The variance of the first 50 even natural numbers is :

Variance of a set of $n$ numbers \left({x}_{1}, {x}_{2} . . . {x}_{i}, . . . {x}_{n}\right)$is given<br /><br />${\sigma}^{2} = \dfrac{1}{n}\mathop{Σ}_{i = 1}^{n}{\left({x}_{i} - \overset{¯}{x}\right)}^{2}$, where$\overset{¯}{x}$is the average of the numbers.<br /><br />We want the variance of \left(2, 4, 6, 8 . . . . . 48, 50, 52, . . . . . 98, 100\right)$

Here $\overset{¯}{x} = 51$.

\left({x}_{i} - \overset{¯}{x}\right)$will be symmetric about$51$,<br /><br />$∴{\sigma}^{2} = \dfrac{2}{50}\left({1}^{2} + {3}^{2} + {5}^{2} + . .. {49}^{2}\right)$<br /><br />Now,${1}^{2} + {3}^{2} + {5}^{2} + . . . + {49}^{2} = \left({1}^{2} + {2}^{2} + {3}^{2} + . . . + {49}^{2} + {50}^{2}\right) - \left({2}^{2} + {4}^{2} + {6}^{2} + . . . + {48}^{2} + {50}^{2}\right)$<br /><br />$= \dfrac{\left(50\right)\left(51\right)\left(101\right)}{6} - 4\left({1}^{2} + {2}^{2} +\right)$<br /><br />$= \dfrac{\left(50\right)\left(51\right)\left(101\right)}{6} - \dfrac{4\left(25\right)\left(26\right)\left(51\right)}{6}$<br /><br />$= 42925 - 22100$<br /><br />$= 20825$<br /><br />$∴ {\sigma}^{2} = \dfrac{2}{50} \times 20825 = 833