Suppose some people are asked to each randomly pick 30 apples and put into a bag.
In this case, the weights of all the bags will surely be different because different apples weigh differently. So for one of the bag, if let $X_1$ be the weight of the first apple, $X_2$ be the second apple, and all the way to $X_{30}$ the weight of the $30th$ apple, then $X$, which is the weight of the bag is $X=X_1+X_2+\cdots+X_{30}$.
Now, if I want to approximate the sampling distribution of $\bar{X}$, I presume that I can simply just say $\bar{X}=\frac{1}{30}\sum _{ i=1 }^{ 30 }{ X_{ i } } $. Is this right?
In the problem, I need to continue to find the sampling distribution for the variation of the weights of the bags from the approximated sampling distribution of $\bar{X}$ that I have found. I am stuck at this part. I don't see how $\bar{X}$ links with the variance of the bag weights.
Best Answer
Look up Central Limit Theorem. Often, the heuristic is to sample from a distribution with a sample size of at least 30. The CLT doesn't require that the $X_i$'s are IID normal random variables, they just need to be IID random variables from some fixed distribution with mean $\mu$ and variance $\sigma^2$. CLT then says that the sampling distribution of the mean of a sample of size $n$ (for large enough $n$, where $n\geq 30$ is one often used heuristic) is approximately normal with mean $\mu$ and variance $\sigma^2/n$, where recall that the $\mu$ and $\sigma^2$ come from the distribution of the underlying population (which, in this case, is the distribution of the weights of apples).
To estimate $\mu$ from the data, we generally use $\bar{X}$. To estimate $\sigma^2$ from data, we generally use $s^2$, which divides by $n-1$ instead of $n$. Using $s^2$ makes the distribution tough to deal with, but here if your population is normally distributed then the sampling distribution is approximately a $t$ distribution with $n-1$ degrees of freedom.