Find Median of Variable Data Given Mean and Binomial Frequencies

To find the median of the given variable $X$, we first determine the value of $n$ using the given mean.

The values of $X$ are given by $x_k = k(k-1)$ for $k = 0, 1, 2, \dots, n$, with corresponding frequencies $f_k = {}^{n}C_{k}$.

The total frequency $N$ is: $N = \sum_{k=0}^{n} f_k = \sum_{k=0}^{n} {}^{n}C_{k} = 2^n$

The sum of the products of values and their respective frequencies is: $\sum_{k=0}^{n} f_k x_k = \sum_{k=0}^{n} k(k-1) {}^{n}C_{k}$

Using the identity $k(k-1) {}^{n}C_{k} = n(n-1) {}^{n-2}C_{k-2}$ for $k \ge 2$, we get: $\sum_{k=0}^{n} k(k-1) {}^{n}C_{k} = n(n-1) \sum_{k=2}^{n} {}^{n-2}C_{k-2} = n(n-1) 2^{n-2}$

The mean $\bar{X}$ is defined as: $\bar{X} = \frac{\sum_{k=0}^{n} f_k x_k}{N} = \frac{n(n-1) 2^{n-2}}{2^n} = \frac{n(n-1)}{4}$

Given that the mean is $60$: $\frac{n(n-1)}{4} = 60$ $n(n-1) = 240$ $n^2 - n - 240 = 0$ $(n - 16)(n + 15) = 0$

Since $n$ must be a positive integer, we have $n = 16$.

Now, for $n = 16$, the total frequency is $N = 2^{16}$. Since $N$ is an even number, the median is the mean of the $\left(\frac{N}{2}\right)^{\text{th}}$ and $\left(\frac{N}{2} + 1\right)^{\text{th}}$ terms, which are the $2^{15\text{th}}$ and $(2^{15} + 1)^{\text{th}}$ terms.

Let us find the cumulative frequency up to $k = 7$ and $k = 8$: Using the symmetry of binomial coefficients: $\sum_{k=0}^{16} {}^{16}C_{k} = \sum_{k=0}^{7} {}^{16}C_{k} + {}^{16}C_{8} + \sum_{k=9}^{16} {}^{16}C_{k} = 2^{16}$ $2 \sum_{k=0}^{7} {}^{16}C_{k} + {}^{16}C_{8} = 2^{16}$ $\sum_{k=0}^{7} {}^{16}C_{k} = 2^{15} - \frac{1}{2} {}^{16}C_{8}$

Since ${}^{16}C_{8} > 0$, the cumulative frequency up to $k = 7$ is strictly less than $2^{15}$: $\text{Cumulative Frequency at } k=7 < 2^{15}$

Adding ${}^{16}C_{8}$ to both sides gives the cumulative frequency up to $k = 8$: $\text{Cumulative Frequency at } k=8 = 2^{15} + \frac{1}{2} {}^{16}C_{8} > 2^{15} + 1$

Thus, both the $2^{15\text{th}}$ and $(2^{15} + 1)^{\text{th}}$ observations correspond to the value at $k = 8$.

Therefore, the median is: $x_8 = 8(8 - 1) = 8 \times 7 = 56$

Hence, the correct option is A.