JEE Main 2026 — 5 April, Evening Shift — Question 34

Given: $X$ takes values $0,0,2,6,12,20,\ldots,n(n-1)$ with frequencies $\binom{n}{0},\binom{n}{1},\binom{n}{2},\ldots,\binom{n}{n}$. Mean $\bar{x} = \frac{\sum_{r=0}^{n} r(r-1)\binom{n}{r}}{2^n}$. Using $r(r-1)\binom{n}{r} = n(n-1)\binom{n-2}{r-2}$ for $r\ge 2$,

we get $\sum_{r=0}^{n} r(r-1)\binom{n}{r} = n(n-1)\sum_{r=2}^{n}\binom{n-2}{r-2} = n(n-1)2^{n-2}$. Thus $\bar{x} = \frac{n(n-1)2^{n-2}}{2^n} = \frac{n(n-1)}{4}$. Set $\frac{n(n-1)}{4}=60 \Rightarrow n^2-n-240=0 \Rightarrow (n-16)(n+15)=0 \Rightarrow n=16$ (since $n>0$). Total frequency $N = \sum_{r=0}^{16}\binom{16}{r}=2^{16}$.

Since $N$ is even, median = average of $\left(\frac{N}{2}\right)^{\text{th}}$ and $\left(\frac{N}{2}+1\right)^{\text{th}}$ terms. $\frac{N}{2}=2^{15}=32768$. Compute cumulative frequencies: $\binom{16}{0}=1$, $\binom{16}{1}=16$, cumulative = 17. $\binom{16}{2}=120$, cumulative = 137. $\binom{16}{3}=560$, cumulative = 697. $\binom{16}{4}=1820$, cumulative = 2517. $\binom{16}{5}=4368$, cumulative = 6885. $\binom{16}{6}=8008$, cumulative = 14893. $\binom{16}{7}=11440$, cumulative = 26333. $\binom{16}{8}=12870$, cumulative = 39203. Since 32768 lies between 26333 and 39203, the $32768^{\text{th}}$ and $32769^{\text{th}}$ terms both correspond to $X=56$. Hence median = $\frac{56+56}{2}=56$.