$$\begin{aligned} & \text { Mean }=\sum x_i P\left(x_i\right) \\ & \frac{46}{9}=4 a+4 a+4 b+12 b+24 b \\ & \frac{46}{9}=8 a+40 b \\ & \frac{23}{9}=4 a+20 b \\ & 36 a+180 b=23 \quad \text{.... (1)} \end{aligned}$$

Sum of probability is 1

$$\Rightarrow 4 a+6 b=1 \quad \text{... (2)}$$

$$\begin{aligned} & \text { Solving (1) and (2) } \\ & a=\frac{1}{12}, b=\frac{1}{9} \\ & \sigma^2=\sum x_i^2 P\left(x_i\right)-\left(\sum x_i P\left(x_i\right)\right)^2 \\ & =4 \times 2 a+16(a+b)+36(2 b)+64(3 b)-\left(\frac{46}{9}\right)^2 \\ & =8(a+2(a+b)+9 b+24 b)-\left(\frac{46}{9}\right)^2 \\ & =8(3 a+35 b)-\left(\frac{46}{9}\right)^2 \\ & =8\left(\frac{3}{12}+\frac{35}{9}\right)-\left(\frac{46}{9}\right)^2 \\ & =8\left(\frac{149}{36}\right)-\left(\frac{46}{9}\right)^2=\frac{566}{81} \\ \end{aligned}$$