Question

Consider the data on x taking the values \(0, 2, 4, 8, \ldots, 2^n\) with frequencies \({}^nC_0, {}^nC_1, {}^nC_2, \ldots, {}^nC_n\) respectively. If the mean of this data is \(\frac{728}{2^n}\), then n is equal to ________ .

Hint:

The values are \(0, 2^1, 2^2, \ldots, 2^n\) with frequencies \({}^nC_k\); use \(\sum {}^nC_k 2^k = 3^n\).

Answer 1

Correct Answer: 6

The values are \(x_0 = 0\) and \(x_k = 2^k\) for \(k = 1, \ldots, n\), with frequencies \(f_k = {}^nC_k\).
Total frequency \(= \sum_{k=0}^{n} {}^nC_k = 2^n\).
\(\sum f_kx_k = \sum_{k=1}^{n} {}^nC_k 2^k = (1+2)^n - 1 = 3^n - 1\).
Mean \(= \frac{3^n - 1}{2^n} = \frac{728}{2^n}\), so \(3^n = 729 = 3^6\).
Hence \(n = 6\).
Answer: 6