Suppose that the longevity of people born in the Americas is normally distributed with $\mu$ and variance $\sigma^2$. A random sample of 10,000 people born in America yields the following results:
$\bar{X}=72.1$ and $S^2=144\ years^2$
Calculate the maximum likelihood estimator of the probability that a person born in America will live to age 50 or more and the estimator of the probability of not reaching 90 years of age.
Considers: $S^2=\frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\bar{X})^2$
I have the following:
If $S^2=144 \Rightarrow S=12 \Rightarrow X\rightarrow N(72.1;12)$
WE CALCULATE THE PROBABILITIES:
$P(X \geq50)=P(\frac{X-72.1}{12}\geq \frac{50-72.1}{12})=P(Z\geq-\frac{221}{120})=P(Z\geq-1.84)=P(Z<1.84)=0.9671$
$P(X<90)=P(Z<\frac{90-72.1}{12})=P(Z<\frac{179}{120})=P(Z<1.49)=0.9319$
After this, I don't know what else to do, I would appreciate your help.
Check the section on Discrete Distribution, Continuous Parameter Space in: https://en.wikipedia.org/wiki/Maximum_likelihood_estimation
Since you are estimating a probability using max likelihood (e.g. $p_{50}\equiv P(X>50)$) you need to think about it differently. Turns out the max likelihood estimate of $p_{50}$ is just the proportion of your sample that was at least 50 years old. Which makes sense. Similarly for the other probability $P(X<90)$.