Distribution of the sum of random variables (are those dependent or independent?)

Question

The weight of the citizens in a city is distributed normally with expected value of 70kg and Standard deviation of 6kg. How is the weight of 9 people is distributed?

Attempt:

Let X be the weight of a person in that city. So $X\sim\mathrm{N}\left(70,36\right)$.

Now Let Y=9X, so as Y is the sum of normally distributed variables, it is also distributed normally. then:

$E\left[Y\right]=E\left[9X\right]=9E\left[X\right]=9\cdot70=630$

$Var\left(Y\right)=Var\left(9X\right)=9^{2}Var\left(X\right)=9^{2}\cdot36=2916$

So we conclude that $Y\sim\mathrm{N}\left(630,2916\right)$

On the other side:

My teacher told me that $Var\left(Y\right)=9\cdot36=324$ as we can conclude it from the Central Limit Theorem. I don't understand how is the CLT related here, as it is a theorem about a limit and also its premise is that the variables are independent no?

He added me the following theorem:

Let $X_1,\ldots,X_n$ be a sample out of a distribution with expected value $\mu$ and variance $\sigma^2$. For large $n$ we have that the sum $\sum_{i=1}^n X_i$ is closely distributed as $N(n\mu,n\sigma^2)$

Please help me figure it out

Thanks

Answer 1

If this is a random sample from the population you should probably assume independence and so your teacher's method

Your solution would be correct if all nine had the same weight

If there was partial dependence (say an individual and their siblings cousins) but not total dependence, then the answer might be somewhere in between

Answer 2

No, the random variable $Y$ describing the sum of weights of 9 people is:

$$Y = X_1+X_2+\dots +X_9 \neq 9X_1$$

For $9X_1$, we assume perfect correlation between the 9 random variables. In reality, they are independent. That is the reason your approach produces a much higher variance than your teachers approach. It's all the covariance terms.

Also, the argument in terms of the central limit theorem, it isn't needed in this case since the individual random variables are normally distributed. So, the distribution of the sum of 9 people is exactly $N(9 \mu, 9 \sigma^2)$