Q268 / 270 · NDA 2024 (Phase 1) · Mathematics

$\begin{array}{|c|c|c|c|c|c|}\hlinex_i & 1 & 2 & 3 & \ldots & n \\\hlinef_i & 1 & 2^{-1} & 2^{-2} & \ldots & 2^{-(n-1)} \\\hline\end{array} What is the mean of the distribution?

  • A
    $\frac{2^{n+1}-n+2}{2^n-1}$
  • B
    $\frac{2^{n+1}-n-2}{2^{n-1}}$
  • Correct answer
    C
    $\frac{2^{n+1}-n-2}{2^n-1}$
  • D
    $\frac{2^{n+1}-n+2}{2^n}$

Solution

$\begin{aligned} & \Sigma f_i=1+\frac{1}{2}+\frac{1}{2^2}+\ldots+\frac{1}{2^{n-1}} \\ & \Sigma f_i=\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}=\frac{2^n-1}{2^{n-1}}\end{aligned}$ $\begin{aligned} \text { Now,} \quad \text { Mean } & =\frac{\Sigma f_t x_t}{\Sigma f_t}=\frac{2^{n+1}-n-2}{2^{n-1}} \times \frac{2^{n-1}}{2^n-1} \\ & =\frac{2^{n+1}-n-2}{2^n-1} \end{aligned}$

Hence, the correct answer is option: 'C'.

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