A variable X takes values 0, 0, 2, 6, 12, 20, , n(n-1) with...

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

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

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:

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

Given that the mean is $60$:

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

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:

$\sum_{r=0}^{7} {}^{16}C_{r} = \dfrac{2^{16} - {}^{16}C_{8}}{2} = 2^{15} - \dfrac{1}{2}\, {}^{16}C_{8}$

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

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

$\sum_{r=0}^{8} {}^{16}C_{r} = 2^{15} + \dfrac{1}{2}\, {}^{16}C_{8}$

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: $56$