The variance of first 50 even natural numbers is

Solution :

\({\sigma^2}\) = \(\sum(x_i – \bar{x})^2\over n\)

\(\bar{X}\) = \(\sum x_i\over n\)

= \(2 + 4 + 6 + 8 + ….. + 100\over 50\)

= 51

\({\sigma^2}\) = \(2^2 + 4^2 + …. + 100^2\over 50\) – \(51^2\)

= 833

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