Since [tex]\mathrm M_Y(t)=e^{6(e^t-1}[/tex], we know that [tex]Y[/tex] follows a Poisson distribution with parameter [tex]\lambda=6[/tex].
Now assuming [tex]\mu,\sigma[/tex] denote the mean and standard deviation of [tex]Y[/tex], respectively, then we know right away that [tex]\mu=6[/tex] and [tex]\sigma=\sqrt6[/tex].
So,
[tex]\mathbb P(|Y-\mu|\le2\sigma)=\mathbb P(6-2\sqrt6\le Y\le6+2\sqrt6)=\dfrac{66366}{175e^6}\approx0.940028[/tex]
