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)=(66366)/(175e^6)\approx0.940028`