A  variable $$X$$ takes values $$0, 0, 2, 6, 12, 20, \ldots, n(n-1)$$ with frequencies $${^{n} C_{0}}, {^{n} C_{1}}, {^{n} C_{2}}, \ldots, {^{n} C_{n}}$$ respectively. If the mean of the data is 60, then the median is :

The random variable $$X$$ assumes the values
$$0,\,0,\,2,\,6,\,12,\,20,\ldots ,\,n(n-1)$$
corresponding to $$k = 0,1,2,\ldots ,n$$ where in general

$$X = k(k-1)\qquad (k = 0,1,2,\ldots ,n)$$

The frequency attached to the value $$k(k-1)$$ is the binomial coefficient $$\binom{n}{k}$$.
Hence the total number of observations is

$$N = \sum_{k=0}^{n}\binom{n}{k}=2^{\,n}$$

Therefore the probability (relative frequency) of getting the value $$k(k-1)$$ is

$$P(X = k(k-1))=\frac{\binom{n}{k}}{2^{\,n}}$$

This is exactly the distribution of a Binomial random variable $$K$$ with parameters $$n$$ and $$p=\tfrac12$$, observed through the transformation $$X=K(K-1)$$.

1. Determination of $$n$$ from the mean

The given mean of $$X$$ is 60. Using the binomial properties,

$$E[X]=E\bigl[K(K-1)\bigr]=n(n-1)p^{2}$$

because for a binomial random variable $$K\sim\text{Bin}(n,p)$$ we have
$$E\bigl[K(K-1)\bigr]=n(n-1)p^{2}$$.

With $$p=\tfrac12$$,

$$E[X]=\frac{n(n-1)}{4}=60$$

$$\Longrightarrow\;n(n-1)=240$$

$$\Longrightarrow\;n^{2}-n-240=0$$

Solving the quadratic equation,

$$n=\frac{1+\sqrt{1+960}}{2}=\frac{1+31}{2}=16$$

(The negative root is discarded.) Hence $$n=16$$.

2. Finding the median

Let $$K$$ again denote the underlying binomial count. With $$n=16,\,p=\tfrac12$$ the distribution of $$K$$ is perfectly symmetric about 8:

$$P(K\le 7)=P(K\ge 9)$$

Consequently, the cumulative probability strictly below $$K=8$$ is less than $$\tfrac12$$, while the cumulative probability up to and including $$K=8$$ is at least $$\tfrac12$$. Thus any median value of $$K$$ is $$8$$.

Since the transformation $$X = K(K-1)$$ is monotonically increasing in $$K$$ for $$K\ge 1$$, the median of $$X$$ is obtained by substituting $$K=8$$:

$$\text{Median of }X = 8(8-1)=8\times 7=56$$

Answer: Option A which is: $$56$$