Question

Find the mean and variance for the data

6, 7, 10, 12, 13, 4, 8, 12

Answer 1

Here $x=6,7,10,12,13,4,8,12$

$\therefore \sum x=6+7+10+12+13+4+8+12=72$

$n=8\therefore \overline{x}=\frac{72}{8}=9$

$\sum x^2=(6{)}^2+(7{)}^2+(10{)}^2+(12{)}^2+(13{)}^2+(4{)}^2+(8{)}^2+(12{)}^2$

$36+49+100+144+169+16+64+144$

= 722

$\therefore \text{Variance}={\sigma }^2=\frac{N\sum x^2-(\sum x{)}^2}{N^2}$

=$\frac{8\times 722-(72{)}^2}{(8{)}^2}$

=$\frac{5776-5184}{64}=\frac{592}{64}=9.25$