Question

variance of first 50 even natural numbers?

Answer 1

First 50 even natural numbers are
2,4,6,8,.......,100
Now Mean (average) of first 50 even natural numbers
$$\begin{gathered} \mu =\frac{2+4+6+8+....+100}{50} \\ \mu =\frac{2(1+2+3+....+50)}{50} \\ \mu =\frac{(1+2+3+....+50)}{25} \\ \mu =\frac{50\left({\displaystyle \frac{50+1}{2}}\right)}{25} \\ \mu =51 \end{gathered}$$
Now Sum of squares of natural even natural numbers
$$\begin{gathered} \underset{n=2}{\overset{100}{\sum x^2}}=2^2+4^2+6^2+........+{100}^2 \\ =2^2(1^2+2^2+3^2+.....+{50}^2) \\ =4\times \frac{50(50+1)(2\times 50+1)}{6} \\ =\frac{4\times 50\times 51\times 101}{6} \\ Nowvariance \\ =\frac{\underset{n=2}{\overset{100}{\sum x^2}}}{n}-{\mu }^2 \\ =\frac{\frac{4\times 50\times 51\times 101}{6}}{50}-{51}^2 \\ =2\times 17\times 101-2601 \\ =3434-2601 \\ =833 \end{gathered}$$