JEE Main 2026 Apr 5 Shift 2, Mathematics Q10: Basic Concepts of Statistics

The $r$-th value is $r(r-1)$ with frequency $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$. Total frequency $=2^n$.

Mean $=\frac{1}{2^n}\sum _{r=0}^{n}r(r-1)\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n(n-1)2^{n-2}}{2^n}=\frac{n(n-1)}{4}$.

Setting this equal to 60: $n(n-1)=240\Rightarrow n=16$.

Total frequency $=2^{16}=65536$ (even), so the median is the average of the $2^{15}$-th and $(2^{15}+1)$-th observations.

Cumulative frequencies: through $r=7$ the CF is ${\sum }_{r=0}^7\left(\genfrac{}{}{0ex}{}{16}{r}\right)=26333$; adding $\left(\genfrac{}{}{0ex}{}{16}{8}\right)=12870$ gives CF = 39203 at $r=8$. So both the 32768-th and 32769-th observations correspond to $r=8$, where the value is $8\cdot 7=56$.

$\therefore$ Median $=56$.