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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Statistics

NDA 2018 · Mathematics · Statistics

The standard deviation σ of the first N natural numbers can be...

The standard deviation σ\sigma of the first NN natural numbers can be obtained using which one of the following formulae?

Options

  1. Aσ=N2−112\sigma=\frac{\mathrm{N}^2-1}{12}
  2. Bσ=N2−112\sigma=\sqrt{\frac{\mathrm{N}^2-1}{12}}
  3. Cσ=N−112\sigma=\sqrt{\frac{\mathrm{N}-1}{12}}
  4. Dσ=N2−16N\sigma=\sqrt{\frac{\mathrm{N}^2-1}{6 \mathrm{ N}}}

Correct answer

B. σ=N2−112\sigma=\sqrt{\frac{\mathrm{N}^2-1}{12}}

Step-by-step solution

σ2=1N∑i=1NXi2−(1N⋅∑i=1NXi)2=1N(12+22+32+….+N2)−(1N(1+2+3+….+N))2=1N⋅N(N+1)(2N+1)6−(1N⋅N(N+1)2)2=N2−112∴σ=N2−112.\begin{aligned} & \sigma^2=\frac{1}{N} \sum_{i=1}^N X_i^2-\left(\frac{1}{N} \cdot \sum_{i=1}^N X_i\right)^2 \\ = & \frac{1}{N}\left(1^2+2^2+3^2+\ldots.+\mathrm{N}^2\right)-\left(\frac{1}{\mathrm{ N}}(1+2+3+\ldots.+\mathrm{N})\right)^2 \\ = & \frac{1}{\mathrm{ N}} \cdot \frac{\mathrm{ N}(\mathrm{ N}+1)(2 \mathrm{ N}+1)}{6}-\left(\frac{1}{\mathrm{ N}} \cdot \frac{\mathrm{ N}(\mathrm{ N}+1)}{2}\right)^2 \\ = & \frac{\mathrm{N}^2-1}{12} \quad \therefore \sigma=\sqrt{\frac{\mathrm{N}^2-1}{12}}.\end{aligned}
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