Probability of being within one standard deviation of mean given only a moment generating function

Question

I am trying to figure out how to approach this:

For starters, I calculated M'(0) and M"(0), which allowed me to find the mean and variance. I got 2 for the mean and 1.6 for the variance. But now, since I don't know what the distribution looks like (and I'm not sure if this MGF can be easily arranged to look like one for a familiar distribution), I am stumped. Is there something very easy here that I am missing?

Answer 1

Hint: this MGF is indeed that of a distribution that should be familiar to you.

Further hint: $9765625 = 5^{10}$.

Answer 2

Then MGF of a binomial is given as $$ \left(Pe^t + q\right)^n.$$ You are given $$M(t) = \frac{\left(e^t+4\right)^{10}}{9765625},$$ but the $9765625 = 5^{10}.$ So $$M(t) = \frac{\left(e^t+4\right)^{10}}{5^{10}}= \left(\frac{e^t +4}{5}\right)^{10}=\left( \frac{1}{5}e^t+\frac{4}{5}\right)^{10}.$$ From this, you get that $p=0.2,~q=0.8,~ n=10$.
Hope this helps you.