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Find out the variance of first 50 even natural number?

I know variance $\sigma^2=\frac{\Sigma(x_i-\bar{x})^2}{50}$

I have also find $\bar{x}=2550/50=51$ by using Arithmetic Progression.

But now what to do further i have no clue.

Freddy
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  • this was the question asked joint entrance exam (JEE)2014 – Freddy May 29 '14 at 23:08
  • It is often a little easier to find the variance by using the formula $\text{Var}(X)=E(X^2)-(E(X))^2$. – André Nicolas May 30 '14 at 00:08
  • @AndréNicolas how can we find variance by that method? – Freddy May 30 '14 at 12:22
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    It is a very similar calculation as the one in the answer posted. All we need is $E(X^2)$, which is $\frac{1}{50}(2^2+4^2+\cdots 100^2)$. So $\frac{4}{50}(1^2+\cdots +50^2)$. You had to calculate that anyway with the other method. – André Nicolas May 30 '14 at 13:07
  • @AndréNicolas this is more easier then the answer given. I think you should post this as answer so that i can accept it as answer:) – Freddy May 30 '14 at 13:31
  • There is no real need, my purpose was that you would know that (always) the variance is $E(X^2)-(E(X))^2$. This happens to be usually an easier computation than the more basic $E((X-\mu)^2)$. Purpose has been accomplished. – André Nicolas May 30 '14 at 13:38

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Assuming your number is right, just expand the brackets in that sum and use the standard summation formulae:

$$\sigma^2=\frac{\Sigma(x_i-\bar{x})^2}{50}=\Sigma(2n-51)^2/50=\frac{4}{50}\Sigma n^2 - \frac{4 \cdot 51}{50}\Sigma n + \frac{51^2}{50}\Sigma 1 $$

$$ \Sigma_{n=1}^{m} n^2 = \frac{1}{6}m \left ( m+1 \right ) \left ( 2m+1 \right ) $$

John Fernley
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