 2026 — Statistics Question with Solution

The given variable $X$ takes values $x_r=r(r-1)$ with frequencies $f_r={}^n{C}_r$ for $r=0,1,2,\dots ,n$.

The mean of the data is given by $\bar{X}=\frac{{\sum }_{r=0}^n{f}_r{x}_r}{{\sum }_{r=0}^n{f}_r}$.

We know that ${\sum }_{r=0}^n{f}_r={\sum }_{r=0}^n{}^n{C}_r=2^n$.

The sum of the products is:

${\sum }_{r=0}^n{f}_r{x}_r={\sum }_{r=0}^{n}r(r-1){}^n{C}_r$

Using the property $r(r-1){}^n{C}_r=n(n-1){}^{n-2}{C}_{r-2}$, we get:

${\sum }_{r=2}^{n}n(n-1){}^{n-2}{C}_{r-2}=n(n-1)2^{n-2}$

Substituting these into the mean formula:

$\bar{X}=\frac{n(n-1)2^{n-2}}{2^n}=\frac{n(n-1)}{4}$

Given that the mean is $60$:

$\frac{n(n-1)}{4}=60\Rightarrow n(n-1)=240$

Since $16\times 15=240$, we have $n=16$.

The total frequency is $N=2^{16}$. Since $N$ is even, the median is the average of the $(2^{15})$-th and $(2^{15}+1)$-th observations.

The cumulative frequency up to $r=7$ is:

${\sum }_{r=0}^7{}^{16}{C}_r=\frac{2^{16}-{}^{16}{C}_8}{2}=2^{15}-\frac{1}{2}{}^{16}{C}_8$

This value is strictly less than $2^{15}$.

The cumulative frequency up to $r=8$ is:

${\sum }_{r=0}^8{}^{16}{C}_r=2^{15}+\frac{1}{2}{}^{16}{C}_8$

This value is strictly greater than $2^{15}$.

Thus, both the $(2^{15})$-th and $(2^{15}+1)$-th observations fall in the class corresponding to $r=8$.

The value of the variable for $r=8$ is $x_8=8(8-1)=56$.

Therefore, the median is $56$.

Answer: $56$