| X | 0 | 2 | 4 | 6 | 8 |
| $\mathrm { P } ( \mathrm { X } )$ | $a$ | $2a$ | $a + b$ | $2b$ | $3b$ |
If the mean of the following probability distribution of a random variable $X$ :
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
X & 0 & 2 & 4 & 6 & 8 \\
\hline
$\mathrm { P } ( \mathrm { X } )$ & $a$ & $2a$ & $a + b$ & $2b$ & $3b$ \\
\hline
\end{tabular}
\end{center}
is $\frac { 46 } { 9 }$, then the variance of the distribution is\\
(1) $\frac { 173 } { 27 }$\\
(2) $\frac { 566 } { 81 }$\\
(3) $\frac { 151 } { 27 }$\\
(4) $\frac { 581 } { 81 }$