Question

If y has moment-generating function m(t) = e 6(e t −1) , what is p(|y − µ| ≤ 2σ)?

Answer 1

Since $\mathrm M_Y(t)=e^{6(e^t-1}$, we know that $Y$ follows a Poisson distribution with parameter $\lambda=6$.

Now assuming $\mu,\sigma$ denote the mean and standard deviation of $Y$, respectively, then we know right away that $\mu=6$ and $\sigma=\sqrt6$.

So,

$\mathbb P(|Y-\mu|\le2\sigma)=\mathbb P(6-2\sqrt6\le Y\le6+2\sqrt6)=\dfrac{66366}{175e^6}\approx0.940028$