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I want to calculate the standard deviation of the following numbers: 30, 45, 45, 60, 75, 80, 90, 100, 110, 120.
As far as I know, that would be

$\sqrt{\frac{1}{10}((30-75.5)^2+(45-75.5)^2+(45-75.5)^2+(60-75.5)^2+(75-75.5)^2+(80-75.5)^2+(90-75.5)^2+(100-75.5)^2+(110-75.5)^2+(120-75.5)^2)}$ = 28.587585

This site: got it right, but according to my book and this site: the result it 30.134.

So what's the right result and how come it differs?

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There are different types of standard deviation, namely the sample and population. – John Nov 13 '13 at 19:42
They are probably asking for the unbiased estimate. $s_{n-1}=\sqrt{\frac{n}{n-1}}s_{n}$ where $s_{n-1}$ is the unbiased estimate – John Nov 13 '13 at 19:49
up vote 1 down vote accepted

As I mentioned in the comments, there are two different types of standard deviation. You are calculating the bias estimate. Your book most probably wants the unbiased estimate. $s_{n-1}=\sqrt{\frac{n}{n-1}}s_{n}$ Use n=10 for the number of values. You can see where the difference comes from

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Thank you. When an exercise says, I should calculate the standard deviation, which one should I go for? I there a "standard" standard deviation? – user1170330 Nov 13 '13 at 20:13
You should go for the unbiased estimate. If you were to calculate it on the calculator there is a special symbol for it so make sure to check it out. – John Nov 13 '13 at 20:17

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