If the variance of the frequency distribution 

$x_i$	2	3	4	5	6	7	8
Frequency $f_i$	3	6	16	$\alpha$	9	5	6

is 3, then $\alpha$ is equal to ______.

$$\begin{gathered} \overline{x}=\frac{\sum _{ }{f}_i{x}_i}{\sum _{ }{f}_i} \\ =\frac{6+18+64+5\alpha +54+35+48}{45+\alpha }=\frac{225+5\alpha }{45+\alpha }=5 \\ \frac{\sum f_i{x}_i^2}{\sum f_i}=\frac{12+54+256+25\alpha +324+245+384}{45+\alpha }=\frac{1275+25\alpha }{45+\alpha } \\ (\sigma {)}^2=\frac{\sum f_i{x}_i^2}{\sum f_i}-(\overline{x}{)}^2 \\ 3=\frac{1275+25\alpha }{45+\alpha }-25 \\ 28(45+\alpha )=1275+25\alpha \Rightarrow 1260+28\alpha =1275+25\alpha \\ 3\alpha =15\alpha \Rightarrow \alpha =5 \end{gathered}$$