Question about central limit theorem? Suppose I draw 10 tickets at random with replacement from a box of tickets, each of which is labeled with a number. The average of the numbers on the tickets is 1, and the SD of the numbers on the tickets is 1. Suppose I repeat this over and over, drawing 10 tickets at a time. Each time, I calculate the sum of the numbers on the 10 tickets I draw. Consider the list of values of the sample sum, one for each sample I draw. This list gets an additional entry every time I draw 10 tickets.
i) As the number of repetitions grows, the average of the list of sums is increasingly likely to be between 9.9 and 10.1.
ii) As the number of repetitions grows, the histogram of the list of sums is likely to be approximated better and better by a normal curve (after converting the list to standard units).
The answer given is that (i) is correct and (ii) is not correct .
I assumed the reason why (i) is correct is the fact that as the number of draws increase the average of the list of sums is going to converge to tickets' expected sum of value.
I can't figure out why (ii) is incorrect. Based on the definitions for CLT : 


*

*The central limit theorem applies to sum of .

*The number of draws should be reasonably large.

*The more lopsided the values are, the more draws needed for
reasonable approximation (compare the approximations of rolling 5 in
roulette to flipping a fair coin).

*It is another type of convergence : as the number of draws grows, the
normal approximation gets better.
Aside from what I can perceive that this case does seem to satisfy the above definitions, doesn't the last definition confirms what (ii) is suggesting to be true?
 A: For (ii) - Not really, what you are doing is drawing the same number of tickets from the same box of tickets every time - this means that your sample size is always $10$.
The CLT requires the sample size to increase to infinity, i.e. you want to sum "infinitely" many random variables and then you can say something about convergence.
In your case you are summing up $10$ variables at each step and this doesn't make the CLT applicable.
A: The sum of the ten tickets has expectation $10$ and variance $10$ and standard deviation $\sqrt{10} \approx 3.16$
To answer part (ii) first, this need not be a normal distribution
For example half the tickets in your box might say $0$ and half $2$, so the sum of ten of them must be an even integer from $0$ to $20$ and a histogram of multiple attempts with breaks at every integer will show a comb type shape, while a normal distribution is a continuous distribution
In part (i), you are not looking at the individual sums of draws of ten, but of the average of their sums.  If you do $n$ draws of ten, then the distribution of the average of the sums has expectation $10$, variance $\frac{10}{n}$, and standard deviation $\sqrt{\frac{10}{n}}$.   Now the Central Limit Theorem kicks in and you can say $\dfrac{\overline{S}_n - 10}{\sqrt{\frac{10}{n}}} \to \mathcal{N}(0,1)$      as $n$ increases 
