Discussion
The standard deviation of 1,2,3,...,100 is
Answer: A
The standard deviation is \(\frac{1}{2}\sqrt{3333}\).
For the first n natural numbers the variance is \(\frac{n^{2} - 1}{12}\).
\[ \begin{aligned} \sigma^{2} &= \frac{100^{2} - 1}{12} = \frac{9999}{12} = \frac{3333}{4} \\ \sigma &= \frac{\sqrt{3333}}{2} \end{aligned} \]
The options with 4, 6 or 8 in front take the wrong root of the denominator 4, and \(\frac{1}{4}\sqrt{1111}\) divides 9999 by 9 instead of 3. The formula follows from \(\sigma^{2} = \frac{\sum x^{2}}{n} - \bar{x}^{2}\) with \(\sum x^{2} = \frac{n(n + 1)(2n + 1)}{6}\) and \(\bar{x} = \frac{n + 1}{2}\).
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