How to calculate variance of first 30 odd numbers ?
The method I used requires finding the mean and subtracting each number with it and then squaring the result . Its getting difficult .
I used this :
Any other way to calculate ?
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Micheal
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1Your method's not too difficult, you just need to know the formulae for sum of squares and such. – Arpan Mar 24 '15 at 13:02
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You can use variance = E(x^2) - [ E(x) ]^2. Computing the last two os only using whole numbers, but a division by n. – coffeemath Mar 24 '15 at 13:09
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1The mean is exactly 30. – Yuval Filmus Mar 24 '15 at 13:09
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Sorry , confused it with another question . – Micheal Mar 24 '15 at 13:14
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@coffeemath E is expectation ? – Micheal Mar 24 '15 at 13:16
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@YuvalFilmus I deleted it . – Micheal Mar 24 '15 at 13:17
1 Answers
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The first 30 odd numbers don't have variance. The uniform distribution over the first 30 odd numbers has variance, given by $$ \frac{1}{30} \sum_{i=0}^{29} (2i+1)^2 - \left(\frac{1}{30} \sum_{i=0}^{29} (2i+1)\right)^2 = \frac{899}{3}. $$ More generally, one calculates that the variance of the uniform distribution over the first $n$ odd numbers is $$ \frac{n^2-1}{3}. $$
Yuval Filmus
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I calculated this using a program, since that was faster and less error-prone than using a formula in this case. But formulas also exist. – Yuval Filmus Mar 24 '15 at 13:54
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But which formula did you apply ? The above eqn doesnt fall into the formula I linked – Micheal Mar 24 '15 at 13:57
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