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This post mentions in passing that the standard deviation of the data set (1, 50) is about 34.65. Wolfram Alpha confirms this, saying it is $\frac{49}{\sqrt{2}}\approx 34.648$. But when I try to calculate this by hand, I get:

$\mu = \frac{1+50}{2}=\frac{51}{2}=25.5$

$\sigma = \sqrt{\frac{(25.5-1)^2+(25.5-50)^2}{2}}=\sqrt{\frac{(24.5)^2+(-24.5)^2}{2}}=\sqrt{\frac{2\cdot(24.5)^2}{2}}=24.5$.

What am I doing wrong?

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  • $\begingroup$ Yeah I'm not sure the original question is what you think it means. But I don't know what it's about either... $\endgroup$ – Adam Rubinson Mar 31 at 20:01
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You actually did standard deviation correct, what you did is called the population standard deviation.

What the original problem was talking about and wolfram alpha said is called sample standard deviation.

Sample standard deviation is calculated with this formula: $s= \sqrt{\frac{\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}}{N-1}}$

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  • $\begingroup$ Thank you for this answer! $\endgroup$ – Rasputin Mar 31 at 20:15
  • $\begingroup$ @Rasputin no problem! $\endgroup$ – I am a person Mar 31 at 20:15
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If $(1,50)$ represents the population, you divide by $n=2$ in the calculation of standard deviation, but if it's a sample from the population (in this case, of size $2$), then you divide by $n-1=2-1$ in this case. The former refers to the population standard deviation and the latter the sample standard deviation.

$$s=\sqrt{\frac{2\cdot (24.5)^2}{2-1}}=34.65$$

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  • $\begingroup$ That makes sense. Thank you for this answer! $\endgroup$ – Rasputin Mar 31 at 20:15

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