If the mean of the discrete distribution 8, 9, 6, 5, x, 4, 6, 5 is 6, then its standard deviation (nearest to two decimal places) is

Mean = 6
$\begin{array}{l}\therefore \frac{8+9+6+5+x+4+6+5}{8}=6\\ \Rightarrow 43+x=48\Rightarrow x=5\end{array}$
So, data are 8 9, 6,  5, 5, 4, 6, 5
xi = 8, 9, 6, 5, 5, 4, 6, 5
$x_i^2=$64, 81, 36, 25, 5, 16, 36 25
$\sum x_i^2=$64+81+36+25+25+16+36+25=308
$\begin{array}{l}SD\left(x\right)=\sqrt{\frac{1}{N}\sum x{p}^2}-{\left(x\right)}^2\\ =\sqrt{\frac{1}{8}\times 30+8+{\left(6\right)}^2}\\ =\sqrt{2.5}=1.58\end{array}$