Understanding the Data Structure

The problem provides a dataset where the values (X) are related to the index $k$ by the formula $X_k = k(k-1)$, and the corresponding frequencies ($f_k$) are given by the binomial coefficients $f_k = {}^{n}C_k$, for $k = 0, 1, 2, \dots, n$.

• Values (X): $0, 0, 2, 6, 12, \dots, n(n-1)$
• Frequencies (f): ${}^{n}C_{0}, {}^{n}C_{1}, {}^{n}C_{2}, {}^{n}C_{3}, {}^{n}C_{4}, \dots, {}^{n}C_{n}$

Calculating the Mean

The formula for the mean ($\bar{X}$) of a frequency distribution is $\bar{X} = \frac{\sum X_k f_k}{\sum f_k}$.

Numerator Calculation:

We need to calculate $\sum_{k=0}^{n} k(k-1) {}^{n}C_k$. Note that the terms for $k=0$ and $k=1$ are zero.

For $k \ge 2$: $k(k-1) {}^{n}C_k = k(k-1) \frac{n!}{k!(n-k)!} = \frac{n!}{(k-2)!(n-k)!}$ $= n(n-1) \frac{(n-2)!}{(k-2)!(n-k)!} = n(n-1) {}^{n-2}C_{k-2}$.

So, the sum becomes: $\sum_{k=2}^{n} n(n-1) {}^{n-2}C_{k-2} = n(n-1) \sum_{k=2}^{n} {}^{n-2}C_{k-2}$

Let $j = k-2$. The sum transforms to: $n(n-1) \sum_{j=0}^{n-2} {}^{n-2}C_j$. Using the binomial theorem property $\sum_{j=0}^{m} {}^{m}C_j = 2^m$, we get: $\sum_{j=0}^{n-2} {}^{n-2}C_j = 2^{n-2}$.

Therefore, the numerator is $n(n-1) 2^{n-2}$.

Denominator Calculation:

The sum of frequencies is $\sum_{k=0}^{n} {}^{n}C_k = 2^n$.

Mean Formula:

$\bar{X} = \frac{n(n-1) 2^{n-2}}{2^n} = \frac{n(n-1)}{2^2} = \frac{n(n-1)}{4}$.

Finding the Value of n

We are given that the mean is 60.

Setting the derived mean formula equal to 60: $\frac{n(n-1)}{4} = 60$ $n(n-1) = 240$

Solving the quadratic equation $n^2 - n - 240 = 0$ or by inspection, we find two consecutive integers whose product is 240. These are 15 and 16.

Since $n$ must be positive, $n = 16$.

Calculating the Median

The total number of observations is $N = \sum_{k=0}^{n} {}^{n}C_k = 2^n$. With $n=16$, $N = 2^{16} = 65536$.

The median is the value corresponding to the $\frac{N}{2}$-th observation.

Median position = $\frac{N}{2} = \frac{2^{16}}{2} = 2^{15} = 32768$.

We need to find the value $X_k = k(k-1)$ for which the cumulative frequency $\sum_{i=0}^{k} {}^{n}C_i$ first reaches or exceeds the median position.

The frequencies ${}^{n}C_k$ are symmetric around $k=n/2$. For $n=16$, the distribution peaks at $k=8$.

Consider the cumulative frequency up to $k=7$: $\sum_{k=0}^{7} {}^{16}C_k$. Due to symmetry, $\sum_{k=0}^{16} {}^{16}C_k = 2^{16}$. And $\sum_{k=0}^{7} {}^{16}C_k = \sum_{k=9}^{16} {}^{16}C_k$. We know $\sum_{k=0}^{16} {}^{16}C_k = \sum_{k=0}^{7} {}^{16}C_k + {}^{16}C_8 + \sum_{k=9}^{16} {}^{16}C_k$. $2^{16} = 2 \times (\sum_{k=0}^{7} {}^{16}C_k) + {}^{16}C_8$. $\sum_{k=0}^{7} {}^{16}C_k = \frac{2^{16} - {}^{16}C_8}{2} = 2^{15} - \frac{1}{2} {}^{16}C_8$.

This cumulative frequency ($2^{15} - \frac{1}{2} {}^{16}C_8$) is less than the median position ($2^{15}$).

Now consider the cumulative frequency up to $k=8$: $\sum_{k=0}^{8} {}^{16}C_k = \sum_{k=0}^{7} {}^{16}C_k + {}^{16}C_8 = (2^{15} - \frac{1}{2} {}^{16}C_8) + {}^{16}C_8 = 2^{15} + \frac{1}{2} {}^{16}C_8$.

This cumulative frequency ($2^{15} + \frac{1}{2} {}^{16}C_8$) is greater than the median position ($2^{15}$).

This indicates that the $32768$-th observation falls within the group corresponding to $k=8$.

The value associated with $k=8$ is $X_8 = 8(8-1) = 8 \times 7 = 56$. Therefore, the median is 56.