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