What is the mean of the distribution?

Consider the following for the next two (02) items that follow : <table> <thead><tr><th>x_i</th><th>1</th><th>2</th><th>3</th><th>...</th><th>n</th></tr></thead> <tbody> <tr><td>f_i</td><td>1</td><td>2^{-1}</td><td>2^{-2}</td><td>...</td><td>2^{-(n-1)}</td></tr> </tbody> </table>

  1. A. \frac{2^{n+1}-n+2}{2^{n}-1}
  2. B. \frac{2^{n+1}-n-2}{2^{n-1}}
  3. C. \frac{2^{n+1}-n-2}{2^{n}-1} ✓
  4. D. \frac{2^{n+1}-n+2}{2^{n}}

Correct Answer: C. \frac{2^{n+1}-n-2}{2^{n}-1}

Explanation

The mean is given by \frac{\sum x_i f_i}{\sum f_i}. We found \sum x_i f_i = \frac{2^{n+1} - n - 2}{2^{n-1}}. The sum of frequencies is a Geometric Progression: \sum f_i = 1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{n-1}} = \frac{1(1 - (1/2)^n)}{1 - 1/2} = 2(1 - \frac{1}{2^n}) = \frac{2^{n+1}-2}{2^n} = \frac{2^n-1}{2^{n-1}}. Dividing the two gives Mean = \frac{\frac{2^{n+1} - n - 2}{2^{n-1}}}{\frac{2^n-1}{2^{n-1}}} = \frac{2^{n+1}-n-2}{2^n-1}.

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