Hints and Comments:
You seem to be mixing up two different limiting properties.
(1) As the sample size $n$ of a survey becomes large the sample proportion of those who like lemons tends to become ever closer to the population probability that a randomly chosen person likes lemons [That is based on the Law of Large Numbers.]
(2) The sum or average of a large number $n$ of random variables (with finite variances) tends to be nearly normal. [That is based on the Central Limit Theorem.] A binomial distribution $\mathsf{Binom}(n, p)$ is the sum of $n$ Bernoulli (0-1) random variables, and so the CLT applies to allow reasonable
normal approximations of binomial distributions, especially when $n$ is large and $p$ is not too near 0 or 1 (near 1/2 is best).
Maybe that is enough to help your sort out your Question. If not, please
deconvolve the LLN and CLT issues an ask about them separately.
Sample proportion: You interview $n=1000$ people and find that $X=331$ of them like lemons. Then the sample proportion $\hat p = X/n$ is the sample proportion.
For large $n$ $\hat p$ is close to the population proportion $p$ who like lemons.
Finding the mean: Suppose $X \sim \mathsf{Binom}(n=3,p=.5).$ Then
the by definition
$$\mu_X = E(X) = \sum_{k=0}^3 kP(S=k) = 0(1/8) + 1(3/8) + 2(3/8) + 3(1/8) = 12/8 = 3/2.$$
For the binomial distribution only it turns out that there is a
convenient formula: $E(X) = np = 3(1/2) = 3/2.$ Somewhat similarly, one
can show that $Var(X) = np(1-p).$
Some other families of distributions have analogous special formulas for expectations and variances. You can take the attitude that there are
confusingly many of them, or you can be glad that when special formulas
exist for the mean and variance of a type of distribution they save a lot
of work.
Suppose 63% of 1200 randomly sampled people (that's 756 of them) like lemons. Then what is the question? The best point estimate of the population proportion $p$ who like lemons is $\hat p = 756/1200 = .63.$ Of course, $\hat p$ isn't
exactly $p$. You may wonder how far off the estimate might be.
Using the normal approximation to the binomial distribution, one can find a 95% confidence interval (CI) for $p.$ The simplest form of CI (OK for $n$ over 1000.) is $\hat p \pm 1.96\sqrt{ \frac{\hat p(1-\hat p)}{1200} }.$ That
computes to the interval $(0.658, 0.702),$ which has $\hat p$ as its center.
