You draw a sample of size 30 from a normally distributed population with a standard deviation of four.
I am not sure if its just because we hadnt had any examples like this one, however I am struggling with how to find the distribution of the mean for this example.
We know that the sum of independent and identically distributed normal random variables i.e. $$\bar{X} = \frac{1}{N} \sum_{i=1}^N X_i$$ is also Normal $$\bar{X} \sim \mathcal{N}(\mu, \frac{\sigma^2}{N})$$ where $N$ is the sample size and $\mu,\sigma^2$ are the true mean and variances. So plugging in values, we get that your samples mean is Normally distributed as follows $$\bar{X} \rightarrow \mathcal{N}(\mu, \frac{4^2}{30})$$ So, the distribution of the mean is again a Normal distribution with mean equal to the true mean $\mu$ and variance equal to $\frac{4^2}{30}$.