Find the MLE of $\theta$ Heights of certain animals are modelled using a normal distribution, with an unknown mean and a known standard deviation $\sigma=5$.
For a random sample of 20 animals, all that is known, is that 5 of them have a height greater than $5$.
Now, let $\theta$ be the probability that an animal's height is greater than $5$. Find the MLE of $\theta$. 
I don't want a solution, please. I only want to be shown a start. 
 A: You only want hints so here are some:
Hints: suppose that the heights of the animals are $H_1,\ldots,H_n$ (where $n=20$) each of which is identically and independently distributed as $N(\mu,\sigma^2)$ ($\mu$ unknown, $\sigma=5$). Let $X_1,\ldots,X_n$ be Bernoulli i.i.d. r.v.'s defined as $\Pr(X_i=1)=\Pr(H_i> 5)=\theta$ which is a function of $\mu$. Let's write $\theta=f(\mu)$. Then, 
$$
S=X_1+\cdots+X_n
$$ 
has a binomial distribution with mean $n\theta$ and variance $n\theta(1-\theta)$. The probability that $S\geq 5$ (or $S=5$, the wording of the problem is a bit unclear) is of course another function $\theta$ and so a function of $\mu$. Write $\Pr(S\geq 5)=h(\theta)=g(\mu)$ (or $\Pr(S=5)=h(\theta)=g(\mu)$ depending on how you interpret the problem). Find $\hat\mu$ that maximises $g$ and your answer is $f(\hat\mu)$. 
In fact, you can directly find $\theta$ that maximises $h$ and so the normality assumption isn't really needed.

Hints (part 2): if $h(\theta)=\Pr(S=5)$, then
$$
h(\theta)=\binom{20}{5}\theta^5(1-\theta)^{15}.
$$
