Mathematics - Statistics Question with Solution | TestHub

Variance of a set of$n$numbers$\left(x_1, x_2 ...x_i, ...x_n\right)$is given

${\sigma }^2=\frac{1}{n}\underset{i=\; 1}{\overset{n}{\Sigma }}{\left(x_i-\overline{x}\right)}^2$, where$\overline{x}$is the average of the numbers.

We want the variance of$\left(2, 4, 6, 8 ..... 48, 50, 52, ..... 98, 100\right)$

Here$\overline{x}=\mathrm{51}$.

$\left(x_i-\overline{x}\right)$will be symmetric about$51$,

$\mathrm{\therefore }{\sigma }^2=\frac{2}{\mathrm{50}}\left(1^2+3^2+5^2+... {\mathrm{49}}^2\right)$

Now,$1^2 + 3^2 + 5^2 + ... + {49}^2 = \left(1^2 + 2^2 + 3^2 + ... + {49}^2 + {50}^2\right) -\left( 2^2 + 4^2 + 6^2 + ... + {48}^2 + {50}^2\right)$

$=\frac{\left(50\right)\left(51\right)\left(101\right)}{6}-4\left(1^2+2^2+\right)$

$=\frac{\left(50\right)\left(51\right)\left(101\right)}{6}-\frac{4\left(25\right)\left(26\right)\left(51\right)}{6}$

$= 42925 - 22100$

$= 20825$

$\therefore {\sigma }^2=\frac{2}{\mathrm{50}}\times \mathrm{20825}=\mathrm{833}$