JEE Main 2026 — Mathematics Probability & Statistics

The given variable $X$X takes values $x_r = r(r-1)$xr​=r(r−1) with frequencies $f_r = {}^{n}C_{r}$fr​=nCr​ for $r = 0, 1, 2, \ldots, n$r=0,1,2,…,n.

The mean of the data is given by $\displaystyle \bar{X} = \dfrac{\sum_{r=0}^{n} f_r x_r}{\sum_{r=0}^{n} f_r}$Xˉ=∑r=0n​fr​∑r=0n​fr​xr​​.

We know that $\displaystyle \sum_{r=0}^{n} f_r = \sum_{r=0}^{n} {}^{n}C_{r} = 2^n$r=0∑n​fr​=r=0∑n​nCr​=2n.

The sum of the products is:

$$\sum_{r=0}^{n} f_r x_r = \sum_{r=0}^{n} r(r-1)\, {}^{n}C_{r}$$r=0∑n​fr​xr​=r=0∑n​r(r−1)nCr​

Using the property $r(r-1)\, {}^{n}C_{r} = n(n-1)\, {}^{n-2}C_{r-2}$r(r−1)nCr​=n(n−1)n−2Cr−2​, we get:

$$\sum_{r=2}^{n} n(n-1)\, {}^{n-2}C_{r-2} = n(n-1)\, 2^{n-2}$$r=2∑n​n(n−1)n−2Cr−2​=n(n−1)2n−2

Substituting these into the mean formula:

$$\bar{X} = \dfrac{n(n-1)\, 2^{n-2}}{2^n} = \dfrac{n(n-1)}{4}$$Xˉ=2nn(n−1)2n−2​=4n(n−1)​

Given that the mean is $60$60:

$$\dfrac{n(n-1)}{4} = 60 \Rightarrow n(n-1) = 240$$4n(n−1)​=60⇒n(n−1)=240

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

The total frequency is $N = 2^{16}$N=216. Since $N$N is even, the median is the average of the $(2^{15})$(215)-th and $(2^{15} + 1)$(215+1)-th observations.

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

$$\sum_{r=0}^{7} {}^{16}C_{r} = \dfrac{2^{16} - {}^{16}C_{8}}{2} = 2^{15} - \dfrac{1}{2}\, {}^{16}C_{8}$$r=0∑7​16Cr​=2216−16C8​​=215−21​16C8​

This value is strictly less than $2^{15}$215.

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

$$\sum_{r=0}^{8} {}^{16}C_{r} = 2^{15} + \dfrac{1}{2}\, {}^{16}C_{8}$$r=0∑8​16Cr​=215+21​16C8​

This value is strictly greater than $2^{15}$215.

Thus, both the $(2^{15})$(215)-th and $(2^{15} + 1)$(215+1)-th observations fall in the class corresponding to $r = 8$r=8.

The value of the variable for $r = 8$r=8 is $x_8 = 8(8-1) = 56$x8​=8(8−1)=56.

Therefore, the median is $56$56.

Answer: $56$56